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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/gHby0M3.jpg)

Formula: $h^2+d^2=R^2$
Known: 
$h=5ft$,
$\frac{dR}{dt}=-2$ $ft/sec$

Want to know:
$\frac{dd}{dt}$ when $R=13 ft$ and $\frac{dR}{dt}=-2$ $ft/sec$
Solve for $d$ when $h=5$ and $R=13$:
$13^2=5^2+d^2$
$169=25+d^2$
$\sqrt144=d^2$
$d=12,-12$
$d=12$ (Since the boat is still in front of the dock, not through it.)


Get derivative of both sides of $R^2=h^2+d^2$:
$2R\frac{dR}{dt}=2h+2d\frac{dd}{dt}$

Substitute known values and solve for $\frac{dd}{dt}$:
$(2)(-2)(13)=(2)(5)+(2)(12)\frac{dd}{dt}$
$-52=10+24\frac{dd}{dt}$
$-62=24\frac{dd}{dt}$
$-\frac{31}{12}=\frac{dd}{dt}$

When the length of rope between the bow of the ship and the pulley is being pulled in at a rate of $2 ft/sec$ and reaches a length of $13ft$, the boat is approaching the dock by $\frac{31}{12}$ or $2.58333$ $ft/sec$.




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**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/yx84x0J.jpg)


Formula: $tan(\theta)=\frac{R}{90}$
Known: The Runner is running at a rate of $24 ft/sec$
So, $\frac{dR}{dt}= -24$
The length from third base to the ump is $90ft$
Want to know: $\frac{d\theta}{dt}$ when $\frac{dR}{dt}=24 ft/sec$ and R (the distance from the runner to third base) is $30 ft$
Derivative of $tan(\theta)=\frac{R}{90}$:
$sec^2(\theta)\frac{d\theta}{dt}$$=\frac{1}{90}\cdot\frac{dR}{dt}$
Solve for $sec^2(\theta)$:
$sec(\theta)=\frac{hyp}{adj}$
$sec(\theta)=\frac{\sqrt(30^2+90^2)}{90}$
$sec(\theta)=\frac{\sqrt(900+8100)}{90}$
$sec(\theta)=\frac{\sqrt(9000)}{90}$
$sec(\theta)=\frac{30\sqrt10}{90}$
$sec(\theta)=\frac{\sqrt10}{3}$
$sec^2(\theta)=(\frac{\sqrt10}{3})^2$
$sec^2(\theta)=\frac{10}{9}$

Substitute known into derivative and solve for $\frac{d\theta}{dt}$:
$\frac{10}{9}\cdot\frac{d\theta}{dt}$$=\frac{1}{90}\cdot-24$
$\frac{d\theta}{dt}$$=-\frac{24}{90}\cdot\frac{9}{10}$
$\frac{d\theta}{dt}$$=-\frac{24}{100}=-\frac{6}{25}$

The angle $\theta$ between the ump's line of sight and third base decreases by $\frac{6}{25}$ $radians/sec$ when the runner is moving at a speed of $24 ft/sec$ and the distance between third base and the runner is $30 feet$.  

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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.



