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009C Sample Midterm 1, Problem 5 Detailed Solution

2018-01-05T21:26:35Z

Kayla Murray:

Find the radius of convergence and interval of convergence of the series.

(a)  <math>\sum_{n=0}^\infty \sqrt{n}x^n</math>

(b)  <math>\sum_{n=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}</math>
<hr>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''Ratio Test'''
|-
|        Let  <math style="vertical-align: -7px">\sum a_n</math>  be a series and  <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
|-
|        Then,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We first use the Ratio Test to determine the radius of convergence.
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}\cdot x^{n+1}}{\sqrt{n}\cdot x^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}}{\sqrt{n}}\cdot x\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(\sqrt{\frac{n+1}{n}}\cdot|x|\bigg)}\\
&&\\
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}}\\
&&\\
& = & \displaystyle{|x|\sqrt{\lim_{n\rightarrow \infty} \frac{n+1}{n}}}\\
&&\\
& = & \displaystyle{|x|\sqrt{1}}\\
&&\\
&=& \displaystyle{|x|.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|The Ratio Test tells us this series is absolutely convergent if  <math style="vertical-align: -5px">|x|<1.</math>
|-
|Hence, the Radius of Convergence of this series is  <math style="vertical-align: -2px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we need to determine the interval of convergence.
|-
|First, note that  <math style="vertical-align: -5px">|x|<1</math>  corresponds to the interval  <math style="vertical-align: -4px">(-1,1).</math>
|-
|To obtain the interval of convergence, we need to test the endpoints of this interval
|-
|for convergence since the Ratio Test is inconclusive when  <math style="vertical-align: -2px">L=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|First, let  <math style="vertical-align: -1px">x=1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty \sqrt{n}.</math>
|-
|We note that
|-
|        <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty.</math>
|-
|Therefore, the series diverges by the  <math style="vertical-align: 0px">n</math>th term test.
|-
|Hence, we do not include  <math style="vertical-align: -1px">x=1</math>  in the interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 5:  
|-
|Now, let  <math style="vertical-align: -1px">x=-1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty (-1)^n \sqrt{n}.</math>
|-
|Since  <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty,</math>
|-
|we have
|-
|        <math>\lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=\text{DNE}.</math>
|-
|Therefore, the series diverges by the  <math style="vertical-align: 0px">n</math>th term test.
|-
|Hence, we do not include  <math style="vertical-align: -1px">x=-1 </math>  in the interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 6:  
|-
|The interval of convergence is  <math style="vertical-align: -4px">(-1,1).</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We first use the Ratio Test to determine the radius of convergence.
|-
|We have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(-1)^{n+1}(x-3)^{n+1}}{2(n+1)+1}\cdot \frac{2n+1}{(-1)^n(x-3)^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)\cdot (x-3)\cdot \frac{2n+1}{2n+3}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(|x-3|\cdot \frac{2n+1}{2n+3}\bigg)}\\
&&\\
& = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\
&&\\
& = & \displaystyle{|x-3|.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|The Ratio Test tells us this series is absolutely convergent if  <math style="vertical-align: -5px">|x-3|<1.</math>
|-
|Hence, the Radius of Convergence of this series is  <math style="vertical-align: -2px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we need to determine the interval of convergence.
|-
|First, note that  <math style="vertical-align: -5px">|x-3|<1</math>  corresponds to the interval  <math style="vertical-align: -4px">(2,4).</math>
|-
|To obtain the interval of convergence, we need to test the endpoints of this interval
|-
|for convergence since the Ratio Test is inconclusive when  <math style="vertical-align: -2px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|First, let  <math style="vertical-align: -1px">x=4.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty (-1)^n \frac{1}{2n+1}.</math>
|-
|This is an alternating series.
|-
|Let  <math style="vertical-align: -16px">b_n=\frac{1}{2n+1}.</math>.
|-
|First, we have
|-
|       <math>\frac{1}{2n+1}\ge 0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|The sequence  <math>\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{2(n+1)+1}<\frac{1}{2n+1}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{2n+1}=0.</math>
|-
|Therefore, this series converges by the Alternating Series Test
|-
|and we include  <math style="vertical-align: -1px">x=4</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 5:  
|-
|Now, let  <math style="vertical-align: -1px">x=2.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty \frac{1}{2n+1}.</math>
|-
|First, we note that  <math>\frac{1}{2n+1}>0</math>  for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|Thus, we can use the Limit Comparison Test.
|-
|We compare this series with the series  <math>\sum_{n=1}^\infty \frac{1}{n},</math>
|-
|which is the harmonic series and divergent.
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \frac{\big(\frac{1}{2n+1}\big)}{\big(\frac{1}{n}\big)}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{2n+1}}\\
&&\\
& = & \displaystyle{\frac{1}{2}.}
\end{array}</math>
|-
|Since this limit is a finite number greater than zero,
|-
|       <math>\sum_{n=0}^\infty \frac{1}{2n+1}</math> 
|-
|diverges by the Limit Comparison Test.
|-
|Therefore, we do not include  <math style="vertical-align: -1px">x=2</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 6:  
|-
|The interval of convergence is  <math style="vertical-align: -4px">(2,4].</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     The radius of convergence is  <math style="vertical-align: -2px">R=1</math>  and the interval of convergence is  <math style="vertical-align: -4px">(-1,1).</math>
|-
|    '''(b)'''     The radius of convergence is  <math style="vertical-align: -2px">R=1</math>  and the interval of convergence is  <math style="vertical-align: -4px">(2,4].</math>
|}
[[009C_Sample_Midterm_1|'''Return to Sample Exam''']]

009C Sample Midterm 1, Problem 5 Detailed Solution

2018-01-05T21:19:17Z

Kayla Murray:

Find the radius of convergence and interval of convergence of the series.

(a)  <math>\sum_{n=0}^\infty \sqrt{n}x^n</math>

(b)  <math>\sum_{n=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}</math>
<hr>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''Ratio Test'''
|-
|        Let  <math style="vertical-align: -7px">\sum a_n</math>  be a series and  <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
|-
|        Then,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We first use the Ratio Test to determine the radius of convergence.
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}x^{n+1}}{\sqrt{n}x^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}}{\sqrt{n}}x\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}|x|}\\
&&\\
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}}\\
&&\\
& = & \displaystyle{|x|\sqrt{\lim_{n\rightarrow \infty} \frac{n+1}{n}}}\\
&&\\
& = & \displaystyle{|x|\sqrt{1}}\\
&&\\
&=& \displaystyle{|x|.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|The Ratio Test tells us this series is absolutely convergent if  <math style="vertical-align: -5px">|x|<1.</math>
|-
|Hence, the Radius of Convergence of this series is  <math style="vertical-align: -2px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we need to determine the interval of convergence.
|-
|First, note that  <math style="vertical-align: -5px">|x|<1</math>  corresponds to the interval  <math style="vertical-align: -4px">(-1,1).</math>
|-
|To obtain the interval of convergence, we need to test the endpoints of this interval
|-
|for convergence since the Ratio Test is inconclusive when  <math style="vertical-align: -2px">L=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|First, let  <math style="vertical-align: -1px">x=1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty \sqrt{n}.</math>
|-
|We note that
|-
|        <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty.</math>
|-
|Therefore, the series diverges by the  <math style="vertical-align: 0px">n</math>th term test.
|-
|Hence, we do not include  <math style="vertical-align: -1px">x=1</math>  in the interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 5:  
|-
|Now, let  <math style="vertical-align: -1px">x=-1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty (-1)^n \sqrt{n}.</math>
|-
|Since  <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty,</math>
|-
|we have
|-
|        <math>\lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=\text{DNE}.</math>
|-
|Therefore, the series diverges by the  <math style="vertical-align: 0px">n</math>th term test.
|-
|Hence, we do not include  <math style="vertical-align: -1px">x=-1 </math>  in the interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 6:  
|-
|The interval of convergence is  <math style="vertical-align: -4px">(-1,1).</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We first use the Ratio Test to determine the radius of convergence.
|-
|We have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(-1)^{n+1}(x-3)^{n+1}}{2(n+1)+1}\frac{2n+1}{(-1)^n(x-3)^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x-3)\frac{2n+1}{2n+3}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} |x-3|\frac{2n+1}{2n+3}}\\
&&\\
& = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\
&&\\
& = & \displaystyle{|x-3|.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|The Ratio Test tells us this series is absolutely convergent if  <math style="vertical-align: -5px">|x-3|<1.</math>
|-
|Hence, the Radius of Convergence of this series is  <math style="vertical-align: -2px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we need to determine the interval of convergence.
|-
|First, note that  <math style="vertical-align: -5px">|x-3|<1</math>  corresponds to the interval  <math style="vertical-align: -4px">(2,4).</math>
|-
|To obtain the interval of convergence, we need to test the endpoints of this interval
|-
|for convergence since the Ratio Test is inconclusive when  <math style="vertical-align: -2px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|First, let  <math style="vertical-align: -1px">x=4.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty (-1)^n \frac{1}{2n+1}.</math>
|-
|This is an alternating series.
|-
|Let  <math style="vertical-align: -16px">b_n=\frac{1}{2n+1}.</math>.
|-
|First, we have
|-
|       <math>\frac{1}{2n+1}\ge 0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|The sequence  <math>\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{2(n+1)+1}<\frac{1}{2n+1}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{2n+1}=0.</math>
|-
|Therefore, this series converges by the Alternating Series Test
|-
|and we include  <math style="vertical-align: -1px">x=4</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 5:  
|-
|Now, let  <math style="vertical-align: -1px">x=2.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty \frac{1}{2n+1}.</math>
|-
|First, we note that  <math>\frac{1}{2n+1}>0</math>  for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|Thus, we can use the Limit Comparison Test.
|-
|We compare this series with the series  <math>\sum_{n=1}^\infty \frac{1}{n},</math>
|-
|which is the harmonic series and divergent.
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \frac{\frac{1}{2n+1}}{\frac{1}{n}}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{2n+1}}\\
&&\\
& = & \displaystyle{\frac{1}{2}.}
\end{array}</math>
|-
|Since this limit is a finite number greater than zero,
|-
|       <math>\sum_{n=0}^\infty \frac{1}{2n+1}</math> 
|-
|diverges by the Limit Comparison Test.
|-
|Therefore, we do not include  <math style="vertical-align: -1px">x=2</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 6:  
|-
|The interval of convergence is  <math style="vertical-align: -4px">(2,4].</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     The radius of convergence is  <math style="vertical-align: -2px">R=1</math>  and the interval of convergence is  <math style="vertical-align: -4px">(-1,1).</math>
|-
|    '''(b)'''     The radius of convergence is  <math style="vertical-align: -2px">R=1</math>  and the interval of convergence is  <math style="vertical-align: -4px">(2,4].</math>
|}
[[009C_Sample_Midterm_1|'''Return to Sample Exam''']]

009C Sample Midterm 1, Problem 4 Detailed Solution

2018-01-05T21:18:03Z

Kayla Murray:

Determine the convergence or divergence of the following series.

 Be sure to justify your answers!

::<math>\sum_{n=1}^\infty \frac{1}{n^23^n}</math>
<hr>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''Direct Comparison Test'''
|-
|        Let  <math>\{a_n\}</math>  and  <math>\{b_n\}</math>  be positive sequences where  <math style="vertical-align: -3px">a_n\le b_n</math>
|-
|        for all  <math style="vertical-align: -3px">n\ge N</math>  for some  <math style="vertical-align: -3px">N\ge 1.</math>
|-
|        '''1.''' If  <math>\sum_{n=1}^\infty b_n</math>  converges, then  <math>\sum_{n=1}^\infty a_n</math>  converges.
|-
|        '''2.''' If  <math>\sum_{n=1}^\infty a_n</math>  diverges, then  <math>\sum_{n=1}^\infty b_n</math>  diverges.
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we note that
|-
|        <math>\frac{1}{n^23^n}>0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|This means that we can use a comparison test on this series.
|-
|Let  <math style="vertical-align: -14px">a_n=\frac{1}{n^23^n}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Let  <math style="vertical-align: -14px">b_n=\frac{1}{n^2}.</math>
|-
|We want to compare the series in this problem with
|-
|        <math>\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty \frac{1}{n^2}.</math>
|-
|This is a  <math style="vertical-align: -4px">p</math>-series with  <math style="vertical-align: -4px">p=2.</math>
|-
|Hence,  <math>\sum_{n=1}^\infty b_n</math>  converges.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Also, we have  <math style="vertical-align: -4px">a_n<b_n</math>  since
|-
|        <math>\frac{1}{n^23^n}<\frac{1}{n^2}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|Therefore, the series  <math>\sum_{n=1}^\infty a_n</math>  converges
|-
|by the Direct Comparison Test.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|        converges (by the Direct Comparison Test)
|-
|
|}
[[009C_Sample_Midterm_1|'''Return to Sample Exam''']]

009C Sample Midterm 1, Problem 3 Detailed Solution

2018-01-05T21:17:32Z

Kayla Murray:

Determine whether the following series converges absolutely,

 conditionally or whether it diverges.

 Be sure to justify your answers!

::<math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math>
<hr>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' A series  <math>\sum a_n</math>  is '''absolutely convergent''' if
|-
|        the series  <math>\sum |a_n|</math>  converges.
|-
|'''2.''' A series  <math>\sum a_n</math>  is '''conditionally convergent''' if
|-
|        the series  <math>\sum |a_n|</math>  diverges and the series  <math>\sum a_n</math>  converges.
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we take the absolute value of the terms in the original series.
|-
|Let  <math style="vertical-align: -14px">a_n=\frac{(-1)^n}{n}.</math>
|-
|Therefore,
|-
|        <math>\begin{array}{rcl}
\displaystyle{\sum_{n=1}^\infty |a_n|} & = & \displaystyle{\sum_{n=1}^\infty \bigg|\frac{(-1)^n}{n}\bigg|}\\
&&\\
& = & \displaystyle{\sum_{n=1}^\infty \frac{1}{n}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|This series is the harmonic series (or  <math style="vertical-align: -5px">p</math>-series with  <math style="vertical-align: -5px">p=1</math> ).
|-
|Thus, it diverges. Hence, the series
|-
|        <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math>
|-
|is not absolutely convergent.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we need to look back at the original series to see
|-
|if it conditionally converges.
|-
|For
|-
|        <math>\sum_{n=1}^\infty \frac{(-1)^n}{n},</math>
|-
|we notice that this series is alternating.
|-
|Let  <math style="vertical-align: -14px"> b_n=\frac{1}{n}.</math>
|-
|First, we have
|-
|       <math>\frac{1}{n}\ge 0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|The sequence  <math style="vertical-align: -4px">\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{n+1}<\frac{1}{n}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n}=0.</math>
|-
|Therefore, the series  <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math>   converges
|-
|by the Alternating Series Test.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|Since the series
|-
|       <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math>  
|-
|converges but does not converge absolutely,
|-
|the series converges conditionally.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|        conditionally convergent (by the p-test and the Alternating Series Test)
|-
|
|}
[[009C_Sample_Midterm_1|'''Return to Sample Exam''']]

009C Sample Midterm 1, Problem 2 Detailed Solution

2018-01-05T21:15:49Z

Kayla Murray:

Consider the infinite series  <math>\sum_{n=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>

(a) Find an expression for the  <math style="vertical-align: 0px">n</math>th partial sum  <math style="vertical-align: -3px">s_n</math>  of the series.

(b) Compute  <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
<hr>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|The  <math style="vertical-align: 0px">n</math>th partial sum,  <math style="vertical-align: -3px">s_n,</math>  for a series  <math>\sum_{n=1}^\infty a_n </math>  is defined as
|-
|
        <math>s_n=\sum_{i=1}^n a_i.</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We need to find a pattern for the partial sums in order to find a formula.
|-
|We start by calculating  <math style="vertical-align: -3px">s_2.</math>  We have
|-
|        <math>s_2=2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Next, we calculate  <math style="vertical-align: -3px">s_3</math>  and  <math style="vertical-align: -3px">s_4.</math>  We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{s_3} & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg)+2\bigg(\frac{1}{2^3}-\frac{1}{2^4}\bigg)}\\
&&\\
& = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^4}\bigg)}
\end{array}</math>
|-
|and
|-
|        <math>\begin{array}{rcl}
\displaystyle{s_4} & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg)+2\bigg(\frac{1}{2^3}-\frac{1}{2^4}\bigg)+2\bigg(\frac{1}{2^4}-\frac{1}{2^5}\bigg)}\\
&&\\
& = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^5}\bigg).}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|If we look at  <math style="vertical-align: -4px">s_2,s_3,</math>  and  <math style="vertical-align: -4px">s_4, </math>  we notice a pattern.
|-
|From this pattern, we get the formula
|-
|        <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|From Part (a), we have
|-
|        <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We now calculate  <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
|-
|We get
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} s_n} & = & \displaystyle{\lim_{n\rightarrow \infty} 2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg)}\\
&&\\
& = & \displaystyle{\frac{2}{2^2}}\\
&&\\
& = & \displaystyle{\frac{1}{2}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg)</math>
|-
|    '''(b)'''     <math>\frac{1}{2}</math>
|}
[[009C_Sample_Midterm_1|'''Return to Sample Exam''']]

009C Sample Midterm 1, Problem 1 Detailed Solution

2018-01-05T21:12:42Z

Kayla Murray:

Does the following sequence converge or diverge?

 If the sequence converges, also find the limit of the sequence.

 Be sure to jusify your answers!

::<math>a_n=\frac{\ln n}{n}</math>
<hr>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''L'Hôpital's Rule, Part 2'''
|-
|
        Let  <math style="vertical-align: -5px">f</math>  and  <math style="vertical-align: -5px">g</math>  be differentiable functions on the open interval  <math style="vertical-align: -5px">(a,\infty)</math>  for some value  <math style="vertical-align: -4px">a,</math> 
|-
|        where  <math style="vertical-align: -5px">g'(x)\ne 0</math>  on  <math style="vertical-align: -5px">(a,\infty)</math>  and  <math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}</math>  returns either  <math style="vertical-align: -15px">\frac{0}{0}</math>  or  <math style="vertical-align: -15px">\frac{\infty}{\infty}.</math> 
|-
|       Then,   <math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, notice that
|-
|        <math>\lim_{n\rightarrow \infty} \ln n =\infty</math>
|-
|and
|-
|        <math>\lim_{n\rightarrow \infty} n=\infty.</math>
|-
|Therefore, the limit has the form  <math style="vertical-align: -11px">\frac{\infty}{\infty},</math>
|-
|which means that we can use L'Hopital's Rule to calculate this limit.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|First, switch to the variable  <math style="vertical-align: 0px">x</math>  so that we have functions and
|-
|can take derivatives. Thus, using L'Hopital's Rule, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \frac{\ln n}{n}} & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln x}{x}}\\
&&\\
& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\big(\frac{1}{x}\big)}{1}}\\
&&\\
& = & \displaystyle{0.}
\end{array}</math>
|-
|Therefore, the sequence converges and the limit of the sequence is  <math style="vertical-align: 0px">0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|        The sequence converges. The limit of the sequence is  <math style="vertical-align: 0px">0.</math>
|}
[[009C_Sample_Midterm_1|'''Return to Sample Exam''']]

009C Sample Midterm 1, Problem 1 Detailed Solution

2018-01-05T21:10:13Z

Kayla Murray:

Does the following sequence converge or diverge?

