009B Sample Final 1, Problem 7 - Revision history

MathAdmin at 01:06, 21 May 2017

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MathAdmin at 16:46, 10 April 2017

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MathAdmin at 22:12, 18 April 2016

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MathAdmin: Created page with "a) Find the length of the curve :::::: $y=\ln (\cos x),~0\leq x \leq \frac{\pi}{3}$ . b) The curve :::::: $y=1-x^2,~0..."$

2016-02-27T05:42:02Z

Created page with "a) Find the length of the curve ::::::<math>y=\ln (\cos x),~~~0\leq x \leq \frac{\pi}{3}</math>. b) The curve ::::::<math>y=1-x^2,~~~0..."

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a) Find the length of the curve

::::::<math>y=\ln (\cos x),~~~0\leq x \leq \frac{\pi}{3}</math>.

b) The curve

::::::<math>y=1-x^2,~~~0\leq x \leq 1</math>

is rotated about the <math style="vertical-align: -3px">y</math>-axis. Find the area of the resulting surface.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' The formula for the length <math style="vertical-align: 0px">L</math> of a curve <math style="vertical-align: -5px">y=f(x)</math> where <math style="vertical-align: -3px">a\leq x \leq b</math> is
|-
|
::<math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>
|-
|'''2.''' <math style="vertical-align: -14px">\int \sec x~dx=\ln|\sec(x)+\tan(x)|+C.</math>
|-
|'''3.''' The surface area <math style="vertical-align: 0px">S</math> of a function <math style="vertical-align: -5px">y=f(x)</math> rotated about the <math style="vertical-align: -4px">y</math>-axis is given by
|-
|
::<math style="vertical-align: -13px">S=\int 2\pi x\,ds</math>, where <math style="vertical-align: -18px">ds=\sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we calculate  <math>\frac{dy}{dx}.</math>
|-
|Since <math style="vertical-align: -12px">y=\ln (\cos x),~\frac{dy}{dx}=\frac{1}{\cos x}(-\sin x)=-\tan x</math>.
|-
|Using the formula given in the Foundations section, we have
|-
|
::<math>L=\int_0^{\pi/3} \sqrt{1+(-\tan x)^2}~dx</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have:
|-
|
::<math>\begin{array}{rcl}
L & = & \displaystyle{\int_0^{\pi/3} \sqrt{1+\tan^2 x}~dx}\\
&&\\
& = & \displaystyle{\int_0^{\pi/3} \sqrt{\sec^2x}~dx}\\
&&\\
& = & \displaystyle{\int_0^{\pi/3} \sec x ~dx}.\\
\end{array}</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Finally,
|-
|
::<math>\begin{array}{rcl}
L& = & \ln |\sec x+\tan x|\bigg|_0^{\frac{\pi}{3}}\\
&&\\
& = & \displaystyle{\ln \bigg|\sec \frac{\pi}{3}+\tan \frac{\pi}{3}\bigg|-\ln|\sec 0 +\tan 0|}\\
&&\\
& = & \displaystyle{\ln |2+\sqrt{3}|-\ln|1|}\\
&&\\
& = & \displaystyle{\ln (2+\sqrt{3})}.
\end{array}</math>
|-
|
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by calculating  <math>\frac{dy}{dx}</math> .
|-
|Since <math style="vertical-align: -13px">y=1-x^2,~ \frac{dy}{dx}=-2x</math>.
|-
|Using the formula given in the Foundations section, we have
|-
|
::<math>S\,=\,\int_0^{1}2\pi x \sqrt{1+(-2x)^2}~dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have <math style="vertical-align: -14px">S=\int_0^{1}2\pi x \sqrt{1+4x^2}~dx.</math>
|-
|We proceed by using trig substitution. Let <math style="vertical-align: -13px">x=\frac{1}{2}\tan \theta</math>. Then, <math style="vertical-align: -12px">dx=\frac{1}{2}\sec^2\theta \,d\theta</math>.
|-
|So, we have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int 2\pi x \sqrt{1+4x^2}~dx} & = & \displaystyle{\int 2\pi \bigg(\frac{1}{2}\tan \theta\bigg)\sqrt{1+\tan^2\theta}\bigg(\frac{1}{2}\sec^2\theta\bigg) d\theta}\\
&&\\
