009C Sample Final 1, Problem 9 - Revision history

Kayla Murray at 01:14, 3 December 2017

2017-12-03T01:14:56Z

Kayla Murray at 23:28, 26 February 2017

2017-02-26T23:28:19Z

Kayla Murray at 23:26, 26 February 2017

2017-02-26T23:26:09Z

Kayla Murray at 00:14, 26 February 2017

2017-02-26T00:14:03Z

Kayla Murray at 01:41, 19 April 2016

2016-04-19T01:41:21Z

Kayla Murray at 18:44, 29 February 2016

2016-02-29T18:44:44Z

Kayla Murray at 00:27, 25 February 2016

2016-02-25T00:27:03Z

Kayla Murray at 18:31, 23 February 2016

2016-02-23T18:31:25Z

Kayla Murray at 20:36, 22 February 2016

2016-02-22T20:36:27Z

Kayla Murray at 20:19, 10 February 2016

2016-02-10T20:19:43Z

@@ Line 5: / Line 5: @@
 <span class="exam">Find the length of the curve.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 1, Problem 9 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 1, Problem 9 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 27: / Line 27: @@
 !Step 1: &nbsp;
 |-
-|First, we need to calculate <math style="vertical-align: -14px">\frac{dr}{d\theta}</math>.
+|First, we need to calculate &nbsp;<math style="vertical-align: -14px">\frac{dr}{d\theta}</math>.
 |-
-|Since <math style="vertical-align: -14px">r=\theta,~\frac{dr}{d\theta}=1.</math>
+|Since &nbsp;<math style="vertical-align: -14px">r=\theta,~\frac{dr}{d\theta}=1.</math>
 |-
 |Using the formula in Foundations, we have
@@ Line 40: / Line 40: @@
 !Step 2: &nbsp;
 |-
-|Now, we proceed using trig substitution. Let <math style="vertical-align: -2px">\theta=\tan x.</math> Then, <math style="vertical-align: -1px">d\theta=\sec^2xdx.</math>
+|Now, we proceed using trig substitution. Let &nbsp;<math style="vertical-align: -2px">\theta=\tan x.</math> &nbsp; Then, &nbsp;<math style="vertical-align: -1px">d\theta=\sec^2xdx.</math>
 |-
 |So, the integral becomes
@@ Line 57: / Line 57: @@
 !Step 3: &nbsp;
 |-
-|Since <math style="vertical-align: -1px">\theta=\tan x,</math> we have <math style="vertical-align: -1px">x=\tan^{-1}\theta .</math>
+|Since &nbsp; <math style="vertical-align: -4px">\theta=\tan x,</math>&nbsp; we have &nbsp;<math style="vertical-align: -1px">x=\tan^{-1}\theta .</math>
 |-
 |So, we have

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' The formula for the arc length <math style="vertical-align: 0px">L</math> of a polar curve <math style="vertical-align: -5px">r=f(\theta)</math> with <math style="vertical-align: -4px">\alpha_1\leq \theta \leq \alpha_2</math> is
+|'''1.''' The formula for the arc length &nbsp;<math style="vertical-align: 0px">L</math>&nbsp; of a polar curve &nbsp;<math style="vertical-align: -5px">r=f(\theta)</math>&nbsp; with &nbsp;<math style="vertical-align: -4px">\alpha_1\leq \theta \leq \alpha_2</math>&nbsp; is
 |-
+|
@@ Line 16: / Line 16: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;You could use trig substitution and let <math style="vertical-align: -1px">x=\tan \theta .</math>
+&nbsp; &nbsp; &nbsp; &nbsp;You could use trig substitution and let &nbsp;<math style="vertical-align: -1px">x=\tan \theta .</math>
 |-
 |'''3.''' Recall that <math>\int \sec^3x~dx=\frac{1}{2}\sec x \tan x +\frac{1}{2}\ln|\sec x +\tan x|+C.</math>