 If the sequence converges, also find the limit of the sequence.

 Be sure to jusify your answers!

::<math>a_n=\frac{\ln n}{n}</math>
<hr>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''L'Hôpital's Rule, Part 2'''
|-
|
        Let  <math style="vertical-align: -5px">f</math>  and  <math style="vertical-align: -5px">g</math>  be differentiable functions on the open interval  <math style="vertical-align: -5px">(a,\infty)</math>  for some value  <math style="vertical-align: -4px">a,</math> 
|-
|        where  <math style="vertical-align: -5px">g'(x)\ne 0</math>  on  <math style="vertical-align: -5px">(a,\infty)</math>  and  <math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}</math>  returns either  <math style="vertical-align: -15px">\frac{0}{0}</math>  or  <math style="vertical-align: -15px">\frac{\infty}{\infty}.</math> 
|-
|       Then,   <math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, notice that
|-
|        <math>\lim_{n\rightarrow \infty} \ln n =\infty</math>
|-
|and
|-
|        <math>\lim_{n\rightarrow \infty} n=\infty.</math>
|-
|Therefore, the limit has the form  <math style="vertical-align: -11px">\frac{\infty}{\infty},</math>
|-
|which means that we can use L'Hopital's Rule to calculate this limit.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|First, switch to the variable  <math style="vertical-align: 0px">x</math>   so that we have functions and
|-
|can take derivatives. Thus, using L'Hopital's Rule, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \frac{\ln n}{n}} & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln x}{x}}\\
&&\\
& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\big(\frac{1}{x}\big)}{1}}\\
&&\\
& = & \displaystyle{0.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|        The sequence converges. The limit of the sequence is  <math style="vertical-align: 0px">0.</math>
|}
[[009C_Sample_Midterm_1|'''Return to Sample Exam''']]

009C Sample Midterm 1, Problem 1 Detailed Solution

2018-01-05T21:10:00Z

Kayla Murray:

Does the following sequence converge or diverge?

 If the sequence converges, also find the limit of the sequence.

 Be sure to jusify your answers!

::<math>a_n=\frac{\ln n}{n}</math>
 

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''L'Hôpital's Rule, Part 2'''
|-
|
        Let  <math style="vertical-align: -5px">f</math>  and  <math style="vertical-align: -5px">g</math>  be differentiable functions on the open interval  <math style="vertical-align: -5px">(a,\infty)</math>  for some value  <math style="vertical-align: -4px">a,</math> 
|-
|        where  <math style="vertical-align: -5px">g'(x)\ne 0</math>  on  <math style="vertical-align: -5px">(a,\infty)</math>  and  <math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}</math>  returns either  <math style="vertical-align: -15px">\frac{0}{0}</math>  or  <math style="vertical-align: -15px">\frac{\infty}{\infty}.</math> 
|-
|       Then,   <math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, notice that
|-
|        <math>\lim_{n\rightarrow \infty} \ln n =\infty</math>
|-
|and
|-
|        <math>\lim_{n\rightarrow \infty} n=\infty.</math>
|-
|Therefore, the limit has the form  <math style="vertical-align: -11px">\frac{\infty}{\infty},</math>
|-
|which means that we can use L'Hopital's Rule to calculate this limit.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|First, switch to the variable  <math style="vertical-align: 0px">x</math>   so that we have functions and
|-
|can take derivatives. Thus, using L'Hopital's Rule, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \frac{\ln n}{n}} & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln x}{x}}\\
&&\\
& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\big(\frac{1}{x}\big)}{1}}\\
&&\\
& = & \displaystyle{0.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|        The sequence converges. The limit of the sequence is  <math style="vertical-align: 0px">0.</math>
|}
[[009C_Sample_Midterm_1|'''Return to Sample Exam''']]

007A Sample Midterm 3, Problem 5 Detailed Solution

2018-01-05T19:11:12Z

Kayla Murray:

At time  <math style="vertical-align: -4px">t,</math>  the position of a body moving along the  <math style="vertical-align: 0px">s-</math>axis is given by  <math style="vertical-align: -2px">s=t^3-6t^2+9t</math> (in meters and seconds).

(a)  Find the times when the velocity of the body is equal to  <math style="vertical-align: 0px">0.</math>

(b)  Find the body's acceleration each time the velocity is  <math style="vertical-align: 0px">0.</math>

(c)  Find the total distance traveled by the body from time  <math style="vertical-align: -1px">t=0</math>  second to  <math style="vertical-align: -1px">t=2</math>  seconds.
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' If  <math style="vertical-align: 0px">s</math>  is the position function of an object and
|-
|
:<math style="vertical-align: 0px">v</math>  is the velocity function of that same object,
|-
|
:then   <math style="vertical-align: 0px">v=s'.</math>
|-
|'''2.''' If  <math style="vertical-align: 0px">v</math>  is the velocity function of an object and
|-
|
:<math style="vertical-align: 0px">a</math>  is the acceleration function of that same object,
|-
|
:then  <math style="vertical-align: 0px">a=v'.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we need to find the velocity function of this body.
|-
|By the Power Rule, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{v} & = & \displaystyle{s'}\\
&&\\
& = & \displaystyle{(t^3-6t^2+9t)'}\\
&&\\
& = & \displaystyle{3t^2-12t+9.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we set the velocity function equal to  <math style="vertical-align: -1px">0</math>  and solve.
|-
|Hence, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{0} & = & \displaystyle{3t^2-12t+9}\\
&&\\
& = & \displaystyle{3(t^2-4t+3)}\\
&&\\
& = & \displaystyle{3(t-1)(t-3).}
\end{array}</math>
|-
|So, the two solutions are  <math style="vertical-align: -1px">t=1</math>  and  <math style="vertical-align: 0px">t=3.</math>
|-
|Therefore, the velocity is zero at 1 second and 3 seconds.
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we need to find the acceleration function of this body.
|-
|Using the Power Rule again, we have
|-
|
       <math>\begin{array}{rcl}
\displaystyle{a} & = & \displaystyle{v'}\\
&&\\
& = & \displaystyle{(3t^2-12t+9)'}\\
&&\\
& = & \displaystyle{6t-12.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we plug in  <math style="vertical-align: -1px">t=1</math>  and  <math style="vertical-align: -1px">t=3.</math> 
|-
|When  <math>t=1,</math>  we get
|-
|
        <math>\begin{array}{rcl}
\displaystyle{a} & = & \displaystyle{6-12}\\
&&\\
& = & \displaystyle{-6 ~\frac{\text{m}^2}{\text{s}}.}
\end{array}</math>
|-
|When  <math>t=3,</math>  we get
|-
|
        <math>\begin{array}{rcl}
\displaystyle{a} & = & \displaystyle{6(3)-12}\\
&&\\
& = & \displaystyle{6 ~\frac{\text{m}^2}{\text{s}}.}
\end{array}</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Since the velocity is  <math style="vertical-align: 0px">0</math>  at 1 second,
|-
|we need to consider the position of this body at 0, 1, and 2 seconds.
|-
|Plugging these values into the position function, we get
|-
|
::<math>s(0)=0,~s(1)=4,~s(2)=2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Hence, the total distance the body traveled is
|-
|
        <math>\begin{array}{rcl}
\displaystyle{[s(1)-s(0)]+ [s(1)-s(2)]} & = & \displaystyle{[4-0]+[4-2]}\\
&&\\
& = & \displaystyle{6 \text{ meters}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''   <math>1 \text{ second, } 3 \text{ seconds}</math>
|-
|   '''(b)'''   <math>-6 ~\frac{\text{m}^2}{\text{s}},~6 ~\frac{\text{m}^2}{\text{s}}</math>
|-
|   '''(c)'''   <math>6 \text{ meters}</math>
|}
[[007A_Sample_Midterm_3|'''Return to Sample Exam''']]

007A Sample Midterm 3, Problem 4 Detailed Solution

2018-01-05T19:06:25Z

Kayla Murray:

Consider the circle  <math style="vertical-align: -4px">x^2+y^2=25.</math>

(a)  Find  <math style="vertical-align: -14px">\frac{dy}{dx}.</math>

(b)  Find the equation of the tangent line at the point  <math style="vertical-align: -4px">(4,-3).</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' What is the result of implicit differentiation of  <math style="vertical-align: -4px">y^2?</math>
|-
|
        It would be  <math style="vertical-align: -13px">2y\cdot \frac{dy}{dx}</math>  by the Chain Rule.
|-
|'''2.''' What two pieces of information do you need to write the equation of a line?
|-
|
        You need the slope of the line and a point on the line.
|-
|'''3.''' What is the slope of the tangent line of a curve?
|-
|
        The slope is  <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using implicit differentiation on the equation  <math style="vertical-align: -4px">x^2+y^2=25,</math>  we get
|-
|
        <math>2x+2y\cdot\frac{dy}{dx}=0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, solve for  <math style="vertical-align: -12px">\frac{dy}{dx}.</math> 
|-
|So, we have
|-
|
       <math>2y\cdot\frac{dy}{dx}=-2x.</math>
|-
|We solve to get
|-
|       <math style="vertical-align: -17px">\frac{dy}{dx}=-\frac{x}{y}.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the slope of the tangent line at the point  <math style="vertical-align: -5px">(4,-3).</math>
|-
|We plug  <math style="vertical-align: -5px">(4,-3)</math>  into the formula for  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  we found in part (a).
|-
|So, we get
|-
|
        <math>\begin{array}{rcl}
\displaystyle{m} & = & \displaystyle{-\bigg(\frac{4}{-3}\bigg)}\\
&&\\
& = & \displaystyle{\frac{4}{3}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have the slope of the tangent line at  <math style="vertical-align: -5px">(4,-3)</math>  and a point.
|-
|Thus, we can write the equation of the line.
|-
|So, the equation of the tangent line at  <math style="vertical-align: -5px">(4,-3)</math>  is
|-
|
        <math>y\,=\,\frac{4}{3}(x-4)-3.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''    <math>\frac{dy}{dx}=-\frac{x}{y}</math>
|-
|    '''(b)'''    <math>y\,=\,\frac{4}{3}(x-4)-3</math>
|}
[[007A_Sample_Midterm_3|'''Return to Sample Exam''']]

007A Sample Midterm 3, Problem 3 Detailed Solution

2018-01-05T19:01:23Z

Kayla Murray:

Find the derivatives of the following functions. '''Do not simplify.'''

(a)  <math style="vertical-align: -16px">f(x)=\frac{(3x-5)(-x^{-2}+4x)}{x^{\frac{4}{5}}}</math>

(b)  <math>g(x)=\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{\pi}</math>  for  <math style="vertical-align: 0px">x>0.</math>

(c)  <math style="vertical-align: -17px">h(x)=\bigg(\frac{3x^2}{x+1}\bigg)^4</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' '''Product Rule'''
|-
|        <math>\frac{d}{dx}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)</math>
|-
|'''2.''' '''Quotient Rule'''
|-
|        <math>\frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}</math>
|-
|'''3.''' '''Chain Rule'''
|-
|        <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using the Quotient Rule, we have
|-
|        <math>f'(x)=\frac{x^{\frac{4}{5}}((3x-5)(-x^{-2}+4x))'-(3x-5)(-x^{-2}+4x)(x^{\frac{4}{5}})'}{(x^{\frac{4}{5}})^2}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we use the Product Rule and Power Rule to get
|-
|
        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^{\frac{4}{5}}((3x-5)(-x^{-2}+4x))'-(3x-5)(-x^{-2}+4x)(x^{\frac{4}{5}})'}{(x^{\frac{4}{5}})^2}}\\
&&\\
& = & \displaystyle{\frac{x^{\frac{4}{5}}[(3x-5)(-x^{-2}+4x)'+(3x-5)'(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{-\frac{1}{5}})}{(x^{\frac{4}{5}})^2}}\\
&&\\
& = & \displaystyle{\frac{x^{\frac{4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{-\frac{1}{5}})}{(x^{\frac{4}{5}})^2}.}
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we have
|-
|        <math>g'(x)=(\sqrt{x})'+\bigg(\frac{1}{\sqrt{x}}\bigg)'+(\sqrt{\pi})'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since  <math style="vertical-align: 0px">\pi</math>  is a constant,  <math style="vertical-align: -3px">\sqrt{\pi}</math>  is also a constant.
|-
|Hence,
|-
|        <math>(\sqrt{\pi})'=0.</math>
|-
|Therefore, using the Power Rule, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{g'(x)} & = & \displaystyle{(\sqrt{x})'+\bigg(\frac{1}{\sqrt{x}}\bigg)'+(\sqrt{\pi})'}\\
&&\\
& = & \displaystyle{\frac{1}{2}x^{-\frac{1}{2}}+-\frac{1}{2}x^{-\frac{3}{2}}+0}\\
&&\\
& = & \displaystyle{\frac{1}{2}x^{-\frac{1}{2}}+-\frac{1}{2}x^{-\frac{3}{2}}.}
\end{array}</math>
|}

'''(c)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, using the Chain Rule, we have
|-
|        <math>h'(x)=4\bigg(\frac{3x^2}{x+1}\bigg)^3 \bigg(\frac{3x^2}{x+1}\bigg)'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Quotient Rule and Power Rule, we get
|-
|        <math>\begin{array}{rcl}
\displaystyle{h'(x)} & = & \displaystyle{4\bigg(\frac{3x^2}{x+1}\bigg)^3 \bigg(\frac{3x^2}{x+1}\bigg)'}\\
&&\\
& = & \displaystyle{4\bigg(\frac{3x^2}{x+1}\bigg)^3 \bigg(\frac{(x+1)(3x^2)'-(3x^2)(x+1)'}{(x+1)^2}\bigg)}\\
&&\\
& = & \displaystyle{4\bigg(\frac{3x^2}{x+1}\bigg)^3 \bigg(\frac{(x+1)(6x)-3x^2}{(x+1)^2}\bigg).}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>f'(x)=\frac{x^{\frac{4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{-\frac{1}{5}})}{(x^{\frac{4}{5}})^2}</math>
|-
|    '''(b)'''     <math>g'(x)=\frac{1}{2}x^{-\frac{1}{2}}+-\frac{1}{2}x^{-\frac{3}{2}}</math>
|-
||    '''(c)'''     <math>h'(x)=4\bigg(\frac{3x^2}{x+1}\bigg)^3 \bigg(\frac{(x+1)(6x)-3x^2}{(x+1)^2}\bigg)</math>
|}
[[007A_Sample_Midterm_3|'''Return to Sample Exam''']]

007A Sample Midterm 3, Problem 1 Detailed Solution

2018-01-05T18:55:41Z

Kayla Murray:

Find the following limits:

(a) If  <math style="vertical-align: -16px">\lim _{x\rightarrow 3} \bigg(\frac{f(x)}{2x}+1\bigg)=2,</math>  find  <math style="vertical-align: -13px">\lim _{x\rightarrow 3} f(x).</math>

(b) Evaluate  <math style="vertical-align: -15px">\lim _{x\rightarrow 2} \frac{2-x}{x^2-4}. </math>

(c) Find  <math style="vertical-align: -19px">\lim _{x\rightarrow 0} \frac{\tan(4x)}{\sin(6x)}. </math>
<hr>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' If  <math style="vertical-align: -13px">\lim_{x\rightarrow a} g(x)\neq 0,</math>  we have
|-
|        <math>\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\displaystyle{\lim_{x\rightarrow a} f(x)}}{\displaystyle{\lim_{x\rightarrow a} g(x)}}.</math>
|-
|'''2.''' Recall
|-
|       <math style="vertical-align: -15px">\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{2} & = & \displaystyle{\lim_{x\rightarrow 3} \bigg(\frac{f(x)}{2x}+1\bigg)}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 3} \frac{f(x)}{2x}+\lim_{x\rightarrow 3} 1}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 3} \frac{f(x)}{2x}+1.}
\end{array}</math>
|-
|Therefore,
|-
|        <math>\lim_{x\rightarrow 3} \frac{f(x)}{2x}=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since  <math style="vertical-align: -13px">\lim_{x\rightarrow 3} 2x=6\ne 0,</math>  we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{1} & = & \displaystyle{\lim_{x\rightarrow 3} \frac{f(x)}{2x}}\\
&&\\
& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 3} f(x)}}{\displaystyle{\lim_{x\rightarrow 3} 2x}}}\\
&&\\
& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 3} f(x)}}{6}.}
\end{array}</math>
|-
|Multiplying both sides by  <math style="vertical-align: -5px">6,</math>  we get
|-
|        <math>\lim_{x\rightarrow 3} f(x)=6.</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 2} \frac{2-x}{x^2-4}} & = & \displaystyle{\lim_{x\rightarrow 2} \frac{2-x}{(x-2)(x+2)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 2} \frac{-(x-2)}{(x-2)(x+2)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 2} \frac{-1}{x+2}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 2} \frac{2-x}{x^2-4}} & = & \displaystyle{\lim_{x\rightarrow 2} \frac{-1}{x+2}}\\
&&\\
& = & \displaystyle{-\frac{1}{4}.}
\end{array}</math>
|}

'''(c)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 0} \frac{\tan(4x)}{\sin(6x)}} & = & \displaystyle{\lim_{x\rightarrow 0} \bigg[\frac{\sin(4x)}{\cos(4x)}\cdot \frac{1}{\sin(6x)}\bigg]}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0} \bigg[\frac{4}{6} \cdot \frac{\sin(4x)}{4x}\cdot \frac{6x}{\sin(6x)}\cdot\frac{1}{\cos(4x)}\bigg]}\\
&&\\
& = & \displaystyle{\frac{4}{6}\lim_{x\rightarrow 0} \bigg[\frac{\sin(4x)}{4x}\cdot \frac{6x}{\sin(6x)}\cdot \frac{1}{\cos(4x)}\bigg].}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 0} \frac{\tan(4x)}{\sin(6x)}} & = & \displaystyle{\frac{4}{6}\lim_{x\rightarrow 0} \bigg[\frac{\sin(4x)}{4x}\cdot\frac{6x}{\sin(6x)}\cdot\frac{1}{\cos(4x)}\bigg]}\\
&&\\
& = & \displaystyle{\frac{4}{6}\bigg(\lim_{x\rightarrow 0} \frac{\sin(4x)}{4x}\bigg)\cdot\bigg(\lim_{x\rightarrow 0} \frac{6x}{\sin(6x)}\bigg)\cdot\bigg(\lim_{x\rightarrow 0} \frac{1}{\cos(4x)}\bigg)}\\
&&\\
& = & \displaystyle{\frac{4}{6} \cdot (1)\cdot (1)\cdot(1)}\\
&&\\
& = & \displaystyle{\frac{2}{3}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>6</math>
|-
|    '''(b)'''     <math>-\frac{1}{4}</math>
|-
|    '''(c)'''     <math>\frac{2}{3}</math>
|}
[[007A_Sample_Midterm_3|'''Return to Sample Exam''']]

007A Sample Midterm 2, Problem 3 Detailed Solution

2018-01-05T18:44:45Z

Kayla Murray:

Find the derivatives of the following functions. '''Do not simplify.'''

