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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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:::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?
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[See Problem 2 and 3 for Solutions and Work]



:::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/UxwMAi7.jpg)

$tan(θ)=\frac{x}{y}$
$h^2=x^2+y^2$

We know that:
$tan(θ)=\frac{30}{90}$
$(θ)=tan^{-1}(\frac{30}{90})=0.322 radians$
$\frac{dx}{dt}=-24\frac{ft}{s}$
$x=30ft$
$y=90ft$

We want to find $\frac{dθ}{dt}$:

Taking the derivative of the above formula:
$tan(θ)=\frac{x}{y}$
$(\frac{dθ}{dt})sec^2(θ)=\frac{(\frac{dx}{dt})(y)-(\frac{dy}{dt})(x)}{y^2}$

$\frac{(\frac{dθ}{dt})}{cos^2(θ)}=\frac{(\frac{dx}{dt})(y)-(\frac{dy}{dt})(x)}{y^2}$

Solve for $\frac{dθ}{dt}$ and plug in known values:
$(\frac{dθ}{dt})=\frac{cos^2(θ)\frac{dx}{dt}(y)-(\frac{dy}{dt})(x)}{y^2}$

$(\frac{dθ}{dt})=\frac{cos^2(0.322)(-24)(90)-(0)(30)}{(90)^2}=\frac{-0.24radians}{s}$

When the player is 30 ft from third base, the angle of the umpire's line of sight is decreasing at a rate of $0.24 \frac{radians}{sec}$. 

:::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?
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![](https://i.imgur.com/Ox90GkR.jpg)


$h^2=x^2+y^2$

We know that: 

Car 1 traveled 100 miles in two hours and car two traveled 140 miles in two hours. 
$\frac{dy}{dt}=70-50=20mph$
This is because this side of the triangle is changing at this rate (the speed of car 2 relative to that of car 1 since both cars are still moving)
$x=30$
$y=40$
$h=50$
We know this since this triangle is just a scaled-up version of a 3-4-5 triangle. 

We want to find $\frac{dh}{dt}$:

Taking the derivative of the above formula:
$2h\frac{dh}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$

Since $x$ is not changing, the first term on the left side of the equal sign cancels out:
$2h\frac{dh}{dt}=2y\frac{dy}{dt}$

Solving for $\frac{dh}{dt}$ and plugging in known values (the 2's on either side cancel out):

$\frac{dh}{dt}=\frac{(40)(20)}{50}=16 mph$

At 2:00 PM, the distance between Car 1 and Car 2 is increasing at a rate of $16\frac{miles}{hr}$.





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