Difference between revisions of "009B Sample Final 2, Problem 4"
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| <math>\rho(x)=25,000e^{-0.15x}</math> | | <math>\rho(x)=25,000e^{-0.15x}</math> | ||
|- | |- | ||
| − | |where <math>x</math> is the distance from the center. | + | |where <math style="vertical-align: 0px">x</math> is the distance from the center. |
|- | |- | ||
|Any slice along a radius gives us a cross section. | |Any slice along a radius gives us a cross section. | ||
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|We can treat this as a solid of revolution, and use the shell method. | |We can treat this as a solid of revolution, and use the shell method. | ||
|- | |- | ||
| − | |We are working on a half disk of radius 7, so we can integrate a cross-section where <math>x</math> goes from 0 to 7 | + | |We are working on a half disk of radius 7, so we can integrate a cross-section where <math style="vertical-align: 0px">x</math> goes from 0 to 7 |
|- | |- | ||
| − | |and the height at each <math>x</math> is our density function, <math>\rho(x).</math> | + | |and the height at each <math style="vertical-align: 0px">x</math> is our density function, <math style="vertical-align: -5px">\rho(x).</math> |
|- | |- | ||
| − | |Normally <math>2\pi R</math> represents once around a circle of radius <math>R,</math> | + | |Normally <math style="vertical-align: 0px">2\pi R</math> represents once around a circle of radius <math style="vertical-align: -5px">R,</math> |
|- | |- | ||
|but in this case we only go half way around. | |but in this case we only go half way around. | ||
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|To solve this, we need to use integration by parts. | |To solve this, we need to use integration by parts. | ||
|- | |- | ||
| − | |Let <math>u=x</math> and <math>dv=e^{-0.15x}dx.</math> | + | |Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^{-0.15x}dx.</math> |
|- | |- | ||
| − | |Then, <math>du=dx</math> and <math>v=-{\displaystyle \frac{e^{-0.15x}}{0.15}.}</math> | + | |Then, <math style="vertical-align: 0px">du=dx</math> and <math style="vertical-align: -14px">v=-{\displaystyle \frac{e^{-0.15x}}{0.15}.}</math> |
|- | |- | ||
|Thus, | |Thus, | ||
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!Final Answer: | !Final Answer: | ||
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| − | | <math>{\displaystyle 25,000\pi\left[\left(-\frac{7e^{-1.05}}{0.15}-\frac{e^{-1.05}}{(0.15)^{2}}\right)+\frac{1}{(0.15)^{2}}\right]\ \approx\ 986,556 | + | | <math>{\displaystyle 25,000\pi\left[\left(-\frac{7e^{-1.05}}{0.15}-\frac{e^{-1.05}}{(0.15)^{2}}\right)+\frac{1}{(0.15)^{2}}\right]\ \approx\ 986,556}</math> |
|} | |} | ||
[[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | ||
Latest revision as of 15:05, 26 May 2017
A city bordered on one side by a lake can be approximated by a semicircle of radius 7 miles, whose city center is on the shoreline. As we move away from the center along a radius the population density of the city can be approximated 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 \rho(x)=25000e^{-0.15x}}
people per square mile. What is the population of the city?
| Foundations: |
|---|
| Many word problems can be confusing, and this is a good example. |
| We know that we are going to integrate over a half-disk of radius 7, but how do we construct the integral? |
| One key could be the expression of our density, |
| 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 \rho(x)=25,000e^{-0.15x}} |
| where is the distance from the center. |
| Any slice along a radius gives us a cross section. |
| If we were revolving ALL the way around the center, this would be typical solid of revolution, |
| and we could find the volume of revolving the center by the usual shell formula |
| 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 V\ =\ \int_{x_{1}}^{x_{2}}2\pi R\cdot h\,dx.} |
| What changes, since we are only doing half of a disk? |
| Also, this particular problem will require integration by parts: |
| 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 \int u\,dv=uv-\int v\,du.} |
Solution:
| Step 1: |
|---|
| We can treat this as a solid of revolution, and use the shell method. |
| We are working on a half disk of radius 7, so we can integrate a cross-section 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 x} goes from 0 to 7 |
| and the height at each 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} is our density 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 \rho(x).} |
| Normally 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 2\pi R} represents once around a circle of radius 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 R,} |
| but in this case we only go half way around. |
| Therefore, we adjust our usual shell method formula to find the population as |
|
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{P} & = & \displaystyle{\int_{x_{1}}^{x_{2}}\pi R\cdot h\,dx}\\ &&\\ & = & \displaystyle{\int_{0}^{7}\pi x\cdot\rho(x)\,dx.} \end{array}} |
| Step 2: |
|---|
| Let's plug in the actual formula for density and solve. 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{P} & = & \displaystyle{\int_{0}^{7}\pi x\cdot25,000e^{-0.15x}\,dx}\\ &&\\ & = & \displaystyle{25,000\pi\int_{0}^{7}xe^{-0.15x}\,dx.} \end{array}} |
| To solve this, we need to use integration by parts. |
| 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=x} and 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 dv=e^{-0.15x}dx.} |
| 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=dx} and 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 v=-{\displaystyle \frac{e^{-0.15x}}{0.15}.}} |
| Thus,
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{P} & = & \displaystyle{25,000\pi\int_{0}^{7}xe^{-0.15x}\,dx}\\ &&\\ & = & \displaystyle{25,000\pi\left[\left.-\frac{xe^{-0.15x}}{0.15}\right|_{0}^{7}+\int_{0}^{7}\frac{e^{-0.15x}}{0.15}\,dx\right]}\\ &&\\ & = & \displaystyle{25,000\pi\left[-\frac{xe^{-0.15x}}{0.15}-\frac{e^{-0.15x}}{(0.15)^{2}}\right]_{0}^{7}}\\ &&\\ & = & \displaystyle{25,000\pi\left[\left(-\frac{7e^{-1.05}}{0.15}-\frac{e^{-1.05}}{(0.15)^{2}}\right)+\frac{1}{(0.15)^{2}}\right]}\\ &&\\ & \approx & 986,556. \end{array}} |
| Note that in a calculator-prohibited test, no one would expect the actual numerical answer. |
| However, you would likely need the line above it to receive full credit. |
| Final Answer: |
|---|
| 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 {\displaystyle 25,000\pi\left[\left(-\frac{7e^{-1.05}}{0.15}-\frac{e^{-1.05}}{(0.15)^{2}}\right)+\frac{1}{(0.15)^{2}}\right]\ \approx\ 986,556}} |