(a)   <math style="vertical-align: -5px">f(x)=x^3(x^{\frac{4}{3}}-1)</math>

(b)   <math style="vertical-align: -14px">f(x)=\frac{x^3+x^{-3}}{1+6x}</math>  where  <math style="vertical-align: 0px">x>0</math>

(c)   <math style="vertical-align: -5px">f(x)=\sqrt{3x^2+5x-7}</math>  where  <math style="vertical-align: 0px">x>0</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' '''Product Rule'''
|-
|        <math>\frac{d}{dx}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)</math>
|-
|'''2.''' '''Quotient Rule'''
|-
|        <math>\frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}</math>
|-
|'''3.''' '''Chain Rule'''
|-
|        <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using the Product Rule, we have
|-
|        <math>f'(x)=(x^3)(x^{\frac{4}{3}}-1)'+(x^3)'(x^{\frac{4}{3}}-1).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Power Rule, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{(x^3)(x^{\frac{4}{3}}-1)'+(x^3)'(x^{\frac{4}{3}}-1)}\\
&&\\
& = & \displaystyle{(x^3)\bigg(\frac{4}{3}x^{\frac{1}{3}}\bigg)+(3x^2)(x^{\frac{4}{3}}-1).}
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using the Quotient Rule, we have
|-
|        <math>f'(x)=\frac{(1+6x)(x^3+x^{-3})'-(x^3+x^{-3})(1+6x)'}{(1+6x)^2}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Power Rule, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{(1+6x)(x^3+x^{-3})'-(x^3+x^{-3})(1+6x)'}{(1+6x)^2}}\\
&&\\
& = & \displaystyle{\frac{(1+6x)(3x^2-3x^{-4})-(x^3+x^{-3})(6)}{(1+6x)^2}.}
\end{array}</math>
|}

'''(c)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write
|-
|       <math>f(x)=(3x^2+5x-7)^{\frac{1}{2}}.</math>
|-
|Using the Chain Rule, we have
|-
|        <math>f'(x)=\frac{1}{2} (3x^2+5x-7)^{-\frac{1}{2}}(3x^2+5x-7)'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Power Rule, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{1}{2} (3x^2+5x-7)^{-\frac{1}{2}}(3x^2+5x-7)'}\\
&&\\
& = & \displaystyle{\frac{1}{2} (3x^2+5x-7)^{-\frac{1}{2}}(6x+5).}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>f'(x)=(x^3)\bigg(\frac{4}{3}x^{\frac{1}{3}}\bigg)+(3x^2)(x^{\frac{4}{3}}-1)</math>
|-
|    '''(b)'''     <math>f'(x)=\frac{(1+6x)(3x^2-3x^{-4})-(x^3+x^{-3})(6)}{(1+6x)^2}</math>
|-
|    '''(c)'''     <math>f'(x)=\frac{1}{2} (3x^2+5x-7)^{-\frac{1}{2}}(6x+5)</math>
|}
[[007A_Sample_Midterm_2|'''Return to Sample Exam''']]

007A Sample Midterm 2, Problem 1 Detailed Solution

2018-01-05T18:30:51Z

Kayla Murray:

Evaluate the following limits.

(a) Find  <math style="vertical-align: -14px">\lim _{x\rightarrow 2} \frac{\sqrt{x^2+12}-4}{x-2}</math>

(b) Find  <math style="vertical-align: -19px">\lim _{x\rightarrow 0} \frac{\sin(3x)}{\sin(7x)} </math>

(c) Evaluate  <math style="vertical-align: -16px">\lim _{x\rightarrow 0} x^2\cos\bigg(\frac{1}{x}\bigg) </math>
<hr>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.'''  <math>\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>
|-
|'''2.''' '''Squeeze Theorem'''
|-
|       Let  <math style="vertical-align: -4px">f,g</math>  and  <math style="vertical-align: 0px">h</math>  be functions on an open interval  <math style="vertical-align: 0px">I</math>  containing  <math style="vertical-align: 0px">c</math> 
|-
|       such that for all  <math style="vertical-align: 0px">x</math>  in  <math style="vertical-align: -5px">I,~f(x)\le g(x)\le h(x).</math>
|-
|       If  <math style="vertical-align: -13px">\lim_{x\rightarrow c} f(x)=L=\lim_{x\rightarrow c} h(x),</math>  then  <math style="vertical-align: -13px">\lim_{x\rightarrow c} g(x)=L.</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by noticing that if we plug in  <math style="vertical-align: 0px">x=2</math>  into
|-
|        <math>\frac{\sqrt{x^2+12}-4}{x-2},</math>
|-
|we get   <math style="vertical-align: -12px">\frac{0}{0}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we multiply the numerator and denominator by the conjugate of the numerator.
|-
|Hence, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim _{x\rightarrow 2} \frac{\sqrt{x^2+12}-4}{x-2}} & = & \displaystyle{\lim_{x\rightarrow 2} \bigg[\frac{(\sqrt{x^2+12}-4)}{(x-2)}\cdot \frac{(\sqrt{x^2+12}+4)}{(\sqrt{x^2+12}+4)}\bigg]}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 2} \frac{(x^2+12)-16}{(x-2)(\sqrt{x^2+12}+4)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 2} \frac{x^2-4}{(x-2)(\sqrt{x^2+12}+4)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 2} \frac{(x-2)(x+2)}{(x-2)(\sqrt{x^2+12}+4)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 2} \frac{x+2}{\sqrt{x^2+12}+4}}\\
&&\\
& = & \displaystyle{\frac{4}{8}}\\
&&\\
& = & \displaystyle{\frac{1}{2}.}
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 0} \frac{\sin(3x)}{\sin(7x)}} & = & \displaystyle{\lim_{x\rightarrow 0} \bigg[\frac{\sin(3x)}{x} \cdot \frac{x}{\sin(7x)}\bigg]}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0} \bigg[\frac{3}{7} \cdot \frac{\sin(3x)}{3x}\cdot \frac{7x}{\sin(7x)}\bigg]}\\
&&\\
& = & \displaystyle{\frac{3}{7}\lim_{x\rightarrow 0} \bigg[\frac{\sin(3x)}{3x}\cdot \frac{7x}{\sin(7x)}\bigg].}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 0} \frac{\sin(3x)}{\sin(7x)}} & = & \displaystyle{\frac{3}{7}\lim_{x\rightarrow 0} \bigg[\frac{\sin(3x)}{3x}\cdot \frac{7x}{\sin(7x)}\bigg]}\\
&&\\
& = & \displaystyle{\frac{3}{7}\bigg(\lim_{x\rightarrow 0} \frac{\sin(3x)}{3x}\bigg)\cdot \bigg(\lim_{x\rightarrow 0} \frac{7x}{\sin(7x)}\bigg)}\\
&&\\
& = & \displaystyle{\frac{3}{7} \cdot (1)\cdot (1)}\\
&&\\
& = & \displaystyle{\frac{3}{7}.}
\end{array}</math>
|}

'''(c)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, recall that
|-
|
::<math>-1\le \cos (\theta) \le 1</math>
|-
|for all  <math style="vertical-align: 0px">\theta.</math>
|-
|Then, for all  <math style="vertical-align: -5px">x\neq 0,</math>
|-
|
::<math>-1\le \cos\bigg(\frac{1}{x}\bigg)\le 1.</math>
|-
|Hence, for all  <math style="vertical-align: -5px">x\neq 0,</math>
|-
|
::<math>-x^2\le x^2\cos\bigg(\frac{1}{x}\bigg) \le x^2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, notice
|-
|
::<math>\lim_{x\rightarrow 0} x^2=0</math>
|-
|and
|-
|
::<math>\lim_{x\rightarrow 0}-x^2=0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Since
|-
|
::<math>\lim_{x\rightarrow 0}x^2=\lim_{x\rightarrow 0} -x^2 =0,</math>
|-
|we have
|-
|
::<math>\lim_{x\rightarrow 0}x^2\cos \bigg(\frac{1}{x}\bigg)=0</math>
|-
|by the Squeeze Theorem.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>\frac{1}{2}</math>
|-
|    '''(b)'''     <math>\frac{3}{7}</math>
|-
|    '''(c)'''     <math>0</math>
|}
[[007A_Sample_Midterm_2|'''Return to Sample Exam''']]

007A Sample Midterm 1, Problem 4 Detailed Solution

2018-01-05T18:21:38Z

Kayla Murray:

Find the derivatives of the following functions. '''Do not simplify.'''

(a)   <math style="vertical-align: -5px">f(x)=\sqrt{x}(x^2+2)</math>

(b)   <math style="vertical-align: -17px">g(x)=\frac{x+3}{x^{\frac{3}{2}}+2}</math> where <math style="vertical-align: 0px">x>0</math>

(c)   <math style="vertical-align: -8px">h(x)=\sqrt{x+\sqrt{x}}</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' '''Product Rule'''
|-
|        <math>\frac{d}{dx}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)</math>
|-
|'''2.''' '''Quotient Rule'''
|-
|        <math>\frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}</math>
|-
|'''3.''' '''Chain Rule'''
|-
|        <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using the Product Rule, we have
|-
|        <math>f'(x)=(\sqrt{x})'(x^2+2)+\sqrt{x}(x^2+2)'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Power Rule, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{(\sqrt{x})'(x^2+2)+\sqrt{x}(x^2+2)'}\\
&&\\
& = & \displaystyle{\bigg(\frac{1}{2}x^{-\frac{1}{2}}\bigg)(x^2+2)+\sqrt{x}(2x).}
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using the Quotient Rule, we have
|-
|        <math>g'(x)=\frac{(x^{\frac{3}{2}}+2)(x+3)'-(x+3)(x^{\frac{3}{2}}+2)'}{(x^{\frac{3}{2}}+2)^2}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Power Rule, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{g'(x)} & = & \displaystyle{\frac{(x^{\frac{3}{2}}+2)(x+3)'-(x+3)(x^{\frac{3}{2}}+2)'}{(x^{\frac{3}{2}}+2)^2}}\\
&&\\
& = & \displaystyle{\frac{(x^{\frac{3}{2}}+2)(1)-(x+3)(\frac{3}{2}x^{\frac{1}{2}})}{(x^{\frac{3}{2}}+2)^2}.}
\end{array}</math>
|}

'''(c)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write
|-
|       <math>h(x)=(x+x^{\frac{1}{2}})^{\frac{1}{2}}.</math>
|-
|Then, using the Chain Rule, we have
|-
|        <math>h'(x)=\frac{1}{2}(x+x^{\frac{1}{2}})^{-\frac{1}{2}}(x+x^{\frac{1}{2}})'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Power Rule, we have
|-
|       <math>h'(x)=\frac{1}{2}(x+x^{\frac{1}{2}})^{-\frac{1}{2}}\bigg(1+\frac{1}{2}x^{-\frac{1}{2}}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>f'(x)=\bigg(\frac{1}{2}x^{-\frac{1}{2}}\bigg)(x^2+2)+\sqrt{x}(2x)</math>
|-
|    '''(b)'''     <math>g'(x)=\frac{(x^{\frac{3}{2}}+2)(1)-(x+3)(\frac{3}{2}x^{\frac{1}{2}})}{(x^{\frac{3}{2}}+2)^2}</math>
|-
|    '''(c)'''     <math>h'(x)=\frac{1}{2}(x+x^{\frac{1}{2}})^{-\frac{1}{2}}\bigg(1+\frac{1}{2}x^{-\frac{1}{2}}\bigg)</math>
|}
[[007A_Sample_Midterm_1|'''Return to Sample Exam''']]

007A Sample Midterm 1, Problem 3 Detailed Solution

2018-01-05T18:18:16Z

Kayla Murray:

Let  <math style="vertical-align: -5px">y=2x^2-3x+1.</math>

(a) Use the '''definition of the derivative''' to compute   <math style="vertical-align: -13px">\frac{dy}{dx}.</math>

(b) Find the equation of the tangent line to  <math style="vertical-align: -4px">y=2x^2-3x+1</math>  at  <math style="vertical-align: -4px">(2,3).</math>
<hr>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|Recall
|-
|        <math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let  <math style="vertical-align: -5px">f(x)=2x^2-3x+1.</math>
|-
|Using the limit definition of the derivative, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{2(x+h)^2-3(x+h)+1-(2x^2-3x+1)}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{2x^2+4xh+2h^2-3x-3h+1-2x^2+3x-1}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{4xh+2h^2-3h}{h}.}
\end{array}</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we simplify to get
|-
|        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{h(4x+2h-3)}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} (4x+2h-3)}\\
&&\\
& = & \displaystyle{4x-3.}
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by finding the slope of the tangent line to  <math style="vertical-align: -5px">f(x)=2x^2-3x+1</math>  at  <math style="vertical-align: -5px">(2,3).</math>
|-
|Using the derivative calculated in part (a), the slope is
|-
|        <math>\begin{array}{rcl}
\displaystyle{m} & = & \displaystyle{f'(2)}\\
&&\\
& = & \displaystyle{4(2)-3}\\
&&\\
& = & \displaystyle{5.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, the tangent line to  <math style="vertical-align: -5px">f(x)=2x^2-3x+1</math>  at  <math style="vertical-align: -5px">(2,3)</math>
|-
|has slope  <math style="vertical-align: 0px">m=5</math>  and passes through the point  <math style="vertical-align: -5px">(2,3).</math>
|-
|Hence, the equation of this line is
|-
|
::<math>y=5(x-2)+3.</math>
|-
|If we simplify, we get
|-
|
::<math>y=5x-7.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>\frac{dy}{dx}=4x-3</math>
|-
|    '''(b)'''     <math>y=5x-7</math>
|}
[[007A_Sample_Midterm_1|'''Return to Sample Exam''']]

007A Sample Midterm 1, Problem 3 Detailed Solution

2018-01-05T18:15:44Z

Kayla Murray:

Let  <math style="vertical-align: -5px">y=2x^2-3x+1.</math>

(a) Use the '''definition of the derivative''' to compute   <math style="vertical-align: -13px">\frac{dy}{dx}.</math>

(b) Find the equation of the tangent line to  <math style="vertical-align: -4px">y=2x^2-3x+1</math>  at  <math style="vertical-align: -4px">(2,3).</math>
<hr>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|Recall
|-
|        <math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let  <math style="vertical-align: -5px">f(x)=2x^2-3x+1.</math>
|-
|Using the limit definition of the derivative, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{2(x+h)^2-3(x+h)+1-(2x^2-3x+1)}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{2x^2+4xh+2h^2-3x-3h+1-2x^2+3x-1}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{4xh+2h^2-3h}{h}.}
\end{array}</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we simplify to get
|-
|        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{h(4x+2h-3)}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} 4x+2h-3}\\
&&\\
& = & \displaystyle{4x-3.}
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by finding the slope of the tangent line to  <math style="vertical-align: -5px">f(x)=2x^2-3x+1</math>  at  <math style="vertical-align: -5px">(2,3).</math>
|-
|Using the derivative calculated in part (a), the slope is
|-
|        <math>\begin{array}{rcl}
\displaystyle{m} & = & \displaystyle{f'(2)}\\
&&\\
& = & \displaystyle{4(2)-3}\\
&&\\
& = & \displaystyle{5.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, the tangent line to  <math style="vertical-align: -5px">f(x)=2x^2-3x+1</math>  at  <math style="vertical-align: -5px">(2,3)</math>
|-
|has slope  <math style="vertical-align: 0px">m=5</math>  and passes through the point  <math style="vertical-align: -5px">(2,3).</math>
|-
|Hence, the equation of this line is
|-
|
::<math>y=5(x-2)+3.</math>
|-
|If we simplify, we get
|-
|
::<math>y=5x-7.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>4x-3</math>
|-
|    '''(b)'''     <math>y=5x-7</math>
|}
[[007A_Sample_Midterm_1|'''Return to Sample Exam''']]

007A Sample Midterm 1, Problem 2 Detailed Solution

2018-01-05T18:14:43Z

Kayla Murray:

Consider the following function  <math style="vertical-align: -5px"> f:</math>
::<math>f(x) = \left\{
\begin{array}{lr}
x^2 & \text{if }x < 1\\
\sqrt{x} & \text{if }x \geq 1
\end{array}
\right.
</math>

(a) Find  <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^-} f(x).</math>

(b) Find  <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^+} f(x).</math>

(c) Find  <math style="vertical-align: -13px"> \lim_{x\rightarrow 1} f(x).</math>

(d) Is  <math style="vertical-align: -5px">f</math>  continuous at  <math style="vertical-align: -1px">x=1?</math>  Briefly explain.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' If  <math style="vertical-align: -15px">\lim_{x\rightarrow a^-} f(x)=\lim_{x\rightarrow a^+} f(x)=c,</math>
|-
|        then  <math style="vertical-align: -12px">\lim_{x\rightarrow a} f(x)=c.</math>
|-
|'''2.'''  <math style="vertical-align: -5px">f(x)</math>  is continuous at  <math style="vertical-align: 0px">x=a</math>  if
|-
|        <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Notice that we are calculating a left hand limit.
|-
|Thus, we are looking at values of  <math style="vertical-align: 0px">x</math>  that are smaller than  <math style="vertical-align: -1px">1.</math>
|-
|Using the definition of  <math style="vertical-align: -5px">f(x),</math>  we have
|-
|        <math>\lim_{x\rightarrow 1^-} f(x)=\lim_{x\rightarrow 1^-} x^2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 1^-} f(x)} & = & \displaystyle{\lim_{x\rightarrow 1^-} x^2}\\
&&\\
& = & \displaystyle{1^2}\\
&&\\
& = & \displaystyle{1.}\\
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Notice that we are calculating a right hand limit.
|-
|Thus, we are looking at values of  <math style="vertical-align: 0px">x</math>  that are bigger than  <math style="vertical-align: -2px">1.</math>
|-
|Using the definition of  <math style="vertical-align: -5px">f(x),</math>  we have
|-
|        <math>\lim_{x\rightarrow 1^+} f(x)=\lim_{x\rightarrow 1^+} \sqrt{x}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 1^+} f(x)} & = & \displaystyle{\lim_{x\rightarrow 1^+} \sqrt{x}}\\
&&\\
& = & \displaystyle{\sqrt{1}}\\
&&\\
& = & \displaystyle{1.}\\
\end{array}</math>
|}

'''(c)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|From (a) and (b), we have
|-
|        <math>\lim_{x\rightarrow 1^-}f(x)=1</math>
|-
|and
|-
|        <math>\lim_{x\rightarrow 1^+}f(x)=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since
|-
|        <math>\lim_{x\rightarrow 1^-}f(x)=\lim_{x\rightarrow 1^+}f(x)=1,</math>
|-
|we have
|-
|        <math>\lim_{x\rightarrow 1}f(x)=1.</math>
|}

