Difference between revisions of "009B Sample Final 1, Problem 7"

Latest revision as of 17:21, 2 December 2017

(a) Find the length of the curve

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=\ln (\cos x),~~~0\leq x \leq \frac{\pi}{3}} .

(b) The curve

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=1-x^2,~~~0\leq x \leq 1}

is rotated about the 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} -axis. Find the area of the resulting surface.

Solution

Detailed Solution

Return to Sample Exam

@@ 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.
+<span class="exam">is rotated about the &nbsp;<math style="vertical-align: -3px">y</math>-axis. Find the area of the resulting surface.
-== Temp1 ==
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations: &nbsp;
-|-
-|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:'''
+<hr>
+[[009B Sample Final 1, Problem 7 Solution|'''<u>Solution</u>''']]
-== Temp2 ==
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009B Sample Final 1, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']]
-!Step 1: &nbsp;
-|-
-|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^{\frac{\pi}{3}} \sqrt{1+(-\tan x)^2}~dx</math>.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we have:
-|-
-|
-::<math>\begin{array}{rcl}
-L & = & \displaystyle{\int_0^{\frac{\pi}{3}} \sqrt{1+\tan^2 x}~dx}\\
-&&\\
-& = & \displaystyle{\int_0^{\frac{\pi}{3}} \sqrt{\sec^2x}~dx}\\
-&&\\
-& = & \displaystyle{\int_0^{\frac{\pi}{3}} \sec x ~dx}\\
-\end{array}</math>
-|-
-|
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 3: &nbsp;
-|-
-|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>
-|-
-|
-|}
-== Temp3 ==
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We start by calculating <math>\frac{dy}{dx}</math>.
-|-
-|Since <math style="vertical-align: -12px">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: &nbsp;
-|-
-|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: &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>.
-|-
-|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: &nbsp;
-|-
-|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>
-|}
-== Temp4 ==
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|'''(a)''' <math>\ln (2+\sqrt{3})</math>
-|-
-|'''(b)''' <math>\frac{\pi}{6}(5\sqrt{5}-1)</math>
-|}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009B Sample Final 1, Problem 7"

Latest revision as of 17:21, 2 December 2017

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