Difference between revisions of "009A Sample Final 1, Problem 4"

From Grad Wiki
Jump to navigation Jump to search
Line 6: Line 6:
 
!Foundations:    
 
!Foundations:    
 
|-
 
|-
−
|'''1.''' What two pieces of information do you need to write the equation of a line?
+
|'''1.''' '''Chain Rule'''
 
|-
 
|-
−
|
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
−
::You need the slope of the line and a point on the line.
+
|-
 +
|'''2.'''
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{d}{dx}(\cos^{-1}(x))=\frac{-1}{\sqrt{1-x^2}}</math>
 
|-
 
|-
−
|'''2.''' What does the Chain Rule state?
+
|'''3.''' The equation of the tangent line to &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at the point &nbsp;<math style="vertical-align: -5px">(a,b)</math>&nbsp; is
 
|-
 
|-
−
|
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">y=m(x-a)+b</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">m=f'(a).</math>
−
::For functions  <math style="vertical-align: -5px">f(x)</math>&thinsp; and <math style="vertical-align: -5px">g(x),</math>&nbsp; <math style="vertical-align: -12px">~\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x).</math>
 
 
|}
 
|}
  
Line 23: Line 25:
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|First, we compute&thinsp; <math style="vertical-align: -13px">\frac{dy}{dx}.</math> We get
+
|First, we compute &nbsp;<math style="vertical-align: -13px">\frac{dy}{dx}.</math>
 +
|-
 +
|Using the Chain Rule, we get
 
|-
 
|-
 
|
 
|
−
::<math>\frac{dy}{dx}\,=\,2x-\sin(\pi(x^2+1))(2\pi x).</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\frac{dy}{dx}} & = & \displaystyle{\frac{-1}{\sqrt{1-(2x)^2}}(2x)'}\\
 +
&&\\
 +
& = & \displaystyle{\frac{-2}{\sqrt{1-4x^2}}.}
 +
\end{array}</math>
 
|}
 
|}
  
Line 34: Line 42:
 
|To find the equation of the tangent line, we first find the slope of the line.  
 
|To find the equation of the tangent line, we first find the slope of the line.  
 
|-
 
|-
−
|Using <math style="vertical-align: -3px">x_0=1</math>&thinsp; in the formula for &thinsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&thinsp; from Step 1, we get
+
|Using &nbsp;<math style="vertical-align: -14px">x_0=\frac{\sqrt{3}}{4}</math>&nbsp; in the formula for &nbsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&nbsp; from Step 1, we get
 
|-
 
|-
 
|
 
|
−
::<math>m=2(1)-\sin(2\pi)2\pi\,=\,2.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{m} & = & \displaystyle{\frac{-2}{\sqrt{1-4(\frac{\sqrt{3}}{4})^2}}}\\
 +
&&\\
 +
& = & \displaystyle{\frac{-2}{\sqrt{\frac{1}{4}}}}\\
 +
&&\\
 +
& = & \displaystyle{-4.}
 +
\end{array}</math>
 +
|}
 +
 
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Step 3: &nbsp;
 
|-
 
|-
−
|To get a point on the line, we plug in <math style="vertical-align: -3px">x_0=1</math>&thinsp; into the equation given.  
+
|To get a point on the line, we plug in &nbsp;<math style="vertical-align: -14px">x_0=\frac{\sqrt{3}}{4}</math>&nbsp; into the equation given.  
 
|-
 
|-
−
|So, we have&thinsp; <math style="vertical-align: -5px">y=1^2+\cos(2\pi)=2.</math>
+
|So, we have
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{y_0} & = & \displaystyle{\cos^{-1}\bigg(2\frac{\sqrt{3}}{4}\bigg)}\\
 +
&&\\
 +
& = & \displaystyle{\cos^{-1}\bigg(\frac{\sqrt{3}}{2}\bigg)}\\
 +
&&\\
 +
& = & \displaystyle{\frac{\pi}{6}.}
 +
\end{array}</math>
 
|-
 
|-
−
|Thus, the equation of the tangent line is&thinsp; <math style="vertical-align: -5px">y=2(x-1)+2.</math>
+
|Thus, the equation of the tangent line is &nbsp; <math style="vertical-align: -14px">y=-4\bigg(x-\frac{\sqrt{3}}{4}\bigg)+\frac{\pi}{6}.</math>
 
|}
 
|}
  
Line 51: Line 78:
 
|-
 
|-
 
|
 
|
−
:<math>\frac{dy}{dx}=2x-\sin(\pi(x^2+1))(2\pi x)</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{dy}{dx}=\frac{-2}{\sqrt{1-4x^2}}</math>
 
|-
 
|-
 
|
 
|
−
:<math>y=2(x-1)+2</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>y=-4\bigg(x-\frac{\sqrt{3}}{4}\bigg)+\frac{\pi}{6}</math>
 
|}
 
|}
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 09:09, 27 February 2017

If  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=\cos^{-1} (2x)} compute  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}}   and find the equation for the tangent line at  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0=\frac{\sqrt{3}}{4}.}

You may leave your answers in point-slope form.

Foundations:  
1. Chain Rule
        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)}
2.
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx}(\cos^{-1}(x))=\frac{-1}{\sqrt{1-x^2}}}
3. The equation of the tangent line to  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   at the point  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a,b)}   is
        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=m(x-a)+b}   where  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m=f'(a).}


Solution:

Step 1:  
First, we compute  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}.}
Using the Chain Rule, we get

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{dy}{dx}} & = & \displaystyle{\frac{-1}{\sqrt{1-(2x)^2}}(2x)'}\\ &&\\ & = & \displaystyle{\frac{-2}{\sqrt{1-4x^2}}.} \end{array}}

Step 2:  
To find the equation of the tangent line, we first find the slope of the line.
Using  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0=\frac{\sqrt{3}}{4}}   in the formula for  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}}   from Step 1, we get

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{m} & = & \displaystyle{\frac{-2}{\sqrt{1-4(\frac{\sqrt{3}}{4})^2}}}\\ &&\\ & = & \displaystyle{\frac{-2}{\sqrt{\frac{1}{4}}}}\\ &&\\ & = & \displaystyle{-4.} \end{array}}

Step 3:  
To get a point on the line, we plug in  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0=\frac{\sqrt{3}}{4}}   into the equation given.
So, we have

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{y_0} & = & \displaystyle{\cos^{-1}\bigg(2\frac{\sqrt{3}}{4}\bigg)}\\ &&\\ & = & \displaystyle{\cos^{-1}\bigg(\frac{\sqrt{3}}{2}\bigg)}\\ &&\\ & = & \displaystyle{\frac{\pi}{6}.} \end{array}}

Thus, the equation of the tangent line is   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=-4\bigg(x-\frac{\sqrt{3}}{4}\bigg)+\frac{\pi}{6}.}


Final Answer:  

       

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=-4\bigg(x-\frac{\sqrt{3}}{4}\bigg)+\frac{\pi}{6}}

Return to Sample Exam

Retrieved from "https://gradwiki.math.ucr.edu/index.php?title=009A_Sample_Final_1,_Problem_4&oldid=2852"