'''(d)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|From (c), we have
|-
|        <math>\lim_{x\rightarrow 1}f(x)=1.</math>
|-
|Also,
|-
|        <math>f(1)=\sqrt{1}=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since
|-
|        <math>\lim_{x\rightarrow 1}f(x)=f(1),</math>
|-
|<math style="vertical-align: -5px">f(x)</math>  is continuous at  <math style="vertical-align: -1px">x=1.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>1</math>
|-
|    '''(b)'''     <math>1</math>
|-
|    '''(c)'''     <math>1</math>
|-
|    '''(d)'''     <math style="vertical-align: -5px">f(x)</math>  is continuous at  <math style="vertical-align: -1px">x=1</math>  since  <math style="vertical-align: -12px">\lim_{x\rightarrow 1}f(x)=f(1).</math>
|}
[[007A_Sample_Midterm_1|'''Return to Sample Exam''']]

007A Sample Midterm 1, Problem 1 Detailed Solution

2018-01-05T18:11:26Z

Kayla Murray:

Find the following limits:

(a) Find  <math style="vertical-align: -13px">\lim _{x\rightarrow 2} g(x),</math>  provided that  <math style="vertical-align: -15px">\lim _{x\rightarrow 2} \bigg[\frac{4-g(x)}{x}\bigg]=5.</math>

(b) Find  <math style="vertical-align: -14px">\lim _{x\rightarrow 0} \frac{\sin(4x)}{5x} </math>

(c) Evaluate  <math style="vertical-align: -14px">\lim _{x\rightarrow -3^+} \frac{x}{x^2-9} </math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
| '''1.''' If  <math style="vertical-align: -12px">\lim_{x\rightarrow a} g(x)\neq 0,</math>  we have
|-
|        <math>\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\displaystyle{\lim_{x\rightarrow a} f(x)}}{\displaystyle{\lim_{x\rightarrow a} g(x)}}.</math>
|-
| '''2.''' Recall
|-
|       <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Since  <math style="vertical-align: -12px">\lim_{x\rightarrow 2} x =2\ne 0,</math>
|-
|we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{5} & = & \displaystyle{\lim _{x\rightarrow 2} \bigg[\frac{4-g(x)}{x}\bigg]}\\
&&\\
& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 2} (4-g(x))}}{\displaystyle{\lim_{x\rightarrow 2} x}}}\\
&&\\
& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 2} (4-g(x))}}{2}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|If we multiply both sides of the last equation by  <math>2,</math>  we get
|-
|        <math>10=\lim_{x\rightarrow 2} (4-g(x)).</math>
|-
|Now, using linearity properties of limits, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{10} & = & \displaystyle{\lim_{x\rightarrow 2} 4 -\lim_{x\rightarrow 2}g(x)}\\
&&\\
& = & \displaystyle{4-\lim_{x\rightarrow 2} g(x).}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Solving for  <math style="vertical-align: -12px">\lim_{x\rightarrow 2} g(x)</math>  in the last equation,
|-
|we get
|-
|
        <math> \lim_{x\rightarrow 2} g(x)=-6.</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write
|-
|        <math>\lim_{x\rightarrow 0} \frac{\sin(4x)}{5x}=\lim_{x\rightarrow 0} \bigg(\frac{4}{5} \cdot \frac{\sin(4x)}{4x}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 0} \frac{\sin(4x)}{5x}} & = & \displaystyle{\frac{4}{5}\lim_{x\rightarrow 0} \frac{\sin(4x)}{4x}}\\
&&\\
& = & \displaystyle{\frac{4}{5}(1)}\\
&&\\
& = & \displaystyle{\frac{4}{5}.}
\end{array}</math>
|}

'''(c)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|When we plug in values close to  <math style="vertical-align: 0px">-3</math>  into   <math style="vertical-align: -12px">\frac{x}{x^2-9},</math>
|-
|we get a small denominator, which results in a large number.
|-
|Thus,
|-
|        <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}</math>
|-
|is either equal to  <math style="vertical-align: -1px">\infty</math>  or  <math style="vertical-align: -1px">-\infty.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|To figure out which one, we factor the denominator to get
|-
|        <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}=\lim_{x\rightarrow -3^+} \frac{x}{(x-3)(x+3)}.</math>
|-
|We are taking a right hand limit. So, we are looking at values of  <math style="vertical-align: 0px">x</math>
|-
|a little bigger than  <math style="vertical-align: 0px">-3.</math>  (You can imagine values like  <math style="vertical-align: 0px">x=-2.9.</math> )
|-
|For these values, the numerator will be negative.
|-
|Also, for these values,  <math style="vertical-align: 0px">x-3</math>  will be negative and  <math style="vertical-align: -1px">x+3</math>  will be positive.
|-
|Therefore, the denominator will be negative.
|-
|Since both the numerator and denominator will be negative (have the same sign),
|-
|        <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}=\infty.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math> -6</math>
|-
|    '''(b)'''     <math>\frac{4}{5}</math>
|-
|    '''(c)'''     <math>\infty</math>
|}
[[007A_Sample_Midterm_1|'''Return to Sample Exam''']]

009C Sample Final 3, Problem 10 Detailed Solution

2017-12-03T23:41:00Z

Kayla Murray:

A curve is described parametrically by
::<math>x=t^2</math>
::<math>y=t^3-t</math>

(a) Sketch the curve for  <math style="vertical-align: -2px">-2\le t \le 2.</math>

(b) Find the equation of the tangent line to the curve at the origin.
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' What two pieces of information do you need to write the equation of a line?
|-
|
       You need the slope of the line and a point on the line.
|-
|'''2.''' What is the slope of the tangent line of a parametric curve?
|-
|
       The slope is  <math style="vertical-align: -21px">m=\frac{dy}{dx}=\frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(a)  
|-
| 
|-
|[[File:9CSF3_10_GP.png|center|277px]]
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we need to find the slope of the tangent line.
|-
|Since   <math style="vertical-align: -14px">\frac{dy}{dt}=3t^2-1</math>   and   <math style="vertical-align: -14px">\frac{dx}{dt}=2t,</math>  we have
|-
|
       <math>\frac{dy}{dx}=\frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}=\frac{3t^2-1}{2t}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, the origin corresponds to  <math style="vertical-align: 0px">x=0</math>  and  <math style="vertical-align: -4px">y=0.</math>
|-
|This gives us two equations. When we solve for  <math style="vertical-align: -4px">t,</math>  we get  <math style="vertical-align: -1px">t=0.</math>
|-
|Plugging in  <math style="vertical-align: -1px">t=0</math>  into
|-
|       <math>\frac{dy}{dx}=\frac{3t^2-1}{2t},</math>
|-
|we see that  <math style="vertical-align: -14px">\frac{dy}{dx}</math>  is undefined at  <math style="vertical-align: -1px">t=0.</math>
|-
|So, there is no tangent line at the origin.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''    See above
|-
|    '''(b)'''    There is no tangent line at the origin.
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 3, Problem 9 Detailed Solution

2017-12-03T23:38:56Z

Kayla Murray:

[[File:9CSF3_9a_GP.png|right|400px]]

A wheel of radius 1 rolls along a straight line, say the  <math>x</math>-axis. A point  <math style="vertical-align: 0px">P</math>  is located halfway between the center of the wheel and the rim; assume  <math style="vertical-align: 0px">P</math>  starts at the point  <math style="vertical-align: -15px">\bigg(0,\frac{1}{2}\bigg).</math>  As the wheel rolls,  <math style="vertical-align: 0px">P</math>  traces a curve. Find parametric equations for the curve.
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|Many concepts in physics involve the notion of a relative frame.
|-
|For example, if I'm in a box dropped from an airplane, I won't be moving relative to the box. However, I'm still heading towards the ground with acceleration  <math style="vertical-align: -5px">10\,\textrm{m/sec}^{2}.</math>  Say it drops for 5 seconds, so the box is going  <math style="vertical-align: -5px">50\,\textrm{m/sec}</math>  when it hits the ground. Even if I jump with all my might and pull off something like  <math style="vertical-align: -5px">5\,\textrm{m/sec}</math>  of upward velocity, I'll still feel the impact of hitting the ground at
|-
|       <math>50-5\,\textrm{m/sec}=45\,\textrm{m/sec}. </math>
|-
|Essentially, equations of motion can often be broken into parts, and then added up.
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|If a wheel of radius one is resting at the origin, its axis will be at the point  <math style="vertical-align: -5px">(1,0). </math>  For this solution, we will assume the point  <math style="vertical-align: 0px">P </math>  is below the axle, although the problem does not state the position of  <math style="vertical-align: 0px">P </math>.  When the wheel rotates clockwise, it will move to the right. Since the length of the arc defined by an angle  <math style="vertical-align: 0px">\theta </math>  on a circle of radius  <math style="vertical-align: 0px">R </math>   is   <math style="vertical-align: -4px">L=R\cdot\theta=1\cdot\theta=\theta, </math>  the wheel will roll forward the length of the arc, which is just  <math style="vertical-align: 0px">\theta. </math>

Moreover, the axle's  <math style="vertical-align: 0px">x </math>  position will change in the same manner, while the height of the axle will always be fixed at  <math style="vertical-align: -4px">y=1 </math>.  This means we can describe the position of the axle as a function of  <math style="vertical-align: -4px">\theta, </math>  or

:: <math>a(\theta)\,=\,(\theta,1). </math>

|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since the wheel is rotating, we also know that our point  <math style="vertical-align: 0px">P </math>  will rotate around the axle. As described in the problem, it is halfway between the rim and the center/axle, so it is  <math style="vertical-align: -5px">1/2 </math>  unit away from the axle, and will rotate clockwise. Using our trig relations (while looking at the image), we find that the position of  <math style="vertical-align: 0px">P </math>  relative to the axle can be described as

:: <math>p(\theta)\,=\,\left(-\frac{1}{2}\sin\theta,-\frac{1}{2}\cos\theta\right). </math>

Notice that when  <math style="vertical-align: -4px">\theta=0, </math>  the point would be at the position

:: <math>p(0)\,=\,\left(-\frac{1}{2}\sin0,-\frac{1}{2}\cos0\right)\,=\,\left(0,-\frac{1}{2}\right), </math>

which is half a unit directly below the axle. This is shown as a gray "ghost" dot in the image, while the black triangle and circle represent the situation at  <math style="vertical-align: -5px">\theta=\pi/4. </math>

|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|We therefore have a frame (the axle) that is moving, and a point  <math style="vertical-align: 0px">P </math>  that is moving relative to the frame. To get the movement relative to the stationary "world", we simply add them up to find

:: <math>P(\theta)\,=\,a(\theta)+p(\theta)\,=\,\left(\theta-\frac{1}{2}\sin\theta,1-\frac{1}{2}\cos\theta\right). </math>

|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|
:: <math>P(\theta)\,=\,\left(\theta-\frac{1}{2}\sin\theta,1-\frac{1}{2}\cos\theta\right). </math>
|  
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]
 
[[File:9CSF3_9Ani_ezgif.gif |center|600px]]
 

'''Contributions to this page were made by [[Contributors|John Simanyi]]'''

009C Sample Final 3, Problem 9 Detailed Solution

2017-12-03T23:35:33Z

Kayla Murray:

[[File:9CSF3_9a_GP.png|right|400px]]

A wheel of radius 1 rolls along a straight line, say the  <math>x</math>-axis. A point  <math style="vertical-align: 0px">P</math>  is located halfway between the center of the wheel and the rim; assume  <math style="vertical-align: 0px">P</math>  starts at the point  <math style="vertical-align: -15px">\bigg(0,\frac{1}{2}\bigg).</math>  As the wheel rolls,  <math style="vertical-align: 0px">P</math>  traces a curve. Find parametric equations for the curve.
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|Many concepts in physics involve the notion of a relative frame. For example, if I'm in a box dropped from an airplane, I won't be moving relative to the box. However, I'm still heading towards the ground with acceleration <math style="vertical-align: -5px">10\,\textrm{m/sec}^{2}. </math>  Say it drops for 5 seconds, so the box is going <math style="vertical-align: -5px">50\,\textrm{m/sec} </math> when it hits the ground. Even if I jump with all my might and pull off something like <math style="vertical-align: -5px">5\,\textrm{m/sec} </math> of upward velocity, I'll still feel the impact of hitting the ground at

     <math>50-5\,\textrm{m/sec}=45\,\textrm{m/sec}. </math>

Essentially, equations of motion can often be broken into parts, and
then added up.
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|If a wheel of radius one is resting at the origin, its axis will be at the point  <math style="vertical-align: -5px">(1,0). </math>  For this solution, we will assume the point  <math style="vertical-align: 0px">P </math>  is below the axle, although the problem does not state the position of  <math style="vertical-align: 0px">P </math>.  When the wheel rotates clockwise, it will move to the right. Since the length of the arc defined by an angle  <math style="vertical-align: 0px">\theta </math>  on a circle of radius  <math style="vertical-align: 0px">R </math>   is   <math style="vertical-align: -4px">L=R\cdot\theta=1\cdot\theta=\theta, </math>  the wheel will roll forward the length of the arc, which is just  <math style="vertical-align: 0px">\theta. </math>

Moreover, the axle's  <math style="vertical-align: 0px">x </math>  position will change in the same manner, while the height of the axle will always be fixed at  <math style="vertical-align: -4px">y=1 </math>.  This means we can describe the position of the axle as a function of  <math style="vertical-align: -4px">\theta, </math>  or

:: <math>a(\theta)\,=\,(\theta,1). </math>

|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since the wheel is rotating, we also know that our point  <math style="vertical-align: 0px">P </math>  will rotate around the axle. As described in the problem, it is halfway between the rim and the center/axle, so it is  <math style="vertical-align: -5px">1/2 </math>  unit away from the axle, and will rotate clockwise. Using our trig relations (while looking at the image), we find that the position of  <math style="vertical-align: 0px">P </math>  relative to the axle can be described as

:: <math>p(\theta)\,=\,\left(-\frac{1}{2}\sin\theta,-\frac{1}{2}\cos\theta\right). </math>

Notice that when  <math style="vertical-align: -4px">\theta=0, </math>  the point would be at the position

:: <math>p(0)\,=\,\left(-\frac{1}{2}\sin0,-\frac{1}{2}\cos0\right)\,=\,\left(0,-\frac{1}{2}\right), </math>

which is half a unit directly below the axle. This is shown as a gray "ghost" dot in the image, while the black triangle and circle represent the situation at  <math style="vertical-align: -5px">\theta=\pi/4. </math>

|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|We therefore have a frame (the axle) that is moving, and a point  <math style="vertical-align: 0px">P </math>  that is moving relative to the frame. To get the movement relative to the stationary "world", we simply add them up to find

:: <math>P(\theta)\,=\,a(\theta)+p(\theta)\,=\,\left(\theta-\frac{1}{2}\sin\theta,1-\frac{1}{2}\cos\theta\right). </math>

|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|
:: <math>P(\theta)\,=\,\left(\theta-\frac{1}{2}\sin\theta,1-\frac{1}{2}\cos\theta\right). </math>
|  
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]
 
[[File:9CSF3_9Ani_ezgif.gif |center|600px]]
 

'''Contributions to this page were made by [[Contributors|John Simanyi]]'''

009C Sample Final 3, Problem 9

2017-12-03T23:35:18Z

Kayla Murray:

A wheel of radius 1 rolls along a straight line, say the  <math>x</math>-axis. A point  <math style="vertical-align: 0px">P</math>  is located halfway between the center of the wheel and the rim; assume  <math style="vertical-align: 0px">P</math>  starts at the point  <math style="vertical-align: -15px">\bigg(0,\frac{1}{2}\bigg).</math>  As the wheel rolls,  <math style="vertical-align: 0px">P</math>  traces a curve. Find parametric equations for the curve.

<hr>
[[009C Sample Final 3, Problem 9 Solution|'''Solution''']]

[[009C Sample Final 3, Problem 9 Detailed Solution|'''Detailed Solution''']]

[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 3

2017-12-03T23:35:06Z

Kayla Murray: /*  Problem 9  */

'''This is a sample, and is meant to represent the material usually covered in Math 9C for the final. An actual test may or may not be similar.'''

'''Click on the  boxed problem numbers  to go to a solution.'''
<div class="noautonum">__TOC__</div>

== [[009C_Sample Final 3,_Problem_1| Problem 1 ]] ==
 Which of the following sequences  <math style="vertical-align: -5px">(a_n)_{n\ge 1}</math>  converges? Which diverges? Give reasons for your answers!

(a)  <math style="vertical-align: -15px">a_n=\bigg(1+\frac{1}{2n}\bigg)^n</math>

(b)  <math style="vertical-align: -15px">a_n=\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n</math>

== [[009C_Sample Final 3,_Problem_2| Problem 2 ]] ==
 Consider the series

::<math>\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}}.</math>

(a) Test if the series converges absolutely. Give reasons for your answer.

(b) Test if the series converges conditionally. Give reasons for your answer.

== [[009C_Sample Final 3,_Problem_3| Problem 3 ]] ==
Test if the following series converges or diverges. Give reasons and clearly state if you are using any standard test.

::<math>\sum_{n=1}^{\infty} \frac{n^3+7n}{\sqrt{1+n^{10}}}</math>

== [[009C_Sample Final 3,_Problem_4| Problem 4 ]] ==
 Determine if the following series converges or diverges. Please give your reason(s).

(a)  <math>\sum_{n=1}^{\infty} \frac{n!}{(2n)!}</math>

(b)  <math>\sum_{n=1}^{\infty} (-1)^n\frac{1}{n+1}</math>

== [[009C_Sample Final 3,_Problem_5| Problem 5 ]] ==
 Consider the function

::<math>f(x)=e^{-\frac{1}{3}x}.</math>

(a) Find a formula for the  <math>n</math>th derivative  <math style="vertical-align: -5px">f^{(n)}(x)</math>  of  <math style="vertical-align: -5px">f</math>  and then find  <math style="vertical-align: -5px">f'(3).</math>

(b) Find the Taylor series for  <math style="vertical-align: -5px">f(x)</math>  at  <math style="vertical-align: -5px">x_0=3,</math>  i.e. write  <math style="vertical-align: -5px">f(x)</math>  in the form

::<math>f(x)=\sum_{n=0}^\infty a_n(x-3)^n.</math>

== [[009C_Sample Final 3,_Problem_6| Problem 6 ]] ==
 Consider the power series

::<math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.</math>

(a) Find the radius of convergence of the above power series.

(b) Find the interval of convergence of the above power series.

(c) Find the closed formula for the function  <math style="vertical-align: -5px">f(x)</math>  to which the power series converges.

(d) Does the series

::<math>\sum_{n=0}^\infty \frac{1}{(n+1)3^{n+1}}</math>

converge?