& = & \displaystyle{\int \frac{\pi}{2} \tan \theta \sec \theta \sec^2\theta d\theta}.\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: 0px">u=\sec \theta</math>. Then, <math style="vertical-align: -1px">du=\sec \theta \tan \theta \,d\theta</math>.
|-
|So, the integral becomes
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int 2\pi x \sqrt{1+4x^2}~dx} & = & \displaystyle{\int \frac{\pi}{2}u^2du}\\
&&\\
& = & \displaystyle{\frac{\pi}{6}u^3+C}\\
&&\\
& = & \displaystyle{\frac{\pi}{6}\sec^3\theta+C}\\
&&\\
& = & \displaystyle{\frac{\pi}{6}(\sqrt{1+4x^2})^3+C}.\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|We started with a definite integral. So, using Step 2 and 3, we have
|-
|
::<math>\begin{array}{rcl}
S & = & \displaystyle{\int_0^1 2\pi x \sqrt{1+4x^2}~dx}\\
&&\\
& = & \displaystyle{\frac{\pi}{6}(\sqrt{1+4x^2})^3}\bigg|_0^1\\
&&\\
& = & \displaystyle{\frac{\pi(\sqrt{5})^3}{6}-\frac{\pi}{6}}\\
&&\\
& = & \displaystyle{\frac{\pi}{6}(5\sqrt{5}-1)}.\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math>\ln (2+\sqrt{3})</math>
|-
|'''(b)'''  <math>\frac{\pi}{6}(5\sqrt{5}-1)</math>
|}
[[009B_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 24: / Line 24: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">S=\int 2\pi x\,ds,</math>&nbsp; where <math style="vertical-align: -18px">ds=\sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">S=\int 2\pi x\,ds,</math>&nbsp; where <math style="vertical-align: -18px">ds=\sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>
 |}
@@ Line 105: / Line 105: @@
 |Now, we have &nbsp;<math style="vertical-align: -14px">S=\int_0^{1}2\pi x \sqrt{1+4x^2}~dx.</math>
 |-
-|We proceed by using trig substitution.
+|We proceed by &nbsp;<math>u</math>-substitution.
 |-
-|Let &nbsp;<math style="vertical-align: -13px">x=\frac{1}{2}\tan \theta.</math>&nbsp; Then, &nbsp;<math style="vertical-align:  -12px">dx=\frac{1}{2}\sec^2\theta \,d\theta.</math>
+|Let &nbsp;<math style="vertical-align: -2px">u=1+4x^2.</math> &nbsp;
 |-
-|So, we have
+|Then, &nbsp; <math style="vertical-align: 0px">du=8xdx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{8}=xdx.</math>
 |-
+|Since the integral is a definite integral, we need to change the bounds of integration.
 |-
-|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
+|Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">u=1+4x^2,</math>&nbsp; we get
 |-
-|Let &nbsp;<math style="vertical-align: 0px">u=\sec \theta.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=\sec \theta \tan \theta \,d\theta.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">u_1=1+4(0)^2=1</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=1+4(1)^2=5.</math>
 |-
-|So, the integral becomes
+|Thus, the integral becomes
 |-
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\int 2\pi x \sqrt{1+4x^2}~dx} & = & \displaystyle{\int \frac{\pi}{2}u^2du}\\
+S& = & \displaystyle{\int_1^5 \frac{2\pi}{8} \sqrt{u}~du}\\
 &&\\
-& = & \displaystyle{\frac{\pi}{6}u^3+C}\\
+& = & \displaystyle{\frac{\pi}{4} \int_1^5 u^{\frac{1}{2}}~du.}
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 4: &nbsp;
+!Step 3: &nbsp;
 |-
-|We started with a definite integral. So, using Step 2 and 3, we have
+|Now, we integrate to get
 |-
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-S & = & \displaystyle{\int_0^1 2\pi x \sqrt{1+4x^2}~dx}\\
+\displaystyle{S} & = & \displaystyle{\frac{\pi}{4}\bigg(\frac{2}{3}u^{\frac{3}{2}}\bigg)\bigg|_{1}^{5}}\\
 &&\\
-& = & \displaystyle{\frac{\pi}{6}(\sqrt{1+4x^2})^3}\bigg|_0^1\\
+& = & \displaystyle{\frac{\pi}{6}u^{\frac{3}{2}}\bigg|_{1}^{5}}\\
 &&\\
-& = & \displaystyle{\frac{\pi(\sqrt{5})^3}{6}-\frac{\pi}{6}}\\
+& = & \displaystyle{\frac{\pi}{6}(5)^{\frac{3}{2}}-\frac{\pi}{6}(1)^{\frac{3}{2}}}\\
 &&\\
 & = & \displaystyle{\frac{\pi}{6}(5\sqrt{5}-1)}.\\

