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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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Let $AC$ be the position of the rope at any time $t$ such that $AC=y$ and $BC=x$ Thus, we can use the pythagorean theorem
\\[
y^2=x^2+5^2
\\] The question gives us the measurement of the pully, the $y$ value, which is 13 ft. This value can be plugged into the pythagorean theorem.
\\[
x^2=13^2-5^2
\\]\\[
x=\sqrt{169-25}
\\]\\[
x=12 ft
\\]Differentiate both sides with respect to $t.$
\\[
2y\frac{dy}{dt}=2x\frac{dx}{dt}+0
\\]\\[
y\frac{dy}{dt}=x\frac{dx}{dt}
\\]\\[
\frac{dy}{dt}=-2 feet/sec
\\]\\[
-y(2)=x\frac{dx}{dt}
\\]\\[
\frac{dx}{dt}=\frac{-2y}{x}
\\]
Since $x$ was found earlier using the pythagorean theorem, and we have the value of $y,$ these values need to be plugged into the derivative expression found.
\\[
\frac{dx}{dt}=\frac{-2y}{x}
\\]\\[
\frac{dx}{dt}=\frac{(-2)(13)}{12}
\\]\\[
\frac{dx}{dt}=-\frac{13}{6}
\\]\\[
\frac{dx}{dt}=-2.167ft/sec
\\]
The boat approaches the dock at a rate of 2.167 feet per 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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The speed of the runner=*24ft/sec*
Time=*t seconds*
So the distance covered is expressed with the equation $24t.$ Thus, $BX=90ft-24tft$ and $BC=90ft$ So, by using $tan,$ we get
\\[
tanθ=\frac{90-24t}{90}
\\] Differientiate on both sides of the eqaution
\\[
sec^2θ\frac{dθ}{dt}=-\frac{24}{90}
\\] When the runner is $30ft$ away from third base,
\\[
tanθ=\frac{30}{90}=\frac{1}{3}
\\]\\[
sec^2θ=1+tan^2θ
\\]\\[
sec^2θ=1+\frac{1}{9}=\frac{10}{9}
\\] Substitute $sec^2θ=\frac{10}{9}$ in $sec^2θ\frac{dθ}{dt}=-\frac{24}{90}$
\\[
\frac{10}{9}\frac{dθ}{dt}=-\frac{24}{90}
\\]\\[
\frac{dθ}{dt}=-0.24 rad/sec
\\] So, the angle is decreasing at 0.24 rad/sec

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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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