== [[009C_Sample Final 3,_Problem_7| Problem 7 ]] ==

A curve is given in polar coordinates by

::<math>r=1+\cos^2(2\theta).</math>

(a) Show that the point with Cartesian coordinates  <math style="vertical-align: -15px">(x,y)=\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg)</math>  belongs to the curve.

(b) Sketch the curve.

(c) In Cartesian coordinates, find the equation of the tangent line at  <math style="vertical-align: -15px">\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).</math>

== [[009C_Sample Final 3,_Problem_8| Problem 8 ]] ==

A curve is given in polar coordinates by  <math style="vertical-align: -2px">r=4+3\sin \theta.</math>

(a) Sketch the curve.

(b) Find the area enclosed by the curve.

== [[009C_Sample Final 3,_Problem_9| Problem 9 ]] ==

A wheel of radius 1 rolls along a straight line, say the  <math>x</math>-axis. A point  <math style="vertical-align: 0px">P</math>  is located halfway between the center of the wheel and the rim; assume  <math style="vertical-align: 0px">P</math>  starts at the point  <math style="vertical-align: -15px">\bigg(0,\frac{1}{2}\bigg).</math>  As the wheel rolls,  <math style="vertical-align: 0px">P</math>  traces a curve. Find parametric equations for the curve.

== [[009C_Sample Final 3,_Problem_10| Problem 10 ]] ==

A curve is described parametrically by
::<math>x=t^2</math>
::<math>y=t^3-t</math>

(a) Sketch the curve for  <math style="vertical-align: -2px">-2\le t \le 2.</math>

(b) Find the equation of the tangent line to the curve at the origin.

009C Sample Final 3, Problem 8 Detailed Solution

2017-12-03T23:32:44Z

Kayla Murray:

A curve is given in polar coordinates by  <math style="vertical-align: -2px">r=4+3\sin \theta.</math>

(a) Sketch the curve.

(b) Find the area enclosed by the curve.
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|The area under a polar curve  <math style="vertical-align: -5px">r=f(\theta)</math>  is given by
|-
|
       <math>\int_{\alpha_1}^{\alpha_2} \frac{1}{2}r^2~d\theta</math>  for appropriate values of  <math>\alpha_1,\alpha_2.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(a)  
|-
| 
|-
|[[File:9CSF3_8.jpg |center|400px]]
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|The area  <math style="vertical-align: 0px">A</math>  enclosed by the curve is
|-
|
        <math>\begin{array}{rcl}
\displaystyle{A} & = & \displaystyle{\int_0^{2\pi} \frac{1}{2}r^2~d\theta}\\
&&\\
& = & \displaystyle{\int_0^{2\pi} \frac{1}{2} (4+3\sin\theta)^2~d\theta.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Using the double angle formula for  <math style="vertical-align: -5px">\cos(2\theta),</math>  we have
|-
|
       <math>\begin{array}{rcl}
\displaystyle{A} & = & \displaystyle{\frac{1}{2}\int_0^{2\pi} (16+24\sin\theta+9\sin^2\theta)~d\theta} \\
&&\\
& = & \displaystyle{\frac{1}{2}\int_0^{2\pi} \bigg(16+24\sin\theta+\frac{9}{2}(1-\cos(2\theta))\bigg)~d\theta}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg[16\theta-24\cos\theta+\frac{9}{2}\theta-\frac{9}{4}\sin(2\theta)\bigg]\bigg|_0^{2\pi}}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg[\frac{41}{2}\theta-24\cos\theta-\frac{9}{4}\sin(2\theta)\bigg]\bigg|_0^{2\pi}.}\\
\end{array}</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Lastly, we evaluate to get
|-
|
       <math>\begin{array}{rcl}
\displaystyle{A} & = &\displaystyle{\frac{1}{2}\bigg[\frac{41}{2}(2\pi)-24\cos(2\pi)-\frac{9}{4}\sin(4\pi)\bigg]-\frac{1}{2}\bigg[\frac{41}{2}(0)-24\cos(0)-\frac{9}{4}\sin(0)\bigg]}\\
&&\\
& = & \displaystyle{\frac{1}{2}(41\pi-24)-\frac{1}{2}(-24)}\\
&&\\
& = & \displaystyle{\frac{41\pi}{2}.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     See above
|-
|    '''(b)'''     <math>\frac{41\pi}{2}</math>
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 3, Problem 8

2017-12-03T23:30:12Z

Kayla Murray:

A curve is given in polar coordinates by  <math style="vertical-align: -2px">r=4+3\sin \theta.</math>

(a) Sketch the curve.

(b) Find the area enclosed by the curve.

<hr>
[[009C Sample Final 3, Problem 8 Solution|'''Solution''']]

[[009C Sample Final 3, Problem 8 Detailed Solution|'''Detailed Solution''']]

[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 3

2017-12-03T23:29:57Z

Kayla Murray: /*  Problem 8  */

'''This is a sample, and is meant to represent the material usually covered in Math 9C for the final. An actual test may or may not be similar.'''

'''Click on the  boxed problem numbers  to go to a solution.'''
<div class="noautonum">__TOC__</div>

== [[009C_Sample Final 3,_Problem_1| Problem 1 ]] ==
 Which of the following sequences  <math style="vertical-align: -5px">(a_n)_{n\ge 1}</math>  converges? Which diverges? Give reasons for your answers!

(a)  <math style="vertical-align: -15px">a_n=\bigg(1+\frac{1}{2n}\bigg)^n</math>

(b)  <math style="vertical-align: -15px">a_n=\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n</math>

== [[009C_Sample Final 3,_Problem_2| Problem 2 ]] ==
 Consider the series

::<math>\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}}.</math>

(a) Test if the series converges absolutely. Give reasons for your answer.

(b) Test if the series converges conditionally. Give reasons for your answer.

== [[009C_Sample Final 3,_Problem_3| Problem 3 ]] ==
Test if the following series converges or diverges. Give reasons and clearly state if you are using any standard test.

::<math>\sum_{n=1}^{\infty} \frac{n^3+7n}{\sqrt{1+n^{10}}}</math>

== [[009C_Sample Final 3,_Problem_4| Problem 4 ]] ==
 Determine if the following series converges or diverges. Please give your reason(s).

(a)  <math>\sum_{n=1}^{\infty} \frac{n!}{(2n)!}</math>

(b)  <math>\sum_{n=1}^{\infty} (-1)^n\frac{1}{n+1}</math>

== [[009C_Sample Final 3,_Problem_5| Problem 5 ]] ==
 Consider the function

::<math>f(x)=e^{-\frac{1}{3}x}.</math>

(a) Find a formula for the  <math>n</math>th derivative  <math style="vertical-align: -5px">f^{(n)}(x)</math>  of  <math style="vertical-align: -5px">f</math>  and then find  <math style="vertical-align: -5px">f'(3).</math>

(b) Find the Taylor series for  <math style="vertical-align: -5px">f(x)</math>  at  <math style="vertical-align: -5px">x_0=3,</math>  i.e. write  <math style="vertical-align: -5px">f(x)</math>  in the form

::<math>f(x)=\sum_{n=0}^\infty a_n(x-3)^n.</math>

== [[009C_Sample Final 3,_Problem_6| Problem 6 ]] ==
 Consider the power series

::<math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.</math>

(a) Find the radius of convergence of the above power series.

(b) Find the interval of convergence of the above power series.

(c) Find the closed formula for the function  <math style="vertical-align: -5px">f(x)</math>  to which the power series converges.

(d) Does the series

::<math>\sum_{n=0}^\infty \frac{1}{(n+1)3^{n+1}}</math>

converge?

== [[009C_Sample Final 3,_Problem_7| Problem 7 ]] ==

A curve is given in polar coordinates by

::<math>r=1+\cos^2(2\theta).</math>

(a) Show that the point with Cartesian coordinates  <math style="vertical-align: -15px">(x,y)=\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg)</math>  belongs to the curve.

(b) Sketch the curve.

(c) In Cartesian coordinates, find the equation of the tangent line at  <math style="vertical-align: -15px">\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).</math>

== [[009C_Sample Final 3,_Problem_8| Problem 8 ]] ==

A curve is given in polar coordinates by  <math style="vertical-align: -2px">r=4+3\sin \theta.</math>

(a) Sketch the curve.

(b) Find the area enclosed by the curve.

== [[009C_Sample Final 3,_Problem_9| Problem 9 ]] ==

A wheel of radius 1 rolls along a straight line, say the  <math>x</math>-axis. A point  <math style="vertical-align: 0px">P</math>  is located halfway between the center of the wheel and the rim. As the wheel rolls,  <math style="vertical-align: 0px">P</math>  traces a curve. Find parametric equations for the curve.

== [[009C_Sample Final 3,_Problem_10| Problem 10 ]] ==

A curve is described parametrically by
::<math>x=t^2</math>
::<math>y=t^3-t</math>

(a) Sketch the curve for  <math style="vertical-align: -2px">-2\le t \le 2.</math>

(b) Find the equation of the tangent line to the curve at the origin.

009C Sample Final 3, Problem 7 Detailed Solution

2017-12-03T23:29:07Z

Kayla Murray:

A curve is given in polar coordinates by

::<math>r=1+\cos^2(2\theta).</math>

(a) Show that the point with Cartesian coordinates  <math style="vertical-align: -15px">(x,y)=\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg)</math>  belongs to the curve.

(b) Sketch the curve.

(c) In Cartesian coordinates, find the equation of the tangent line at  <math style="vertical-align: -15px">\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' What two pieces of information do you need to write the equation of a line?
|-
|
       You need the slope of the line and a point on the line.
|-
|'''2.''' How do you calculate  <math style="vertical-align: -5px">y'</math>  for a polar curve  <math style="vertical-align: -5px">r=f(\theta)?</math>
|-
|
       Since  <math style="vertical-align: -5px">x=r\cos(\theta)</math>   and  <math style="vertical-align: -5px">y=r\sin(\theta),</math>  we have
|-
|
       <math>y'=\frac{dy}{dx}=\frac{(\frac{dr}{d\theta})\sin\theta+r\cos\theta}{(\frac{dr}{d\theta})\cos\theta-r\sin\theta}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we need to convert this Cartesian point into polar.
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{r} & = & \displaystyle{\sqrt{x^2+y^2}}\\
&&\\
& = & \displaystyle{\sqrt{\frac{2}{4}+\frac{2}{4}}}\\
&&\\
& = & \displaystyle{\sqrt{1}}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|-
|Also, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\tan \theta } & = & \displaystyle{\frac{y}{x}}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|-
|So,  <math style="vertical-align: -15px">\theta=\frac{\pi}{4}.</math>
|-
|Now, this point in polar is  <math style="vertical-align: -15px">\bigg(1,\frac{\pi}{4}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we plug in  <math style="vertical-align: -15px">\theta=\frac{\pi}{4}</math>  into our polar equation.
|-
|We get
|-
|        <math>\begin{array}{rcl}
\displaystyle{r} & = & \displaystyle{1+\cos^2\bigg(\frac{2\pi}{4}\bigg)}\\
&&\\
& = & \displaystyle{1+(0)^2}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|-
|So, the point  <math style="vertical-align: -5px">(x,y)</math>  belongs to the curve.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(b)  
|-
| 
|-
|[[File:9CSF3_7.jpg |center|441px]]
|}
'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Since  <math style="vertical-align: -5px">r=1+\cos^2(2\theta),</math>
|-
|
       <math>\frac{dr}{d\theta}=-4\cos(2\theta)\sin(2\theta).</math>
|-
|Since
|-
|
       <math>y'=\frac{dy}{dx}=\frac{(\frac{dr}{d\theta})\sin\theta+r\cos\theta}{(\frac{dr}{d\theta})\cos\theta-r\sin\theta},</math>
|-
|we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\frac{dy}{dx}} & = & \displaystyle{\frac{-4\cos(2\theta)\sin(2\theta)\sin\theta+(1+\cos^2(2\theta))\cos\theta}{-4\cos(2\theta)\sin(2\theta)\cos\theta-(1+\cos^2(2\theta))\sin\theta}.}\\
\end{array}</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, recall from part (a) that the given point in polar coordinates is  <math style="vertical-align: -15px">\bigg(1,\frac{\pi}{4}\bigg).</math>
|-
|Therefore, the slope of the tangent line at this point is
|-
|        <math>\begin{array}{rcl}
\displaystyle{m} & = & \displaystyle{\frac{-4\cos(\frac{\pi}{2})\sin(\frac{\pi}{2})\sin(\frac{\pi}{4})+(1+\cos^2(\frac{\pi}{2}))\cos(\frac{\pi}{4})}{-4\cos(\frac{\pi}{2})\sin(\frac{\pi}{2})\cos(\frac{\pi}{4})-(1+\cos^2(\frac{\pi}{2}))\sin(\frac{\pi}{4})}}\\
&&\\
& = & \displaystyle{\frac{0+(1)(\frac{\sqrt{2}}{2})}{0-(1)(\frac{\sqrt{2}}{2})}}\\
&&\\
& = & \displaystyle{-1.}
\end{array}</math>
|-
|Therefore, the equation of the tangent line at the point  <math style="vertical-align: -5px">(x,y)</math>  is
|-
|       <math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     See above.
|-
|    '''(b)'''     See above.
|-
|    '''(c)'''     <math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}</math>
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 3, Problem 7 Detailed Solution

2017-12-03T23:25:39Z

Kayla Murray:

A curve is given in polar coordinates by

::<math>r=1+\cos^2(2\theta).</math>

(a) Show that the point with Cartesian coordinates  <math style="vertical-align: -15px">(x,y)=\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg)</math>  belongs to the curve.

(b) Sketch the curve.

(c) In Cartesian coordinates, find the equation of the tangent line at  <math style="vertical-align: -15px">\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' What two pieces of information do you need to write the equation of a line?
|-
|
       You need the slope of the line and a point on the line.
|-
|'''2.''' How do you calculate   <math style="vertical-align: -5px">y'</math>   for a polar curve  <math style="vertical-align: -5px">r=f(\theta)?</math>
|-
|
       Since   <math style="vertical-align: -5px">x=r\cos(\theta),~y=r\sin(\theta),</math>  we have
|-
|
       <math>y'=\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we need to convert this Cartesian point into polar.
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{r} & = & \displaystyle{\sqrt{x^2+y^2}}\\
&&\\
& = & \displaystyle{\sqrt{\frac{2}{4}+\frac{2}{4}}}\\
&&\\
& = & \displaystyle{\sqrt{1}}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|-
|Also, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\tan \theta } & = & \displaystyle{\frac{y}{x}}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|-
|So,  <math style="vertical-align: -15px">\theta=\frac{\pi}{4}.</math>
|-
|Now, this point in polar is  <math style="vertical-align: -15px">\bigg(1,\frac{\pi}{4}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we plug in  <math style="vertical-align: -15px">\theta=\frac{\pi}{4}</math>  into our polar equation.
|-
|We get
|-
|        <math>\begin{array}{rcl}
\displaystyle{r} & = & \displaystyle{1+\cos^2\bigg(\frac{2\pi}{4}\bigg)}\\
&&\\
& = & \displaystyle{1+(0)^2}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|-
|So, the point  <math style="vertical-align: -5px">(x,y)</math>  belongs to the curve.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(b)  
|-
| 
|-
|[[File:9CSF3_7.jpg |center|441px]]
|}
'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Since  <math style="vertical-align: -5px">r=1+\cos^2(2\theta),</math>
|-
|
       <math>\frac{dr}{d\theta}=-4\cos(2\theta)\sin(2\theta).</math>
|-
|Since
|-
|
       <math>y'=\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta},</math>
|-
|we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\frac{dy}{dx}} & = & \displaystyle{\frac{-4\cos(2\theta)\sin(2\theta)\sin\theta+(1+\cos^2(2\theta))\cos\theta}{-4\cos(2\theta)\sin(2\theta)\cos\theta-(1+\cos^2(2\theta))\sin\theta}.}\\
\end{array}</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, recall from part (a) that the given point in polar coordinates is  <math style="vertical-align: -15px">\bigg(1,\frac{\pi}{4}\bigg).</math>
|-
|Therefore, the slope of the tangent line at this point is
|-
|        <math>\begin{array}{rcl}
\displaystyle{m} & = & \displaystyle{\frac{-4\cos(\frac{\pi}{2})\sin(\frac{\pi}{2})\sin(\frac{\pi}{4})+(1+\cos^2(\frac{\pi}{2}))\cos(\frac{\pi}{4})}{-4\cos(\frac{\pi}{2})\sin(\frac{\pi}{2})\cos(\frac{\pi}{4})-(1+\cos^2(\frac{\pi}{2}))\sin(\frac{\pi}{4})}}\\
&&\\
& = & \displaystyle{\frac{0+(1)(\frac{\sqrt{2}}{2})}{0-(1)(\frac{\sqrt{2}}{2})}}\\
&&\\
& = & \displaystyle{-1.}
\end{array}</math>
|-
|Therefore, the equation of the tangent line at the point  <math style="vertical-align: -5px">(x,y)</math>  is
|-
|       <math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     See above.
|-
|    '''(b)'''     See above.
|-
|    '''(c)'''     <math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}</math>
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 3, Problem 7

2017-12-03T23:25:25Z

Kayla Murray:

009C Sample Final 3

2017-12-03T23:25:10Z

Kayla Murray: /*  Problem 7  */

'''This is a sample, and is meant to represent the material usually covered in Math 9C for the final. An actual test may or may not be similar.'''

'''Click on the  boxed problem numbers  to go to a solution.'''
<div class="noautonum">__TOC__</div>

== [[009C_Sample Final 3,_Problem_1| Problem 1 ]] ==
 Which of the following sequences  <math style="vertical-align: -5px">(a_n)_{n\ge 1}</math>  converges? Which diverges? Give reasons for your answers!

(a)  <math style="vertical-align: -15px">a_n=\bigg(1+\frac{1}{2n}\bigg)^n</math>

(b)  <math style="vertical-align: -15px">a_n=\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n</math>

== [[009C_Sample Final 3,_Problem_2| Problem 2 ]] ==
 Consider the series

::<math>\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}}.</math>

(a) Test if the series converges absolutely. Give reasons for your answer.

(b) Test if the series converges conditionally. Give reasons for your answer.

== [[009C_Sample Final 3,_Problem_3| Problem 3 ]] ==
Test if the following series converges or diverges. Give reasons and clearly state if you are using any standard test.

::<math>\sum_{n=1}^{\infty} \frac{n^3+7n}{\sqrt{1+n^{10}}}</math>

== [[009C_Sample Final 3,_Problem_4| Problem 4 ]] ==
 Determine if the following series converges or diverges. Please give your reason(s).

