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?
:::

$Formula: Pythagorean Theorem$
$h^2+y^2=x^2$
$(5)^2+y^2=x^2$
$25+y^2=x^2$
$We$ $know$: $\frac{dy}{dt}=\frac{2ft}{\sec}$
$We$ $Want$:
$\frac{dx}{dt}\ when\ y=13\ and\ x=12\ because\ when\ we\ plug\ in\ :$
$13^{2}=x^{2}+25^{2}$
$169=x^{2}+25$
$\sqrt{144}=\sqrt{x^{2}}$
$12=x$
$Derivative$
$25+y^2=x^2$
$0+2y\cdot\frac{dy}{dt}=2x\cdot\frac{dx}{dt}$
$When$ $y=13$ and $x=12$ $substitute:$
$2\left(13\right)\left(2\right)=2\left(12\right)\cdot\frac{dx}{dt}$
$\frac{dx}{dt}=\frac{52}{24}=\frac{13}{6}\approx2.167$

The boat aproaches the dock at a rate of $\frac{13}{6}\approx2.167$ ft/sec when the length of rope from bow to the pulley is 13ft.

:::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?
:::

:::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?
:::
Formula Pythogorean Theorem
$h^{2}=x^{2}+y^{2}$
$We Know:$
$h^{2}=30^{2}+\left(y-x\right)^{2}$
$\frac{dx}{dt}=50\ \frac{mi}{hr}$
$\frac{dy}{dt}=70\ \frac{mi}{hr}$

$We Want:$
$\frac{dh}{dt\ }at\ 2:00\ when$
$x=\left(50\right)\left(2\right)=100$
$y=\left(70\right)\left(2\right)=140$
$h^{2}=\left(140-100\right)^{2}+30^{2}=1600+900=2500=\sqrt{h^{2}}=\sqrt{2500}=h=50$

Differentiate:
$h^{2}=30^{2}+\left(y-x\right)^{2}$
$2h\cdot\frac{dh}{dt}=0+2\left(y-x\right)\left(\frac{dy}{dt}-\frac{dx}{dt}\right)$

At 2:00 we get
$2\left(50\right)\cdot\frac{dh}{dt}=2\left(140-100\right)\left(70-50\right)$
$\frac{dh}{dt}=\frac{1600}{100}=16\ \frac{mi}{hr}$

At 2:00 the distance between the cars is changing at a rate of 16mi/hr.

---

To submit this assignment click on the Publish button . Then copy the url of the final document and submit it in Canvas.