009B Sample Final 1, Problem 7 - Revision history

Kayla Murray at 00:21, 3 December 2017

2017-12-03T00:21:38Z

Kayla Murray at 00:58, 21 May 2017

2017-05-21T00:58:12Z

Kayla Murray at 21:04, 20 May 2017

2017-05-20T21:04:23Z

Kayla Murray at 19:50, 18 March 2017

2017-03-18T19:50:23Z

Kayla Murray at 21:27, 4 March 2017

2017-03-04T21:27:43Z

Kayla Murray at 17:40, 27 February 2017

2017-02-27T17:40:24Z

Kayla Murray at 02:13, 19 February 2017

2017-02-19T02:13:19Z

Grad at 05:37, 26 February 2016

2016-02-26T05:37:49Z

Grad: /* Temp2 */

2016-02-26T05:37:20Z

Temp2

Grad: /* Temp3 */

2016-02-26T05:34:20Z

Temp3

@@ Line 9: / Line 9: @@
 <span class="exam">is rotated about the &nbsp;<math style="vertical-align: -3px">y</math>-axis. Find the area of the resulting surface.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Final 1, Problem 7 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009B Sample Final 1, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ 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 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 17: / Line 17: @@
 &nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>
 |-
-|'''2.'''
+|'''2.''' Recall
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\int \sec x~dx=\ln|\sec(x)+\tan(x)|+C.</math>
@@ Line 37: / Line 37: @@
 |First, we calculate &nbsp;<math>\frac{dy}{dx}.</math>
 |-
-|Since &nbsp;<math style="vertical-align: -12px">y=\ln (\cos x),~\frac{dy}{dx}=\frac{1}{\cos x}(-\sin x)=-\tan x</math>.
+|Since &nbsp;<math style="vertical-align: -5px">y=\ln (\cos x),</math>
 |-
 |Using the formula given in the Foundations section, we have
@@ Line 88: / Line 90: @@
 |We start by calculating &nbsp;<math>\frac{dy}{dx}.</math>
 |-
-|Since &nbsp;<math style="vertical-align: -13px">y=1-x^2,~ \frac{dy}{dx}=-2x.</math>
+|Since &nbsp;<math style="vertical-align: -5px">y=1-x^2,</math>
 |-
 |Using the formula given in the Foundations section, we have

@@ Line 11: / Line 11: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
 |'''1.''' The formula for the length &nbsp;<math style="vertical-align: 0px">L</math>&nbsp; of a curve &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; where &nbsp;<math style="vertical-align: -3px">a\leq x \leq b</math>&nbsp; is

@@ 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 &nbsp;<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 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
+|'''1.''' The formula for the length &nbsp;<math style="vertical-align: 0px">L</math>&nbsp; of a curve &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; where &nbsp;<math style="vertical-align: -3px">a\leq x \leq b</math>&nbsp; is
 |-
+|
-::<math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<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>
+|'''2.'''
 |-
-|'''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
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\int \sec x~dx=\ln|\sec(x)+\tan(x)|+C.</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>
+&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>
 |}
@@ Line 35: / Line 37: @@
 !Step 1: &nbsp;
 |-
-|First, we calculate&thinsp; <math>\frac{dy}{dx}.</math>
+|First, we calculate &nbsp;<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 &nbsp;<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>.
+&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_0^{\pi/3} \sqrt{1+(-\tan x)^2}~dx.</math>
 |}
@@ Line 48: / Line 50: @@
 !Step 2: &nbsp;
 |-
-|Now, we have:
+|Now, we have
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 L & = & \displaystyle{\int_0^{\pi/3} \sqrt{1+\tan^2 x}~dx}\\
 &&\\
@@ Line 68: / Line 70: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 L& = & \ln |\sec x+\tan x|\bigg|_0^{\frac{\pi}{3}}\\
 &&\\
@@ Line 86: / Line 88: @@
 !Step 1: &nbsp;
 |-
-|We start by calculating&thinsp; <math>\frac{dy}{dx}</math>&thinsp;.
+|We start by calculating &nbsp;<math>\frac{dy}{dx}.</math>
 |-
-|Since <math style="vertical-align: -13px">y=1-x^2,~ \frac{dy}{dx}=-2x</math>.
+|Since &nbsp;<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>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>S\,=\,\int_0^{1}2\pi x \sqrt{1+(-2x)^2}~dx.</math>
 |}
@@ Line 99: / Line 101: @@
 !Step 2: &nbsp;
 |-
-|Now, we have <math style="vertical-align: -14px">S=\int_0^{1}2\pi x \sqrt{1+4x^2}~dx.</math>
+|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. 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.
 |-
 |So, we have
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<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}\\
 &&\\
@@ Line 116: / Line 120: @@
 !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 &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
 |So, the integral becomes
 |-
+|
-::<math>\begin{array}{rcl}
+&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}\\
 &&\\
@@ Line 138: / Line 144: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 S & = & \displaystyle{\int_0^1 2\pi x \sqrt{1+4x^2}~dx}\\
 &&\\
@@ Line 153: / Line 159: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp;<math>\ln (2+\sqrt{3})</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math>\ln (2+\sqrt{3})</math>
 |-
-|'''(b)''' &nbsp;<math>\frac{\pi}{6}(5\sqrt{5}-1)</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &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
+<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
+<span class="exam">(b) The curve
-::::::<math>y=1-x^2,~~~0\leq x \leq 1</math>
+::<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.
@@ Line 26: / Line 26: @@
 ::<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:'''
@@ Line 147: / Line 148: @@
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 35: / Line 35: @@
 !Step 1: &nbsp;
 |-
-|First, we calculate <math>\frac{dy}{dx}</math>.
+|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>.
@@ Line 42: / Line 42: @@
 |-
+|
-::<math>L=\int_0^{\frac{\pi}{3}} \sqrt{1+(-\tan x)^2}~dx</math>.
+::<math>L=\int_0^{\pi/3} \sqrt{1+(-\tan x)^2}~dx</math>.
 |}
@@ Line 52: / Line 52: @@
+|
 ::<math>\begin{array}{rcl}
-L & = & \displaystyle{\int_0^{\frac{\pi}{3}} \sqrt{1+\tan^2 x}~dx}\\
+L & = & \displaystyle{\int_0^{\pi/3} \sqrt{1+\tan^2 x}~dx}\\
 &&\\
-& = & \displaystyle{\int_0^{\frac{\pi}{3}} \sqrt{\sec^2x}~dx}\\
+& = & \displaystyle{\int_0^{\pi/3} \sqrt{\sec^2x}~dx}\\
 &&\\
-& = & \displaystyle{\int_0^{\frac{\pi}{3}} \sec x ~dx}\\
+& = & \displaystyle{\int_0^{\pi/3} \sec x ~dx}.\\
 \end{array}</math>
 |-
@@ Line 75: / Line 75: @@
 & = & \displaystyle{\ln |2+\sqrt{3}|-\ln|1|}\\
 &&\\
-& = & \displaystyle{\ln (2+\sqrt{3})}
+& = & \displaystyle{\ln (2+\sqrt{3})}.
 \end{array}</math>
 |-
+|
 |}
 == Temp3 ==
 '''(b)'''