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title: 'Math 181 Miniproject 9: Related Rates'

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Math 181 Miniproject 9: Related Rates.md
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tags: MATH 181
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Math 181 Miniproject 9: Related Rates
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**Overview:**   This miniproject focuses on a central application of calculus, namely *related rates*. These problems augment and extend the kinds of problems you have worked with in WeBWorK and class discussions.

**Prerequisites:** Section 3.5 of *Active Calculus.*

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For this miniproject, select EXACTLY TWO of the following and give complete and correct solutions that abide by the specifications for student work. Include a labeled picture with each solution.  Full calculus justification of your conclusions is required.

**Problem 1.** A sailboat is sitting at rest near its dock. A rope attached to the bow of the boat is drawn in over a pulley that stands on a post on the end of the dock that is 5 feet higher than the bow. If the rope is being pulled in at a rate of 2 feet per second, how fast is the boat approaching the dock when the length of rope from bow to pulley is 13 feet?
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![](https://i.imgur.com/Nl0Qdez.png)

We use the following equation $b^2+5^2=r^2$ this equation then relates the variables representing the horizontal distance of the boat and the length of the rope.
we know $r=13$ and $\frac{dr}{dt}=2$
we want to find $b$ and $\frac{db}{dt}$
Use the Pythagorean Theorem.
$b^2+5^2=13^2$
$b^2+25=169$
$b=12$
and now that we have enough information we can find $\frac{db}{dt}$
$2b\frac{db}{dt}=2r\ \frac{dr}{dt}$
$2(12)\frac{db}{dt}=2(13)(2)$
$24\frac{db}{dt}=52$
$\frac{db}{dt}=2.1667$  or $\frac{52}{24}$
So the boat is approaching the dock at an approxomation of $2.1667m/s$ 





:::info
**Problem 2.** A baseball diamond is a square with sides 90 feet long. Suppose a baseball player is advancing from second to third base at the rate of 24 feet per second, and an umpire is standing on home plate. Let $\theta$ be the angle between the third baseline and the line of sight from the umpire to the runner. How fast is $\theta$ changing when the runner is 30 feet from third base?
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![](https://i.imgur.com/7yXJjWN.png)
we know that $\frac{dx}{dt}=24ft/s$ and $y=90ft$
we want $\frac{d\theta}{dt}$ when $x=30ft$
$\tan\left(\theta\right)=\frac{x}{90}$
when $x=30$ we get
$\sec^{2}\theta\cdot\frac{d\theta}{dt}=\frac{1}{90}\cdot\frac{dx}{dt}$
$\sec\theta=\frac{\sqrt{90^{2}+30^{2}}}{90}$
$\sec^{2}\theta=\frac{90^{2}+30^{2}}{90^{2}}$
$=\frac{10}{9}$
$\frac{10}{9}\cdot\frac{d\theta}{dt}=\frac{1}{9}\left(-24\right)$
$\frac{d\theta}{dt}=-\frac{24}{100}$
$\frac{d\theta}{dt}=\frac{-6}{25}  r/s$
So when the runner is $30ft$ from the third base then $\theta$ is lessening at a rate of $6/25r/s$



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**Problem 3.**  Point $A$ is 30 miles west of point $B$. At noon a car starts driving South from point $A$ at a rate of 50 mi/h and a car starts driving South from point $B$ at a rate of 70 mi/h. At 2:00 how quickly is the distance between the cars changing?
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To submit this assignment click on the Publish button ![Publish button icon](https://i.imgur.com/Qk7vi9V.png). Then copy the url of the final document and submit it in Canvas.



