Math 181 Miniproject 9: Related Rates.md
---
---
tags: MATH 181
---
Math 181 Miniproject 9: Related Rates
===
**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.*
---
:::info
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?
:::
$\frac{dx}{dt}=\frac{2ft}{sec}$
$x^2+y^2=5^2$
$x^2+13^2=5^2$
$x^2=5^2+13^2$
$x^2=25+169$
$\sqrt{x^2}=\sqrt{194}$
$x=\sqrt{194}$
$2*\frac{dx}{dt}=2y\frac{dy}{dt}$
$2\sqrt{194}\frac{dx}{dt}=2(13)\frac{dy}{dt}$
$\frac{2*\sqrt194*(2)}{2(13)}=\frac{dy}{dt}$
$\frac{dy}{dt}=2.1428\frac{ft}{sec}$
The boat is approaching 2.1428$\frac{ft}{sec}$
:::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?
:::
$tan\theta=\frac{x}{90}$
$sec^2\theta\frac{d\theta}{dt}=\frac{1}{90}\frac{dx}{dt}$
when $x=30$
$tan\theta=\frac{1}{3}=\frac{30}{90}$
$sec^2\theta=1+tan^2\theta=1+\frac{1}{9}=\frac{10}{9}$
$\frac{10}{9}\frac{d\theta}{dt}=\frac{1}{90}\frac{dx}{dt}$
$]\frac{dx}{dt}=-24\frac{ft}{sec}$
$\frac{d\theta}{dt}=\frac{-24}{100}=-.24\frac{rad}{sec}$
$\frac{d\theta}{dt}=(-.24)\frac{180}{pi}\frac{degree}{sec}$
$-\frac{43.2}{pi}\frac{degree}{sec}$
$-13.751\frac{degree}{sec}$
The angle is decreasing at a rate of $\frac{43.2}{pi}\frac{degree}{sec}$ or 13.751 $\frac{degree}{sec}$
:::info
**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?
:::
---
To submit this assignment click on the Publish button . Then copy the url of the final document and submit it in Canvas.