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

From Grad Wiki
Jump to navigation Jump to search
Line 80: Line 80:
 
!Step 1:    
 
!Step 1:    
 
|-
 
|-
−
|
+
|We start by calculating <math>\frac{dy}{dx}</math>.
 +
|-
 +
|Since <math>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;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 +
|-
 +
|Now, we have <math>S=\int_0^{1}2\pi x \sqrt{1+4x^2}~dx</math>
 +
|-
 +
|We proceed by using trig substitution. Let <math>x=\frac{1}{2}\tan \theta</math>. Then, <math>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;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 3: &nbsp;
 
!Step 3: &nbsp;
 +
|-
 +
|Now, we use <math>u</math>-substitution. Let <math>u=\sec \theta</math>. Then, <math>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>
 
|}
 
|}
  
Line 106: Line 147:
 
|'''(a)''' <math>\ln (2+\sqrt{3})</math>
 
|'''(a)''' <math>\ln (2+\sqrt{3})</math>
 
|-
 
|-
−
|'''(b)'''  
+
|'''(b)''' <math>\frac{\pi}{6}(5\sqrt{5}-1)</math>
 
|}
 
|}
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 16:31, 4 February 2016

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.

Foundations:  
The formula for the length 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 L} of a 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=f(x)} 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 a\leq x \leq 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 L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx} .
integral of 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 \sec x}
The surface area 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 S} of a function 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=f(x)} 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 is given by
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 S=\int 2\pi x ds} 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 ds=\sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}} .

Solution:

(a)

Step 1:  
First, we calculate 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}} .
Since 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),~\frac{dy}{dx}=\frac{1}{\cos x}(-\sin x)=-\tan x} .
Using the formula given in the Foundations section, 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 L=\int_0^{\frac{\pi}{3}} \sqrt{1+(-\tan x)^2}~dx} .
Step 2:  
Now, 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} 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}}
Step 3:  
Finally,
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} 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}}

(b)

Step 1:  
We start by calculating 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}} .
Since 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,~ \frac{dy}{dx}=-2x} .
Using the formula given in the Foundations section, 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 S=\int_0^{1}2\pi x \sqrt{1+(-2x)^2}~dx} .
Step 2:  
Now, 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 S=\int_0^{1}2\pi x \sqrt{1+4x^2}~dx}
We proceed by using trig substitution. Let 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=\frac{1}{2}\tan \theta} . Then, 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 dx=\frac{1}{2}\sec^2\theta d\theta} .
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{\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}}
Step 3:  
Now, we use 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 u} -substitution. Let 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 u=\sec \theta} . Then, 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 du=\sec \theta \tan \theta d\theta} .
So, the integral becomes
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{\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}}
Step 4:  
We started with a definite integral. So, using Step 2 and 3, 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} 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}}
Final Answer:  
(a) 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 \ln (2+\sqrt{3})}
(b) 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{\pi}{6}(5\sqrt{5}-1)}

Return to Sample Exam

Retrieved from "https://gradwiki.math.ucr.edu/index.php?title=009B_Sample_Final_1,_Problem_7&oldid=1333"