(a)  <math>\sum_{n=1}^{\infty} \frac{n!}{(2n)!}</math>

(b)  <math>\sum_{n=1}^{\infty} (-1)^n\frac{1}{n+1}</math>

== [[009C_Sample Final 3,_Problem_5| Problem 5 ]] ==
 Consider the function

::<math>f(x)=e^{-\frac{1}{3}x}.</math>

(a) Find a formula for the  <math>n</math>th derivative  <math style="vertical-align: -5px">f^{(n)}(x)</math>  of  <math style="vertical-align: -5px">f</math>  and then find  <math style="vertical-align: -5px">f'(3).</math>

(b) Find the Taylor series for  <math style="vertical-align: -5px">f(x)</math>  at  <math style="vertical-align: -5px">x_0=3,</math>  i.e. write  <math style="vertical-align: -5px">f(x)</math>  in the form

::<math>f(x)=\sum_{n=0}^\infty a_n(x-3)^n.</math>

== [[009C_Sample Final 3,_Problem_6| Problem 6 ]] ==
 Consider the power series

::<math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.</math>

(a) Find the radius of convergence of the above power series.

(b) Find the interval of convergence of the above power series.

(c) Find the closed formula for the function  <math style="vertical-align: -5px">f(x)</math>  to which the power series converges.

(d) Does the series

::<math>\sum_{n=0}^\infty \frac{1}{(n+1)3^{n+1}}</math>

converge?

== [[009C_Sample Final 3,_Problem_7| Problem 7 ]] ==

A curve is given in polar coordinates by

::<math>r=1+\cos^2(2\theta).</math>

(a) Show that the point with Cartesian coordinates  <math style="vertical-align: -15px">(x,y)=\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg)</math>  belongs to the curve.

(b) Sketch the curve.

(c) In Cartesian coordinates, find the equation of the tangent line at  <math style="vertical-align: -15px">\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).</math>

== [[009C_Sample Final 3,_Problem_8| Problem 8 ]] ==

A curve is given in polar coordinates by  <math style="vertical-align: -2px">r=4+3\sin \theta</math>
::<math>0\leq \theta \leq 2\pi</math>

(a) Sketch the curve.

(b) Find the area enclosed by the curve.

== [[009C_Sample Final 3,_Problem_9| Problem 9 ]] ==

A wheel of radius 1 rolls along a straight line, say the  <math>x</math>-axis. A point  <math style="vertical-align: 0px">P</math>  is located halfway between the center of the wheel and the rim. As the wheel rolls,  <math style="vertical-align: 0px">P</math>  traces a curve. Find parametric equations for the curve.

== [[009C_Sample Final 3,_Problem_10| Problem 10 ]] ==

A curve is described parametrically by
::<math>x=t^2</math>
::<math>y=t^3-t</math>

(a) Sketch the curve for  <math style="vertical-align: -2px">-2\le t \le 2.</math>

(b) Find the equation of the tangent line to the curve at the origin.

009C Sample Final 3, Problem 6 Detailed Solution

2017-12-03T23:24:44Z

Kayla Murray:

Consider the power series

::<math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.</math>

(a) Find the radius of convergence of the above power series.

(b) Find the interval of convergence of the above power series.

(c) Find the closed formula for the function  <math style="vertical-align: -5px">f(x)</math>  to which the power series converges.

(d) Does the series

::<math>\sum_{n=0}^\infty \frac{1}{(n+1)3^{n+1}}</math>

converge?
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' '''Ratio Test'''
|-
|        Let  <math style="vertical-align: -7px">\sum a_n</math>  be a series and  <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>  Then,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|-
|'''2.''' '''Direct Comparison Test'''
|-
|        Let  <math>\{a_n\}</math>  and  <math>\{b_n\}</math>  be positive sequences where  <math style="vertical-align: -3px">a_n\le b_n</math>
|-
|        for all  <math style="vertical-align: -3px">n\ge N</math>  for some  <math style="vertical-align: -3px">N\ge 1.</math>
|-
|        '''1.''' If  <math>\sum_{n=1}^\infty b_n</math>  converges, then  <math>\sum_{n=1}^\infty a_n</math>  converges.
|-
|        '''2.''' If  <math>\sum_{n=1}^\infty a_n</math>  diverges, then  <math>\sum_{n=1}^\infty b_n</math>  diverges.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We use the Ratio Test to determine the radius of convergence.
|-
|We have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(-1)^{n+1}(x)^{n+2}}{(n+2)}\frac{n+1}{(-1)^n(x)^{n+1}}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x)\frac{n+1}{n+2}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} |x|\frac{n+1}{n+2}}\\
&&\\
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \frac{n+1}{n+2}}\\
&&\\
& = & \displaystyle{|x|.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|The Ratio Test tells us this series is absolutely convergent if  <math style="vertical-align: -5px">|x|<1.</math>
|-
|Hence, the Radius of Convergence of this series is  <math style="vertical-align: -1px">R=1.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, note that  <math style="vertical-align: -5px">|x|<1</math>  corresponds to the interval  <math style="vertical-align: -4px">(-1,1).</math>
|-
|To obtain the interval of convergence, we need to test the endpoints of this interval
|-
|for convergence since the Ratio Test is inconclusive when  <math style="vertical-align: -1px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|First, let  <math style="vertical-align: -1px">x=1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty (-1)^n \frac{1}{n+1}.</math>
|-
|This is an alternating series.
|-
|Let  <math style="vertical-align: -15px">b_n=\frac{1}{n+1}.</math>.
|-
|First, we have
|-
|       <math>\frac{1}{n+1}\ge 0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|The sequence  <math>\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{n+2}<\frac{1}{n+1}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{n+1}=0.</math>
|-
|Therefore, this series converges by the Alternating Series Test
|-
|and we include  <math style="vertical-align: -1px">x=1</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, let  <math style="vertical-align: -1px">x=-1.</math>
|-
|Then, the series becomes
|-
|        <math>\begin{array}{rcl}
\displaystyle{\sum_{n=0}^\infty \frac{(-1)^{2n+1}}{n+1}} & = & \displaystyle{\sum_{n=1}^\infty \frac{-1}{n+1}}\\
&&\\
& = & \displaystyle{(-1)\sum_{n=1}^\infty \frac{1}{n+1}.}
\end{array}</math>
|-
|Now, we note that
|-
|        <math>\frac{1}{n+1}>0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|This means that we can use the limit comparison test on this series.
|-
|Let  <math style="vertical-align: -16px">a_n=\frac{1}{n+1}.</math>
|-
|Let  <math style="vertical-align: -14px">b_n=\frac{1}{n}.</math>
|-
|Then,  <math style="vertical-align: -20px">\sum_{n=1}^\infty b_n</math>  diverges since it is the harmonic series.
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \frac{a_n}{b_n}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{(\frac{1}{n+1})}{(\frac{1}{n})}}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{n+1}}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} 1.}
\end{array}</math>
|-
|Therefore, the series
|-
|        <math>\sum_{n=1}^{\infty} \frac{1}{n+1}</math>
|-
|diverges by the Limit Comparison Test.
|-
|Therefore, we do not include  <math style="vertical-align: -1px">x=-1</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|The interval of convergence is  <math style="vertical-align: -4px">(-1,1].</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let
|-
|       <math>f(x)=\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.</math>
|-
|Then,
|-
|        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{d}{dx}\bigg(\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}\bigg)}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{d}{dx}\bigg( (-1)^n \frac{x^{n+1}}{n+1}\bigg)}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty (-1)^n x^n}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty (-x)^n}\\
&&\\
& = & \displaystyle{\frac{1}{1-(-x)}}\\
&&\\
& = & \displaystyle{\frac{1}{1+x}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Thus,
|-
|        <math>\begin{array}{rcl}
\displaystyle{f(x)} & = & \displaystyle{\int \frac{1}{1+x}~dx}\\
&&\\
& = & \displaystyle{\ln(1+x)+C.}
\end{array}</math>
|-
|Since there is no constant term in the series <math style="vertical-align: -20px">\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1},</math>
|-
|       <math>C=0.</math>
|-
|Hence,
|-
|        <math>f(x)=\ln(1+x).</math>
|}

'''(d)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we note that
|-
|        <math>\frac{1}{(n+1)3^{n+1}}>0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|This means that we can use a comparison test on this series.
|-
|Let  <math style="vertical-align: -19px">a_n=\frac{1}{(n+1)3^{n+1}}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Let  <math style="vertical-align: -14px">b_n=\frac{1}{3^{n+1}}.</math>
|-
|We want to compare the series in this problem with
|-
|        <math>\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty \frac{1}{3}\bigg(\frac{1}{3}\bigg)^n.</math>
|-
|This is a geometric series with  <math style="vertical-align: -13px">r=\frac{1}{3}.</math>
|-
|Since   <math>|r|<1,</math>  the series  <math style="vertical-align: -20px">\sum_{n=1}^\infty b_n</math>  converges.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Also, we have  <math style="vertical-align: -4px">a_n<b_n</math>  since
|-
|        <math>\frac{1}{(n+1)3^{n+1}}<\frac{1}{3^{n+1}}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|Therefore, the series  <math>\sum_{n=1}^\infty a_n</math>  converges
|-
|by the Direct Comparison Test.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     The radius of convergence is  <math style="vertical-align: -1px">R=1.</math>
|-
|    '''(b)'''     <math>(-1,1]</math>
|-
|    '''(c)'''     <math>f(x)=\ln(1+x)</math>
|-
|    '''(d)'''     converges (by the Direct Comparison Test)
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 3, Problem 6

2017-12-03T23:21:21Z

Kayla Murray:

009C Sample Final 3

2017-12-03T23:21:01Z

Kayla Murray: /*  Problem 6  */

'''This is a sample, and is meant to represent the material usually covered in Math 9C for the final. An actual test may or may not be similar.'''

'''Click on the  boxed problem numbers  to go to a solution.'''
<div class="noautonum">__TOC__</div>

== [[009C_Sample Final 3,_Problem_1| Problem 1 ]] ==
 Which of the following sequences  <math style="vertical-align: -5px">(a_n)_{n\ge 1}</math>  converges? Which diverges? Give reasons for your answers!

(a)  <math style="vertical-align: -15px">a_n=\bigg(1+\frac{1}{2n}\bigg)^n</math>

(b)  <math style="vertical-align: -15px">a_n=\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n</math>

== [[009C_Sample Final 3,_Problem_2| Problem 2 ]] ==
 Consider the series

::<math>\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}}.</math>

(a) Test if the series converges absolutely. Give reasons for your answer.

(b) Test if the series converges conditionally. Give reasons for your answer.

== [[009C_Sample Final 3,_Problem_3| Problem 3 ]] ==
Test if the following series converges or diverges. Give reasons and clearly state if you are using any standard test.

::<math>\sum_{n=1}^{\infty} \frac{n^3+7n}{\sqrt{1+n^{10}}}</math>

== [[009C_Sample Final 3,_Problem_4| Problem 4 ]] ==
 Determine if the following series converges or diverges. Please give your reason(s).

(a)  <math>\sum_{n=1}^{\infty} \frac{n!}{(2n)!}</math>

(b)  <math>\sum_{n=1}^{\infty} (-1)^n\frac{1}{n+1}</math>

== [[009C_Sample Final 3,_Problem_5| Problem 5 ]] ==
 Consider the function

::<math>f(x)=e^{-\frac{1}{3}x}.</math>

(a) Find a formula for the  <math>n</math>th derivative  <math style="vertical-align: -5px">f^{(n)}(x)</math>  of  <math style="vertical-align: -5px">f</math>  and then find  <math style="vertical-align: -5px">f'(3).</math>

(b) Find the Taylor series for  <math style="vertical-align: -5px">f(x)</math>  at  <math style="vertical-align: -5px">x_0=3,</math>  i.e. write  <math style="vertical-align: -5px">f(x)</math>  in the form

::<math>f(x)=\sum_{n=0}^\infty a_n(x-3)^n.</math>

== [[009C_Sample Final 3,_Problem_6| Problem 6 ]] ==
 Consider the power series

::<math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.</math>

(a) Find the radius of convergence of the above power series.

(b) Find the interval of convergence of the above power series.

(c) Find the closed formula for the function  <math style="vertical-align: -5px">f(x)</math>  to which the power series converges.

(d) Does the series

::<math>\sum_{n=0}^\infty \frac{1}{(n+1)3^{n+1}}</math>

converge?

== [[009C_Sample Final 3,_Problem_7| Problem 7 ]] ==

A curve is given in polar coordinates by

::<math>r=1+\cos^2(2\theta)</math>

(a) Show that the point with Cartesian coordinates  <math style="vertical-align: -15px">(x,y)=\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg)</math>  belongs to the curve.

(b) Sketch the curve.

(c) In Cartesian coordinates, find the equation of the tangent line at  <math style="vertical-align: -15px">\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).</math>

== [[009C_Sample Final 3,_Problem_8| Problem 8 ]] ==

A curve is given in polar coordinates by  <math style="vertical-align: -2px">r=4+3\sin \theta</math>
::<math>0\leq \theta \leq 2\pi</math>

(a) Sketch the curve.

(b) Find the area enclosed by the curve.

== [[009C_Sample Final 3,_Problem_9| Problem 9 ]] ==

A wheel of radius 1 rolls along a straight line, say the  <math>x</math>-axis. A point  <math style="vertical-align: 0px">P</math>  is located halfway between the center of the wheel and the rim. As the wheel rolls,  <math style="vertical-align: 0px">P</math>  traces a curve. Find parametric equations for the curve.

== [[009C_Sample Final 3,_Problem_10| Problem 10 ]] ==

A curve is described parametrically by
::<math>x=t^2</math>
::<math>y=t^3-t</math>

(a) Sketch the curve for  <math style="vertical-align: -2px">-2\le t \le 2.</math>

(b) Find the equation of the tangent line to the curve at the origin.

009C Sample Final 3

2017-12-03T23:20:45Z

Kayla Murray: /*  Problem 5  */

'''This is a sample, and is meant to represent the material usually covered in Math 9C for the final. An actual test may or may not be similar.'''

'''Click on the  boxed problem numbers  to go to a solution.'''
<div class="noautonum">__TOC__</div>

== [[009C_Sample Final 3,_Problem_1| Problem 1 ]] ==
 Which of the following sequences  <math style="vertical-align: -5px">(a_n)_{n\ge 1}</math>  converges? Which diverges? Give reasons for your answers!

(a)  <math style="vertical-align: -15px">a_n=\bigg(1+\frac{1}{2n}\bigg)^n</math>

(b)  <math style="vertical-align: -15px">a_n=\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n</math>

== [[009C_Sample Final 3,_Problem_2| Problem 2 ]] ==
 Consider the series

::<math>\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}}.</math>

(a) Test if the series converges absolutely. Give reasons for your answer.

(b) Test if the series converges conditionally. Give reasons for your answer.

== [[009C_Sample Final 3,_Problem_3| Problem 3 ]] ==
Test if the following series converges or diverges. Give reasons and clearly state if you are using any standard test.

::<math>\sum_{n=1}^{\infty} \frac{n^3+7n}{\sqrt{1+n^{10}}}</math>

== [[009C_Sample Final 3,_Problem_4| Problem 4 ]] ==
 Determine if the following series converges or diverges. Please give your reason(s).

(a)  <math>\sum_{n=1}^{\infty} \frac{n!}{(2n)!}</math>

(b)  <math>\sum_{n=1}^{\infty} (-1)^n\frac{1}{n+1}</math>

== [[009C_Sample Final 3,_Problem_5| Problem 5 ]] ==
 Consider the function

::<math>f(x)=e^{-\frac{1}{3}x}.</math>

(a) Find a formula for the  <math>n</math>th derivative  <math style="vertical-align: -5px">f^{(n)}(x)</math>  of  <math style="vertical-align: -5px">f</math>  and then find  <math style="vertical-align: -5px">f'(3).</math>

(b) Find the Taylor series for  <math style="vertical-align: -5px">f(x)</math>  at  <math style="vertical-align: -5px">x_0=3,</math>  i.e. write  <math style="vertical-align: -5px">f(x)</math>  in the form

::<math>f(x)=\sum_{n=0}^\infty a_n(x-3)^n.</math>

== [[009C_Sample Final 3,_Problem_6| Problem 6 ]] ==
 Consider the power series

::<math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}</math>

(a) Find the radius of convergence of the above power series.

(b) Find the interval of convergence of the above power series.

(c) Find the closed formula for the function  <math style="vertical-align: -5px">f(x)</math>  to which the power series converges.

(d) Does the series

::<math>\sum_{n=0}^\infty \frac{1}{(n+1)3^{n+1}}</math>

converge?

== [[009C_Sample Final 3,_Problem_7| Problem 7 ]] ==

A curve is given in polar coordinates by

::<math>r=1+\cos^2(2\theta)</math>

(a) Show that the point with Cartesian coordinates  <math style="vertical-align: -15px">(x,y)=\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg)</math>  belongs to the curve.

(b) Sketch the curve.

(c) In Cartesian coordinates, find the equation of the tangent line at  <math style="vertical-align: -15px">\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).</math>

== [[009C_Sample Final 3,_Problem_8| Problem 8 ]] ==

A curve is given in polar coordinates by  <math style="vertical-align: -2px">r=4+3\sin \theta</math>
::<math>0\leq \theta \leq 2\pi</math>

(a) Sketch the curve.

(b) Find the area enclosed by the curve.

== [[009C_Sample Final 3,_Problem_9| Problem 9 ]] ==

A wheel of radius 1 rolls along a straight line, say the  <math>x</math>-axis. A point  <math style="vertical-align: 0px">P</math>  is located halfway between the center of the wheel and the rim. As the wheel rolls,  <math style="vertical-align: 0px">P</math>  traces a curve. Find parametric equations for the curve.

== [[009C_Sample Final 3,_Problem_10| Problem 10 ]] ==

A curve is described parametrically by
::<math>x=t^2</math>
::<math>y=t^3-t</math>

(a) Sketch the curve for  <math style="vertical-align: -2px">-2\le t \le 2.</math>

(b) Find the equation of the tangent line to the curve at the origin.