@@ Line 7: / Line 7: @@
 ::::::<math>y=1-x^2,~~~0\leq x \leq 1</math>
-::<span class="exam">is rotated about the <math style="vertical-align: -3px">y</math>-axis. Find the area of the resulting surface.
+:::<span class="exam">is rotated about the <math style="vertical-align: -3px">y</math>-axis. Find the area of the resulting surface.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 1: / Line 1: @@
-<span class="exam">a) Find the length of the curve
+::<span class="exam">a) Find the length of the curve
 ::::::<math>y=\ln (\cos x),~~~0\leq x \leq \frac{\pi}{3}</math>.
-<span class="exam">b) The curve
+::<span class="exam">b) The curve
 ::::::<math>y=1-x^2,~~~0\leq x \leq 1</math>
-<span class="exam">is rotated about the <math style="vertical-align: -3px">y</math>-axis. Find the area of the resulting surface.
+::<span class="exam">is rotated about the <math style="vertical-align: -3px">y</math>-axis. Find the area of the resulting surface.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 155: / Line 155: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp;<math>\ln (2+\sqrt{3})</math>
+|&nbsp;&nbsp; '''(a)''' &nbsp;<math>\ln (2+\sqrt{3})</math>
 |-
-|'''(b)''' &nbsp;<math>\frac{\pi}{6}(5\sqrt{5}-1)</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp;<math>\frac{\pi}{6}(5\sqrt{5}-1)</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 ::<span class="exam">a) Find the length of the curve
-::::::<math>y=\ln (\cos x),~~~0\leq x \leq \frac{\pi}{3}</math>.
+::::::<math>y=\ln (\cos x),~~~0\leq x \leq \frac{\pi}{3}.</math>
 ::<span class="exam">b) The curve
@@ Line 14: / Line 14: @@
 |Recall:
 |-
 |'''1.''' The formula for the length <math style="vertical-align: 0px">L</math> of a curve <math style="vertical-align: -5px">y=f(x)</math> where <math style="vertical-align: -3px">a\leq x \leq b</math> is
 |-
+|
-::<math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>
+:::<math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>
 |-
 |'''2.''' <math style="vertical-align: -14px">\int \sec x~dx=\ln|\sec(x)+\tan(x)|+C.</math>
 |-
 |'''3.''' The surface area <math style="vertical-align: 0px">S</math> of a function <math style="vertical-align: -5px">y=f(x)</math> rotated about the <math style="vertical-align: -4px">y</math>-axis is given by
 |-
+|
-::<math style="vertical-align: -13px">S=\int 2\pi x\,ds</math>, where <math style="vertical-align: -18px">ds=\sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}.</math>
+:::<math style="vertical-align: -13px">S=\int 2\pi x\,ds</math>, where <math style="vertical-align: -18px">ds=\sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}.</math>
 |}
@@ Line 36: / Line 39: @@
 |First, we calculate&thinsp; <math>\frac{dy}{dx}.</math>
 |-
-|Since <math style="vertical-align: -12px">y=\ln (\cos x),~\frac{dy}{dx}=\frac{1}{\cos x}(-\sin x)=-\tan x</math>.
+|Since <math style="vertical-align: -5px">y=\ln (\cos x),</math>
 |-
 |Using the formula given in the Foundations section, we have
 |-
+|
-::<math>L=\int_0^{\pi/3} \sqrt{1+(-\tan x)^2}~dx</math>.
+::<math>L=\int_0^{\pi/3} \sqrt{1+(-\tan x)^2}~dx.</math>
 |}
@@ Line 76: / Line 82: @@
 & = & \displaystyle{\ln (2+\sqrt{3})}.
 \end{array}</math>
 |}
@@ Line 85: / Line 89: @@
 !Step 1: &nbsp;
 |-
-|We start by calculating&thinsp; <math>\frac{dy}{dx}</math>&thinsp;.
+|We start by calculating&thinsp; <math>\frac{dy}{dx}.</math>
 |-
-|Since <math style="vertical-align: -13px">y=1-x^2,~ \frac{dy}{dx}=-2x</math>.
+|Since <math style="vertical-align: -13px">y=1-x^2,~ \frac{dy}{dx}=-2x.</math>
 |-
 |Using the formula given in the Foundations section, we have
@@ Line 100: / Line 104: @@
 |Now, we have <math style="vertical-align: -14px">S=\int_0^{1}2\pi x \sqrt{1+4x^2}~dx.</math>
 |-
-|We proceed by using trig substitution. Let <math style="vertical-align: -13px">x=\frac{1}{2}\tan \theta</math>. Then, <math style="vertical-align:  -12px">dx=\frac{1}{2}\sec^2\theta \,d\theta</math>.
+|We proceed by using trig substitution. Let <math style="vertical-align: -13px">x=\frac{1}{2}\tan \theta.</math> Then, <math style="vertical-align:  -12px">dx=\frac{1}{2}\sec^2\theta \,d\theta.</math>
 |-
 |So, we have
@@ Line 115: / Line 119: @@
 !Step 3: &nbsp;
 |-
-|Now, we use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: 0px">u=\sec \theta</math>. Then, <math style="vertical-align: -1px">du=\sec \theta \tan \theta \,d\theta</math>.
+|Now, we use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: 0px">u=\sec \theta.</math> Then, <math style="vertical-align: -1px">du=\sec \theta \tan \theta \,d\theta.</math>
 |-
 |So, the integral becomes