@@ Line 1: / Line 1: @@
 <span class="exam">A curve is given in polar coordinates by
-::::::<span class="exam"><math>r=\theta</math>
+::<span class="exam"><math>r=\theta</math>
-::::::<span class="exam"><math>0\leq \theta \leq 2\pi</math>
+::<span class="exam"><math>0\leq \theta \leq 2\pi</math>
 <span class="exam">Find the length of the curve.
@@ Line 11: / Line 11: @@
 |-
+|
-::<math>L=\int_{\alpha_1}^{\alpha_2} \sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}d\theta.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_{\alpha_1}^{\alpha_2} \sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}d\theta.</math>
 |-
 |'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sqrt{1+x^2}~dx?</math>
 |-
+|
-::You could use trig substitution and let <math style="vertical-align: -1px">x=\tan \theta .</math>
+&nbsp; &nbsp; &nbsp; &nbsp;You could use trig substitution and let <math style="vertical-align: -1px">x=\tan \theta .</math>
 |-
 |'''3.''' Recall that <math>\int \sec^3x~dx=\frac{1}{2}\sec x \tan x +\frac{1}{2}\ln|\sec x +\tan x|+C.</math>
 |}
 '''Solution:'''
@@ Line 33: / Line 34: @@
 |-
+|
-::<math>L=\int_0^{2\pi}\sqrt{\theta^2+1}d\theta.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_0^{2\pi}\sqrt{\theta^2+1}d\theta.</math>
 |}
@@ Line 44: / Line 45: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{L} & = & \displaystyle{\int_{\theta=0}^{\theta=2\pi}\sqrt{\tan^2x+1}\sec^2xdx}\\
 &&\\
@@ Line 61: / Line 62: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{L} & = & \displaystyle{\frac{1}{2}\sec (\tan^{-1}(\theta)) \theta +\frac{1}{2}\ln|\sec (\tan^{-1}(\theta)) +\theta|\bigg|_{0}^{2\pi}}\\
 &&\\
@@ Line 67: / Line 68: @@
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; <math>\frac{1}{2}\sec(\tan^{-1}(2\pi))2\pi+\frac{1}{2}\ln|\sec(\tan^{-1}(2\pi))+2\pi|</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{2}\sec(\tan^{-1}(2\pi))2\pi+\frac{1}{2}\ln|\sec(\tan^{-1}(2\pi))+2\pi|</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 26: / Line 26: @@
 !Step 1: &nbsp;
 |-
-|First, we need to calculate <math style="vertical-align: -14px">\frac{dr}{d\theta}</math>. Since <math style="vertical-align: -14px">r=\theta,~\frac{dr}{d\theta}=1.</math>
+|First, we need to calculate <math style="vertical-align: -14px">\frac{dr}{d\theta}</math>.
 |-
 |Using the formula in Foundations, we have
@@ Line 69: / Line 71: @@
 !Final Answer: &nbsp;
 |-
-|<math>\frac{1}{2}\sec(\tan^{-1}(2\pi))2\pi+\frac{1}{2}\ln|\sec(\tan^{-1}(2\pi))+2\pi|</math>
+|&nbsp;&nbsp; <math>\frac{1}{2}\sec(\tan^{-1}(2\pi))2\pi+\frac{1}{2}\ln|\sec(\tan^{-1}(2\pi))+2\pi|</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 13: / Line 13: @@
 ::<math>L=\int_{\alpha_1}^{\alpha_2} \sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}d\theta</math>.
 |-
-|'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sqrt{1+x^2}~dx</math> ? You could use trig substitution and let <math style="vertical-align: -1px">x=\tan \theta </math>.
+|'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sqrt{1+x^2}~dx</math> ?
 |-
 |'''3.''' Recall that <math>\int \sec^3x~dx=\frac{1}{2}\sec x \tan x +\frac{1}{2}\ln|\sec x +\tan x|+C</math>.

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
-|The formula for the arc length <math>L</math> of a polar curve <math>r=f(\theta)</math> with <math>\alpha_1\leq \theta \leq \alpha_2</math> is
+|The formula for the arc length <math style="vertical-align: 0px">L</math> of a polar curve <math style="vertical-align: -5px">r=f(\theta)</math> with <math style="vertical-align: -4px">\alpha_1\leq \theta \leq \alpha_2</math> is
 |-
 |<math>L=\int_{\alpha_1}^{\alpha_2} \sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}d\theta</math>.
 |-
 |Review trig substitution.
@@ Line 20: / Line 21: @@
 !Step 1: &nbsp;
 |-
-|First, we need to calculate <math>\frac{dr}{d\theta}</math>. Since <math>r=\theta,~\frac{dr}{d\theta}=1</math>.
+|First, we need to calculate <math style="vertical-align: -14px">\frac{dr}{d\theta}</math>. Since <math style="vertical-align: -14px">r=\theta,~\frac{dr}{d\theta}=1</math>.
 |-
 |Using the formula in Foundations, we have
 |-
 |<math>L=\int_0^{2\pi}\sqrt{\theta^2+1}d\theta</math>.
 |}
@@ Line 30: / Line 32: @@
 !Step 2: &nbsp;
 |-
-|Now, we proceed using trig substitution. Let <math>\theta=\tan x</math>. Then, <math>d\theta=\sec^2xdx</math>.
+|Now, we proceed using trig substitution. Let <math style="vertical-align: -2px">\theta=\tan x</math>. Then, <math style="vertical-align: -1px">d\theta=\sec^2xdx</math>.
 |-
 |So, the integral becomes
@@ Line 47: / Line 49: @@
 !Step 3: &nbsp;
 |-
-|Since <math>\theta=\tan x</math>, we have <math>x=\tan^{-1}\theta</math>.
+|Since <math style="vertical-align: -1px">\theta=\tan x</math>, we have <math style="vertical-align: -1px">x=\tan^{-1}\theta</math>.
 |-
 |So, we have
@@ Line 64: / Line 66: @@
 !Final Answer: &nbsp;
 |-
-|<math>L=\frac{1}{2}\sec(\tan^{-1}(2\pi))2\pi+\frac{1}{2}\ln|\sec(\tan^{-1}(2\pi))+2\pi|</math>
+|<math>\frac{1}{2}\sec(\tan^{-1}(2\pi))2\pi+\frac{1}{2}\ln|\sec(\tan^{-1}(2\pi))+2\pi|</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]