009C Sample Final 3, Problem 5

2017-12-03T23:20:26Z

Kayla Murray:

009C Sample Final 3, Problem 5 Detailed Solution

2017-12-03T23:20:08Z

Kayla Murray:

Consider the function

::<math>f(x)=e^{-\frac{1}{3}x}.</math>

(a) Find a formula for the  <math>n</math>th derivative  <math style="vertical-align: -5px">f^{(n)}(x)</math>  of  <math style="vertical-align: -5px">f</math>  and then find  <math style="vertical-align: -5px">f'(3).</math>

(b) Find the Taylor series for  <math style="vertical-align: -5px">f(x)</math>  at  <math style="vertical-align: -5px">x_0=3,</math>  i.e. write  <math style="vertical-align: -5px">f(x)</math>  in the form

::<math>f(x)=\sum_{n=0}^\infty a_n(x-3)^n.</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|The Taylor polynomial of  <math style="vertical-align: -5px">f(x)</math>  at  <math style="vertical-align:0px">a</math>  is
|-
|
       <math>\sum_{n=0}^{\infty}c_n(x-a)^n</math>  where  <math style="vertical-align: -14px">c_n=\frac{f^{(n)}(a)}{n!}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We have
|-
|       <math>f'(x)=\bigg(-\frac{1}{3}\bigg)e^{-\frac{1}{3}x},</math>
|-
|       <math>f''(x)=\bigg(-\frac{1}{3}\bigg)^2 e^{-\frac{1}{3}x},</math>
|-
|and
|-
|       <math>f^{(3)}(x)=\bigg(-\frac{1}{3}\bigg)^3e^{-\frac{1}{3}x}.</math>
|-
|If we compare these three equations, we notice a pattern.
|-
|Thus,
|-
|       <math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^ne^{-\frac{1}{3}x}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since
|-
|       <math>f'(x)=\bigg(-\frac{1}{3}\bigg)e^{-\frac{1}{3}x},</math>
|-
|we have
|-
|       <math>f'(3)=\bigg(-\frac{1}{3}\bigg)e^{-1}.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Since
|-
|       <math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^3e^{-\frac{1}{3}x},</math>
|-
|we have
|-
|       <math>f^{(n)}(3)=\bigg(-\frac{1}{3}\bigg)^ne^{-1}.</math>
|-
|Therefore, the coefficients of the Taylor series are
|-
|       <math>c_n=\frac{\bigg(-\frac{1}{3}\bigg)^ne^{-1}}{n!}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Therefore, the Taylor series for  <math style="vertical-align: -5px">f(x)</math>  at  <math style="vertical-align: -3px">x_0=3</math>  is
|-
|       <math>\sum_{n=0}^\infty \bigg(-\frac{1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''    <math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^ne^{-\frac{1}{3}x},~f'(3)=\bigg(-\frac{1}{3}\bigg)e^{-1}</math>
|-
|    '''(b)'''    <math>\sum_{n=0}^\infty \bigg(-\frac{1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n</math>
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 3, Problem 4 Detailed Solution

2017-12-03T23:17:18Z

Kayla Murray:

Determine if the following series converges or diverges. Please give your reason(s).

(a)  <math>\sum_{n=1}^{\infty} \frac{n!}{(2n)!}</math>

(b)  <math>\sum_{n=1}^{\infty} (-1)^n\frac{1}{n+1}</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' '''Ratio Test'''
|-
|        Let  <math style="vertical-align: -7px">\sum a_n</math>  be a series and  <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>  Then,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|-
|'''2.''' If a series absolutely converges, then it also converges.
|-
|'''3.''' '''Alternating Series Test'''
|-
|        Let  <math>\{a_n\}</math>  be a positive, decreasing sequence where  <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} a_n=0.</math>
|-
|        Then,  <math>\sum_{n=1}^\infty (-1)^na_n</math>  and  <math>\sum_{n=1}^\infty (-1)^{n+1}a_n</math>  converge.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by using the Ratio Test.
|-
|We have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)!}{(2(n+1))!}\cdot \frac{(2n)!}{n!}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)n!}{(2n+2)(2n+1)(2n)!} \cdot \frac{(2n)!}{n!}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n+1}{(2n+2)(2n+1)}}\\
&&\\
& = & \displaystyle{0.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since
|-
|        <math>\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=0<1,</math>
|-
|the series is absolutely convergent by the Ratio Test.
|-
|Therefore, the series converges.
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|For
|-
|        <math>\sum_{n=1}^\infty (-1)^n\frac{1}{n+1},</math>
|-
|we notice that this series is alternating.
|-
|Let  <math style="vertical-align: -16px"> b_n=\frac{1}{n+1}.</math>
|-
|First, we have
|-
|       <math>\frac{1}{n+1}\ge 0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|The sequence  <math style="vertical-align: -5px">\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{n+2}<\frac{1}{n+1}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n+1}=0.</math>
|-
|Therefore,
|-
|       <math>\sum_{n=1}^\infty (-1)^n\frac{1}{n+1}</math>  
|-
|converges by the Alternating Series Test.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    converges (by the Ratio Test)
|-
|   '''(b)'''    converges (by the Alternating Series Test)
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 3, Problem 2 Detailed Solution

2017-12-03T23:14:34Z

Kayla Murray:

Consider the series

::<math>\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}}.</math>

(a) Test if the series converges absolutely. Give reasons for your answer.

(b) Test if the series converges conditionally. Give reasons for your answer.
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' A series  <math>\sum a_n</math>  is '''absolutely convergent''' if
|-
|        the series  <math>\sum |a_n|</math>  converges.
|-
|'''2.''' A series  <math>\sum a_n</math>  is '''conditionally convergent''' if
|-
|        the series  <math>\sum |a_n|</math>  diverges and the series  <math>\sum a_n</math>  converges.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we take the absolute value of the terms in the original series.
|-
|Let  <math style="vertical-align: -20px">a_n=\frac{(-1)^n}{\sqrt{n}}.</math>
|-
|Therefore,
|-
|        <math>\begin{array}{rcl}
\displaystyle{\sum_{n=1}^\infty |a_n|} & = & \displaystyle{\sum_{n=1}^\infty \bigg|\frac{(-1)^n}{\sqrt{n}}\bigg|}\\
&&\\
& = & \displaystyle{\sum_{n=1}^\infty \frac{1}{\sqrt{n}}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|This series is a  <math style="vertical-align: -5px">p</math>-series with  <math style="vertical-align: -14px">p=\frac{1}{2}.</math> 
|-
|Therefore, it diverges.
|-
|Hence, the series
|-
|        <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math>
|-
|is not absolutely convergent.
|-
|
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|For
|-
|        <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}},</math>
|-
|we notice that this series is alternating.
|-
|Let  <math style="vertical-align: -20px"> b_n=\frac{1}{\sqrt{n}}.</math>
|-
|First, we have
|-
|       <math>\frac{1}{\sqrt{n}}\ge 0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|The sequence  <math style="vertical-align: -4px">\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}=0.</math>
|-
|Therefore, the series  <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math>   converges
|-
|by the Alternating Series Test.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since the series  <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math>   is not absolutely convergent but convergent,
|-
|this series is conditionally convergent.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    not absolutely convergent (by the  <math style="vertical-align: -5px">p</math>-series test)
|-
|   '''(b)'''    conditionally convergent (by the Alternating Series Test)
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 3, Problem 1 Detailed Solution

2017-12-03T23:11:43Z

Kayla Murray:

Which of the following sequences  <math style="vertical-align: -5px">(a_n)_{n\ge 1}</math>  converges? Which diverges? Give reasons for your answers!

(a)  <math style="vertical-align: -15px">a_n=\bigg(1+\frac{1}{2n}\bigg)^n</math>

(b)  <math style="vertical-align: -15px">a_n=\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''L'Hôpital's Rule, Part 1'''
|-
|
        Let  <math style="vertical-align: -12px">\lim_{x\rightarrow c}f(x)=0</math>  and  <math style="vertical-align: -12px">\lim_{x\rightarrow c}g(x)=0,</math>  where  <math style="vertical-align: -5px">f</math>  and  <math style="vertical-align: -5px">g</math>  are differentiable functions
|-
|       on an open interval  <math style="vertical-align: 0px">I</math>  containing  <math style="vertical-align: -5px">c,</math>  and  <math style="vertical-align: -5px">g'(x)\ne 0</math>  on  <math style="vertical-align: 0px">I</math>  except possibly at  <math style="vertical-align: 0px">c.</math> 
|-
|       Then,   <math style="vertical-align: -18px">\lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let
        <math>\begin{array}{rcl}
\displaystyle{y} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(1+\frac{1}{2n}\bigg)^n.}
\end{array}</math>
|-
|We then take the natural log of both sides to get
|-
|        <math>\ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(1+\frac{1}{2n}\bigg)^n\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We can interchange limits and continuous functions.
|-
|Therefore, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\ln y} & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(1+\frac{1}{2n}\bigg)^n}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(1+\frac{1}{2n}\bigg)}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2n}\bigg)}{(\frac{1}{n})}.}
\end{array}</math>
|-
|Now, this limit has the form  <math style="vertical-align: -13px">\frac{0}{0}.</math>
|-
|Hence, we can use L'Hopital's Rule to calculate this limit.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2n}\bigg)}{(\frac{1}{n})}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2x}\bigg)}{(\frac{1}{x})}}\\
&&\\
& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{2x}{2x+1}\cdot\big(-\frac{1}{2x^2}\big)}{\big(-\frac{1}{x^2}\big)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{1}{2}\bigg(\frac{2x}{2x+1}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|Since  <math style="vertical-align: -5px">\ln y= 1/2,</math>  we know
|-
|        <math>y=e^{1/2}.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we have
|-
|       <math>\lim_{n\rightarrow \infty}\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n=\lim_{n\rightarrow \infty} (-1)^n\bigg(\frac{1+n}{n}\bigg)^n.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, let
|-
|        <math>\begin{array}{rcl}
\displaystyle{y} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{1+n}{n}\bigg)^n.}
\end{array}</math>
|-
|We then take the natural log of both sides to get
|-
|        <math>\ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(\frac{1+n}{n}\bigg)^n\bigg).</math>
|-
|We can interchange limits and continuous functions.
|-
|Therefore, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\ln y} & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(\frac{1+n}{n}\bigg)^n}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(\frac{1+n}{n}\bigg)}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{1+n}{n}\bigg)}{(\frac{1}{n})}.}
\end{array}</math>
|-
|Now, this limit has the form  <math style="vertical-align: -13px">\frac{0}{0}.</math>
|-
|Hence, we can use L'Hopital's Rule to calculate this limit.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{1+n}{n}\bigg)}{(\frac{1}{n})}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{1+x}{x}\bigg)}{(\frac{1}{x})}}\\
&&\\
& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{x}{1+x}\cdot\big(-\frac{1}{x^2}\big)}{\big(-\frac{1}{x^2}\big)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{x}{1+x}}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|Since  <math style="vertical-align: -4px">\ln y= 1,</math>  we know
|-
|        <math>y=e.</math>
|-
|Since
|-
|       <math>\lim_{n\rightarrow \infty} \bigg(\frac{1+n}{n}\bigg)^n\neq 0,</math>
|-
|we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} a_n} & = & \displaystyle{\lim_{n\rightarrow \infty} (-1)^n\bigg(\frac{1+n}{n}\bigg)^n}\\
&&\\
& = & \displaystyle{\text{DNE}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''    <math>e^{1/2}</math>
|-
|    '''(b)'''    <math>\text{DNE}</math>
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 10 Detailed Solution

2017-12-03T23:05:15Z

Kayla Murray:

Find the length of the curve given by
::<math>x=t^2</math>
::<math>y=t^3</math>
::<math>1\leq t \leq 2</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|The arc length  <math style="vertical-align: 0px">L</math>  of a parametric curve with  <math style="vertical-align: -4px">\alpha \leq t \leq \beta </math>  is given by
|-
|
       <math>L=\int_{\alpha}^{\beta} \sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2}~dt.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we need to calculate  <math style="vertical-align: -14px">\frac{dx}{dt}</math>  and  <math style="vertical-align: -14px">\frac{dy}{dt}.</math>
|-
|Since  <math style="vertical-align: -14px">x=t^2,~\frac{dx}{dt}=2t.</math>
|-
|Since  <math style="vertical-align: -14px">y=t^3,~\frac{dy}{dt}=3t^2.</math>
|-
|Using the arc length formula, we have
|-
|
       <math>L=\int_1^2 \sqrt{(2t)^2+(3t^2)^2}~dt.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|
       <math>\begin{array}{rcl}
\displaystyle{L} & = & \displaystyle{\int_1^2 \sqrt{4t^2+9t^4}~dt}\\
&&\\
& = & \displaystyle{\int_1^2 \sqrt{t^2(4+9t^2)}~dt}\\
&&\\
& = & \displaystyle{\int_1^2 t\sqrt{4+9t^2}~dt.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we use  <math>u</math>-substitution.
|-
|Let  <math style="vertical-align: -2px">u=4+9t^2.</math>
|-
|Then,  <math style="vertical-align: -2px">du=18tdt</math>  and  <math style="vertical-align: -15px">\frac{du}{18}=tdt.</math>
|-
|Also, since this is a definite integral, we need to change the bounds of integration.
|-
|We have
|-
|       <math style="vertical-align: -5px">u_1=4+9(1)^2=13</math>  and  <math style="vertical-align: -5px">u_2=4+9(2)^2=40.</math>
|-
|Hence,
|-
|       <math>\begin{array}{rcl}
\displaystyle{L} & = & \displaystyle{\frac{1}{18}\int_{13}^{40} \sqrt{u}~du}\\
&&\\
& = & \displaystyle{\frac{1}{18}\cdot\frac{2}{3} u^{\frac{3}{2}}\bigg|_{13}^{40}}\\
&&\\
& = & \displaystyle{\frac{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|        <math>\frac{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 10

2017-12-03T23:03:01Z

Kayla Murray:

Find the length of the curve given by
::<math>x=t^2</math>
::<math>y=t^3</math>
::<math>1\leq t \leq 2</math>

<hr>
[[009C Sample Final 2, Problem 10 Solution|'''Solution''']]

[[009C Sample Final 2, Problem 10 Detailed Solution|'''Detailed Solution''']]

[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2

2017-12-03T23:02:49Z

Kayla Murray: /*  Problem 10  */

'''This is a sample, and is meant to represent the material usually covered in Math 9C for the final. An actual test may or may not be similar.'''

'''Click on the  boxed problem numbers  to go to a solution.'''
<div class="noautonum">__TOC__</div>

== [[009C_Sample Final 2,_Problem_1| Problem 1 ]] ==
Test if the following sequences converge or diverge. Also find the limit of each convergent sequence.

(a)  <math style="vertical-align: -16px">a_n=\frac{\ln(n)}{\ln(n+1)}</math>

(b)  <math style="vertical-align: -15px">a_n=\bigg(\frac{n}{n+1}\bigg)^n</math>

== [[009C_Sample Final 2,_Problem_2| Problem 2 ]] ==
 For each of the following series, find the sum if it converges. If it diverges, explain why.

(a)  <math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math>

(b)  <math>\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}</math>

== [[009C_Sample Final 2,_Problem_3| Problem 3 ]] ==
Determine if the following series converges or diverges. Please give your reason(s).

(a)  <math>\sum_{n=1}^{\infty} \frac{n!}{(2n)!}</math>

(b)  <math>\sum_{n=1}^{\infty} (-1)^n \frac{1}{n+1}</math>

== [[009C_Sample Final 2,_Problem_4| Problem 4 ]] ==
(a) Find the radius of convergence for the power series

::<math>\sum_{n=1}^{\infty} (-1)^n \frac{x^n}{n}.</math>

(b) Find the interval of convergence of the above series.

== [[009C_Sample Final 2,_Problem_5| Problem 5 ]] ==
 Find the Taylor Polynomials of order 0, 1, 2, 3 generated by  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  at  <math style="vertical-align: -14px">x=\frac{\pi}{4}.</math>

== [[009C_Sample Final 2,_Problem_6| Problem 6 ]] ==
(a) Express the indefinite integral  <math style="vertical-align: -13px">\int \sin(x^2)~dx</math>  as a power series.

(b) Express the definite integral  <math style="vertical-align: -14px">\int_0^1 \sin(x^2)~dx</math>  as a number series.

== [[009C_Sample Final 2,_Problem_7| Problem 7 ]] ==

(a) Consider the function  <math style="vertical-align: -16px">f(x)=\bigg(1-\frac{1}{2}x\bigg)^{-2}.</math>  Find the first three terms of its Binomial Series.

(b) Find its radius of convergence.

== [[009C_Sample Final 2,_Problem_8| Problem 8 ]] ==

Find  <math>n</math>  such that the Maclaurin polynomial of degree  <math>n</math>  of  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  approximates  <math style="vertical-align: -13px">\cos \bigg(\frac{\pi}{3}\bigg)</math>  within 0.0001 of the actual value.

== [[009C_Sample Final 2,_Problem_9| Problem 9 ]] ==

A curve is given in polar coordinates by
::<math>r=\sin(2\theta).</math>

(a) Sketch the curve.

(b) Compute  <math style="vertical-align: -14px">y'=\frac{dy}{dx}.</math>

(c) Compute  <math style="vertical-align: -14px">y''=\frac{d^2y}{dx^2}.</math>

== [[009C_Sample Final 2,_Problem_10| Problem 10 ]] ==

Find the length of the curve given by
::<math>x=t^2</math>
::<math>y=t^3</math>
::<math>1\leq t \leq 2</math>

009C Sample Final 2, Problem 9 Detailed Solution

2017-12-03T23:01:33Z

Kayla Murray:

A curve is given in polar coordinates by
::<math>r=\sin(2\theta).</math>

(a) Sketch the curve.

(b) Compute  <math style="vertical-align: -14px">y'=\frac{dy}{dx}.</math>

(c) Compute  <math style="vertical-align: -14px">y''=\frac{d^2y}{dx^2}.</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|How do you calculate  <math style="vertical-align: -5px">y'</math>  for a polar curve  <math style="vertical-align: -5px">r=f(\theta)?</math>
|-
|
       Since  <math style="vertical-align: -5px">x=r\cos(\theta)</math>  and  <math style="vertical-align: -5px">y=r\sin(\theta),</math>  we have
|-
|
       <math>y'=\frac{dy}{dx}=\frac{(\frac{dr}{d\theta})\sin\theta+r\cos\theta}{(\frac{dr}{d\theta})\cos\theta-r\sin\theta}.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(a)  
|-
| 
|-
|[[File:009C_SF2_9_GP.png|center|400px]]
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Since  <math style="vertical-align: -5px">r=\sin(2\theta),</math>
|-
|
       <math>\frac{dr}{d\theta}=2\cos(2\theta).</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since
|-
|
       <math>y'=\frac{dy}{dx}=\frac{(\frac{dr}{d\theta})\sin\theta+r\cos\theta}{(\frac{dr}{d\theta})\cos\theta-r\sin\theta},</math>
|-
|we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{y'} & = & \displaystyle{\frac{2\cos(2\theta)\sin\theta+\sin(2\theta)\cos\theta}{2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta}}\\
&&\\
& = & \displaystyle{\frac{2\cos^2\theta \sin\theta-\sin^3\theta}{\cos^3\theta-2\sin^2\theta\cos\theta}}
\end{array}</math>
|-
|since
|-
|       <math>\sin(2\theta)=2\sin\theta\cos\theta,~\cos(2\theta)=\cos^2\theta-\sin^2\theta.</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We have   <math>\frac{d^2y}{dx^2}=\frac{(\frac{dy'}{d\theta})}{(\frac{dr}{d\theta})\cos\theta-r\sin\theta}.</math>
|-
|So, first we need to find   <math>\frac{dy'}{d\theta}.</math>
|-
|We have
|-
|
       <math>\begin{array}{rcl}
\displaystyle{\frac{dy'}{d\theta}} & = & \displaystyle{\frac{d}{d\theta}\bigg(\frac{2\cos^2\theta \sin\theta-\sin^3\theta}{\cos^3\theta-2\sin^2\theta\cos\theta}\bigg)}\\
&&\\
& = & \displaystyle{\frac{(\cos^3\theta-2\sin^2\theta\cos\theta)(-4\cos\theta\sin^2\theta+2\cos^3\theta-3\sin^2\theta\cos\theta)-(2\cos^2\theta\sin\theta-\sin^3\theta)(-3\cos^2\theta\sin\theta-4\sin \theta\cos^2\theta+2\sin^3\theta)}{(\cos^3\theta-2\sin^2\theta\cos\theta)^2}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the resulting formula for   <math>\frac{dy'}{d\theta},</math>  we get
|-
|
       <math>\frac{d^2y}{dx^2}=\frac{(\cos^3\theta-2\sin^2\theta\cos\theta)(-4\cos\theta\sin^2\theta+2\cos^3\theta-3\sin^2\theta\cos\theta)-(2\cos^2\theta\sin\theta-\sin^3\theta)(-3\cos^2\theta\sin\theta-4\sin \theta\cos^2\theta+2\sin^3\theta)}{(\cos^3\theta-2\sin^2\theta\cos\theta)^2(2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta)}.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''    See above
|-
|    '''(b)'''    <math style="vertical-align: -18px">y'=\frac{2\cos^2\theta \sin\theta-\sin^3\theta}{\cos^3\theta-2\sin^2\theta\cos\theta}</math>
|-
|    '''(c)'''    
|-
|<math>\frac{d^2y}{dx^2}=\frac{(\cos^3\theta-2\sin^2\theta\cos\theta)(-4\cos\theta\sin^2\theta+2\cos^3\theta-3\sin^2\theta\cos\theta)-(2\cos^2\theta\sin\theta-\sin^3\theta)(-3\cos^2\theta\sin\theta-4\sin \theta\cos^2\theta+2\sin^3\theta)}{(\cos^3\theta-2\sin^2\theta\cos\theta)^2(2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta)}</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 8

2017-12-03T22:57:19Z

Kayla Murray:

009C Sample Final 2

2017-12-03T22:56:52Z

Kayla Murray: /*  Problem 8  */

'''This is a sample, and is meant to represent the material usually covered in Math 9C for the final. An actual test may or may not be similar.'''

