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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**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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For this problem, we know that the player dx/dt = 24 ft/s

We want to know $\frac{d\theta}{dt}$

We can use the formula $\tan\theta=\ \frac{x}{y}$--- Next we can take the derivative of $\tan\theta$

$\sec^{2}\theta\ \left(\frac{d\theta}{dt}\right)=\ \frac{\left(\left(\frac{dx}{dt}\right)y-x\left(\frac{dy}{dt}\right)\right)}{y^{2}}$ ----- From this point, We can plug in the given information:

$\sec^{2}\theta\ \left(\frac{d\theta}{dt}\right)=\ \frac{\left(\left(-24\right)\left(90\right)-\left(30\right)\left(0\right)\right)}{90^{2}}$

$\sec^{2}\theta\ \left(\frac{d\theta}{dt}\right)=-\ \frac{\left(2160\left(\frac{ft}{s}\right)\right)}{8100}$ $=-0.2666$

$\frac{d\theta}{dt}=-\frac{0.2666}{\sec^{2}\theta}$

Before I can proceeded, I need to find theta in our original l triangle.

$\tan\theta=\ \frac{opp}{adj}$

$\tan\theta=\ \frac{30}{90}$ $=1/3$

$\theta=\ \arctan\left(\frac{1}{3}\right)$

$\theta=\ 0.32175055$ Now that I found theta, I can use this an plug it back into $\frac{d\theta}{dt}$

$\frac{d\theta}{dt}=-\frac{0.2666}{\sec^{2}\left(0.32175055\right)}$ $= -0.2399 ft/s$

We can say that when the baseball player is 30ft from 3rd base, angle theta is closing at a rate of 0.2399 feet per second.

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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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$a^{2}+b^{2}=c^{2}$

We know that db/dt= 70miles/hr
We know that da/dt= 0
We want to know dc/dt

$2a\left(\frac{da}{dt}\right)+2b\left(\frac{db}{dt}\right)=2c\left(\frac{dc}{dt}\right)$

$2\left(30\right)\left(0\right)+2\left(40\right)\left(70\left(\frac{mi}{hr}\right)\right)=2\left(50\right)\left(\frac{dc}{dt}\right)$

$80\left(70\left(\frac{mi}{hr}\right)\right)=100\left(\frac{dc}{dt}\right)$

$\frac{dc}{dt}=\ 56\ \frac{mi}{hr}$

We can say that at 2pm, the two cars are gaining distance from eachother at a rate of 56 mi/hrs.

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