'''Click on the  boxed problem numbers  to go to a solution.'''
<div class="noautonum">__TOC__</div>

== [[009C_Sample Final 2,_Problem_1| Problem 1 ]] ==
Test if the following sequences converge or diverge. Also find the limit of each convergent sequence.

(a)  <math style="vertical-align: -16px">a_n=\frac{\ln(n)}{\ln(n+1)}</math>

(b)  <math style="vertical-align: -15px">a_n=\bigg(\frac{n}{n+1}\bigg)^n</math>

== [[009C_Sample Final 2,_Problem_2| Problem 2 ]] ==
 For each of the following series, find the sum if it converges. If it diverges, explain why.

(a)  <math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math>

(b)  <math>\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}</math>

== [[009C_Sample Final 2,_Problem_3| Problem 3 ]] ==
Determine if the following series converges or diverges. Please give your reason(s).

(a)  <math>\sum_{n=1}^{\infty} \frac{n!}{(2n)!}</math>

(b)  <math>\sum_{n=1}^{\infty} (-1)^n \frac{1}{n+1}</math>

== [[009C_Sample Final 2,_Problem_4| Problem 4 ]] ==
(a) Find the radius of convergence for the power series

::<math>\sum_{n=1}^{\infty} (-1)^n \frac{x^n}{n}.</math>

(b) Find the interval of convergence of the above series.

== [[009C_Sample Final 2,_Problem_5| Problem 5 ]] ==
 Find the Taylor Polynomials of order 0, 1, 2, 3 generated by  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  at  <math style="vertical-align: -14px">x=\frac{\pi}{4}.</math>

== [[009C_Sample Final 2,_Problem_6| Problem 6 ]] ==
(a) Express the indefinite integral  <math style="vertical-align: -13px">\int \sin(x^2)~dx</math>  as a power series.

(b) Express the definite integral  <math style="vertical-align: -14px">\int_0^1 \sin(x^2)~dx</math>  as a number series.

== [[009C_Sample Final 2,_Problem_7| Problem 7 ]] ==

(a) Consider the function  <math style="vertical-align: -16px">f(x)=\bigg(1-\frac{1}{2}x\bigg)^{-2}.</math>  Find the first three terms of its Binomial Series.

(b) Find its radius of convergence.

== [[009C_Sample Final 2,_Problem_8| Problem 8 ]] ==

Find  <math>n</math>  such that the Maclaurin polynomial of degree  <math>n</math>  of  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  approximates  <math style="vertical-align: -13px">\cos \bigg(\frac{\pi}{3}\bigg)</math>  within 0.0001 of the actual value.

== [[009C_Sample Final 2,_Problem_9| Problem 9 ]] ==

A curve is given in polar coordinates by
::<math>r=\sin(2\theta).</math>

(a) Sketch the curve.

(b) Compute  <math style="vertical-align: -14px">y'=\frac{dy}{dx}.</math>

(c) Compute  <math style="vertical-align: -14px">y''=\frac{d^2y}{dx^2}.</math>

== [[009C_Sample Final 2,_Problem_10| Problem 10 ]] ==

Find the length of the curve given by
::<math>x=t^2</math>
::<math>y=t^3</math>
::<math>0\leq t \leq 2</math>

009C Sample Final 2, Problem 8 Detailed Solution

2017-12-03T22:56:20Z

Kayla Murray:

Find  <math>n</math>  such that the Maclaurin polynomial of degree  <math>n</math>  of  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  approximates  <math style="vertical-align: -13px">\cos \bigg(\frac{\pi}{3}\bigg)</math>  within 0.0001 of the actual value.
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''Taylor's Theorem'''
|-
|        Let  <math style="vertical-align: -5px">f</math>  be a function whose  <math style="vertical-align: -4px">(n+1)^{\mathrm{th}}</math> derivative exists on an interval  <math style="vertical-align: -4px">I,</math>  and let  <math style="vertical-align: 0px">c</math>  be in  <math style="vertical-align: 0px">I.</math>
|-
|        Then, for each  <math style="vertical-align: 0px">x</math>  in  <math style="vertical-align: -4px">I,</math>  there exists  <math style="vertical-align: -3px">z_x</math>  between  <math style="vertical-align: 0px">x</math>  and  <math style="vertical-align: 0px">c</math>  such that
|-
|        <math>f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}(x-c)^2+\cdots+\frac{f^{(n)}(c)}{n!}(x-c)^n+R_n(x),</math>
|-
|        where  <math style="vertical-align: -18px">R_n(x)=\frac{f^{n+1}(z_x)}{(n+1)!}(x-c)^{n+1}.</math>
|-
|        Also,  <math style="vertical-align: -17px">|R_n(x)|\le \frac{\max |f^{n+1}(z)|}{(n+1)!}|(x-c)^{n+1}|.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using Taylor's Theorem, we have that the error in approximating  <math style="vertical-align: -13px">\cos \bigg(\frac{\pi}{3}\bigg)</math>  with
|-
|the Maclaurin polynomial of degree  <math style="vertical-align: 0px">n</math>  is  <math style="vertical-align: -16px">R_n\bigg(\frac{\pi}{3}\bigg)</math>  where
|-
|        <math>\bigg|R_n\bigg(\frac{\pi}{3}\bigg)\bigg|\le \frac{\max |f^{n+1}(z)|}{(n+1)!}\bigg|\bigg(\frac{\pi}{3}-0\bigg)^{n+1}\bigg|.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We note that
|-
|       <math style="vertical-align: -5px">|f^{n+1}(z)|=|\cos(z)|\le 1</math>  or  <math style="vertical-align: -5px">|f^{n+1}(z)|=|\sin(z)|\le 1.</math>
|-
|Therefore, we have
|-
|        <math>\bigg|R_n\bigg(\frac{\pi}{3}\bigg)\bigg|\le \frac{1}{(n+1)!}\bigg(\frac{\pi}{3}\bigg)^{n+1}.</math>
|-
|Now, we have the following table.
|-
|<table border="1" cellspacing="0" cellpadding="6" align = "center">
<tr>
<td align = "center"><math> n</math></td>
<td align = "center"><math> \approx\frac{1}{(n+1)!}\bigg(\frac{\pi}{3}\bigg)^{n+1}</math></td>
</tr>
<tr>
<td align = "center"><math>1</math></td>
<td align = "center"><math> 0.548311 </math></td>
</tr>
<tr>
<td align = "center"><math>2</math></td>
<td align = "center"><math> 0.191396</math></td>
</tr>
<tr>
<td align = "center"><math>3</math></td>
<td align = "center"><math> 0.050107 </math></td>
</tr>
<tr>
<td align = "center"><math>4</math></td>
<td align = "center"><math> 0.01049 </math></td>
</tr>
<tr>
<td align = "center"><math>5</math></td>
<td align = "center"><math> 0.00183 </math></td>
</tr>
<tr>
<td align = "center"><math>6</math></td>
<td align = "center"><math> 0.000274 </math></td>
</tr>
<tr>
<td align = "center"><math>7</math></td>
<td align = "center"><math> 0.0000358 </math></td>
</tr>
</table>
|-
|So,  <math style="vertical-align: 0px">n=7</math>  is the smallest value of  <math style="vertical-align: 0px">n</math>  where the error is less than or equal to 0.0001.
|-
|Therefore, for  <math style="vertical-align: 0px">n=7</math>  the Maclaurin polynomial approximates  <math style="vertical-align: -13px">\cos \bigg(\frac{\pi}{3}\bigg)</math>  within 0.0001 of the actual value.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>n=7</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 7 Detailed Solution

2017-12-03T22:52:27Z

Kayla Murray:

(a) Consider the function  <math style="vertical-align: -16px">f(x)=\bigg(1-\frac{1}{2}x\bigg)^{-2}.</math>  Find the first three terms of its Binomial Series.

(b) Find its radius of convergence.
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' The Taylor polynomial of  <math style="vertical-align: -5px">f(x)</math>  at  <math style="vertical-align: -1px">a</math>  is
|-
|
       <math>\sum_{n=0}^{\infty}c_n(x-a)^n</math>  where  <math style="vertical-align: -14px">c_n=\frac{f^{(n)}(a)}{n!}.</math>
|-
| '''2.''' '''Ratio Test'''
|-
|        Let  <math style="vertical-align: -7px">\sum a_n</math>  be a series and  <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>  Then,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by finding the coefficients of the Maclaurin series for  <math>f(x)=\frac{1}{(1-\frac{1}{2}x)^2}.</math>
|-
|We make a table to find the coefficients of the Maclaurin series.
|-
|
<table border="1" cellspacing="0" cellpadding="11" align = "center">
<tr>
<td align = "center"><math> n</math></td>
<td align = "center"><math> f^{(n)}(x) </math></td>
<td align = "center"><math> f^{(n)}(0) </math></td>
<td align = "center"><math> \frac{f^{(n)}(0)}{n!} </math></td>
</tr>
<tr>
<td align = "center"><math>0</math></td>
<td align = "center"><math> \frac{1}{(1-\frac{1}{2}x)^2} </math></td>
<td align = "center"><math> 1</math></td>
<td align = "center"><math> 1</math></td>
</tr>
<tr>
<td align = "center"><math>1</math></td>
<td align = "center"><math> \frac{1}{(1-\frac{1}{2}x)^3}</math></td>
<td align = "center"><math> 1 </math></td>
<td align = "center"><math> 1</math></td>
</tr>
<tr>
<td align = "center"><math>2</math></td>
<td align = "center"><math> \frac{\frac{3}{2}}{(1-\frac{1}{2}x)^4} </math></td>
<td align = "center"><math> \frac{3}{2} </math></td>
<td align = "center"><math> \frac{3}{4} </math></td>
</tr>
</table>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the first three terms of the Binomial Series are
|-
|        <math>1+x+\frac{3}{4}x^2.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By taking the derivative of the known series
|-
|       <math>\frac{1}{1-x}\,=\,1+x+x^2+\cdots,</math>
|-
|we find that the Maclaurin series of  <math>\frac{1}{(1-x)^2}</math>  is
|-
|       <math>\sum_{n=0}^\infty (n+1)x^n.</math>
|-
|Letting  <math style="vertical-align: -15px"> \frac{x}{2}</math>  play the role of  <math style="vertical-align: -4px">x,</math>  the Maclaurin series of  <math>\frac{1}{(1-\frac{1}{2}x)^2}</math>  is
|-
|       <math>\sum_{n=0}^\infty (n+1)\bigg(\frac{1}{2}x\bigg)^n=\sum_{n=0}^\infty \frac{(n+1)x^n}{2^n}.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we use the Ratio Test to determine the radius of convergence of this power series.
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+2)x^{n+1}}{2^{n+1}} \frac{2^n}{(n+1)x^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{|x|}{2} \frac{n+2}{n+1}}\\
&&\\
& = & \displaystyle{\frac{|x|}{2}\lim_{n\rightarrow \infty}\frac{n+2}{n+1}}\\
&&\\
& = & \displaystyle{\frac{|x|}{2}.}
\end{array}</math>
|-
|Now, the Ratio Test says this series converges if  <math style="vertical-align: -14px">\frac{|x|}{2}<1.</math> 
|-
|So,  <math style="vertical-align: -6px">|x|<2.</math>
|-
|Hence, the radius of convergence is  <math style="vertical-align: 0px">R=2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>1+x+\frac{3}{4}x^2</math>
|-
|   '''(b)'''    The radius of convergence is  <math style="vertical-align: 0px">R=2.</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 7 Detailed Solution

2017-12-03T22:50:02Z

Kayla Murray:

(a) Consider the function  <math style="vertical-align: -16px">f(x)=\bigg(1-\frac{1}{2}x\bigg)^{-2}.</math>  Find the first three terms of its Binomial Series.

(b) Find its radius of convergence.
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' The Taylor polynomial of  <math style="vertical-align: -5px">f(x)</math>  at  <math style="vertical-align: -1px">a</math>  is
|-
|
       <math>\sum_{n=0}^{\infty}c_n(x-a)^n</math>  where  <math style="vertical-align: -14px">c_n=\frac{f^{(n)}(a)}{n!}.</math>
|-
| '''2.''' '''Ratio Test'''
|-
|        Let  <math style="vertical-align: -7px">\sum a_n</math>  be a series and  <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>  Then,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by finding the coefficients of the Maclaurin series for  <math>f(x)=\frac{1}{(1-\frac{1}{2}x)^2}.</math>
|-
|We make a table to find the coefficients of the Maclaurin series.
|-
|
<table border="1" cellspacing="0" cellpadding="11" align = "center">
<tr>
<td align = "center"><math> n</math></td>
<td align = "center"><math> f^{(n)}(x) </math></td>
<td align = "center"><math> f^{(n)}(0) </math></td>
<td align = "center"><math> \frac{f^{(n)}(0)}{n!} </math></td>
</tr>
<tr>
<td align = "center"><math>0</math></td>
<td align = "center"><math> \frac{1}{(1-\frac{1}{2}x)^2} </math></td>
<td align = "center"><math> 1</math></td>
<td align = "center"><math> 1</math></td>
</tr>
<tr>
<td align = "center"><math>1</math></td>
<td align = "center"><math> \frac{1}{(1-\frac{1}{2}x)^3}</math></td>
<td align = "center"><math> 1 </math></td>
<td align = "center"><math> 1</math></td>
</tr>
<tr>
<td align = "center"><math>2</math></td>
<td align = "center"><math> \frac{\frac{3}{2}}{(1-\frac{1}{2}x)^4} </math></td>
<td align = "center"><math> \frac{3}{2} </math></td>
<td align = "center"><math> \frac{3}{4} </math></td>
</tr>
</table>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the first three terms of the Binomial Series are
|-
|        <math>1+x+\frac{3}{4}x^2.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By taking the derivative of the known series
|-
|   <math>\frac{1}{1-x}\,=\,1+x+x^2+\cdots,</math>
|-
|we find that the Maclaurin series of  <math>\frac{1}{(1-x)^2}</math>  is
|-
|       <math>\sum_{n=0}^\infty (n+1)x^n.</math>
|-
|Letting <math style="vertical-align: -5px"> x/2 </math>   play the role of <math style="vertical-align: -4px">x,</math> the Maclaurin series of  <math>\frac{1}{(1-\frac{1}{2}x)^2}</math>  is
|-
|       <math>\sum_{n=0}^\infty (n+1)\bigg(\frac{1}{2}x\bigg)^n=\sum_{n=0}^\infty \frac{(n+1)x^n}{2^n}.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we use the Ratio Test to determine the radius of convergence of this power series.
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+2)x^{n+1}}{2^{n+1}} \frac{2^n}{(n+1)x^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{|x|}{2} \frac{n+2}{n+1}}\\
&&\\
& = & \displaystyle{\frac{|x|}{2}\lim_{n\rightarrow \infty}\frac{n+2}{n+1}}\\
&&\\
& = & \displaystyle{\frac{|x|}{2}.}
\end{array}</math>
|-
|Now, the Ratio Test says this series converges if  <math style="vertical-align: -14px">\frac{|x|}{2}<1.</math> 
|-
|So,  <math style="vertical-align: -6px">|x|<2.</math>
|-
|Hence, the radius of convergence is  <math style="vertical-align: 0px">R=2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>1+x+\frac{3}{4}x^2</math>
|-
|   '''(b)'''    The radius of convergence is  <math style="vertical-align: 0px">R=2.</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 6 Detailed Solution

2017-12-03T22:46:43Z

Kayla Murray:

(a) Express the indefinite integral  <math style="vertical-align: -13px">\int \sin(x^2)~dx</math>  as a power series.

(b) Express the definite integral  <math style="vertical-align: -14px">\int_0^1 \sin(x^2)~dx</math>  as a number series.
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|What is the power series of  <math style="vertical-align: -1px">\sin x?</math>
|-
|       The power series of  <math style="vertical-align: -1px"> \sin x</math>  is   <math>\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|The power series of  <math style="vertical-align: -1px"> \sin x</math>  is   <math>\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}.</math>
|-
|So, the power series of  <math style="vertical-align: -5px">\sin(x^2)</math>   is
|-
| 
|-
|        <math>\begin{array}{rcl}
\displaystyle{\sin(x^2)} & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n(x^2)^{2n+1}}{(2n+1)!}}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!}.}
\end{array}</math>
|-
| 
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, to express the indefinite integral as a power series, we have
|-
| 
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int \sin(x^2)~dx} & = & \displaystyle{\int \sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!}~dx}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \int \frac{(-1)^nx^{4n+2}}{(2n+1)!}~dx}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}.}
\end{array}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|From part (a), we have
|-
|        <math>\int \sin(x^2)~dx=\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}.</math>
|-
|Now, we have
|-
|        <math>\int_0^1 \sin(x^2)~dx=\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}\bigg|_0^1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Hence, we have
|-
| 
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int_0^1 \sin(x^2)~dx} & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n (1)^{4n+3}}{(4n+3)(2n+1)!}-\sum_{n=0}^\infty \frac{(-1)^n (0)^{4n+3}}{(4n+3)(2n+1)!}}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}-0}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}.}
\end{array}</math>
|-
| 
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}</math>
|-
|    '''(b)'''     <math>\sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]