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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A helpful begining to any problem is to draw a picture:

As we can see, the information provided leads us to a right triangle.

A helpful step in a right triangle problem is to use the pythagorian theorem to find the value of all of the sides.

$a^2+b^2=c^2$ or in this case, $a^2 + 5^2 = 13^2$

$a=12$

Now, we want to identify what information we know, and which information we are trying to find.

We know the rate at which the rope is pulling the boat is 2 feet per second

so: $\frac{dc}{dt}=2ft/sec$

We want the rate at which the boat is heading towards the dock,

in this case: $\frac{da}{dt}$

the formula we can use to solve for $\frac{da}{dt}$ in terms of a triangle is the derivative of the pythagorean theorem

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

taking the derivitve we get:

$2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}$

Now the goal is to substitute in our known values, and solve for $\frac{da}{dt}$

Two important things to note in the context of this problem is that since side B of this triangle is a vertical line, its not moving, so its rate of change left or right is fixed at the value 0
and also that since the rope is being pulled in, the rate of change in this case is technically going to be negative.

so $\frac{db}{dt}=0$ and is thus subbed out of the equation entirely

which leaves:

$2(12)\frac{da}{dt}=2(13)(-2ft/sec)$

$24\frac{da}{dt}=-52ft/sec$

$\frac{da}{dt}=\frac{-52}{24} ft/sec$

simplfified:

$\frac{da}{dt}=\frac{-13}{6} ft/sec$

knowing this we can say:

When the length of the rope attached from the bow of the boat to pulley is 13 feet long, the ship is being pulled in at a rate of $\frac{-13}{6}$ or $2.166$ feet per second.

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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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:::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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As with every word problem, the first step is to decipher the information enough to draw a picture.


A lot is happening in this picture, the cars are both traveling south in a straight path paralell to eachother spaced 30 miles apart at different speeds.

the first car is driving at a speed of $50$ miles per hour, so $\frac{dA}{dt}=50 mi/hr$

the second car is driving at a speed of $70$ miles per hour, so $\frac{dB}{dt}=70 mi/hr$

at 2:00pm, two hours have passed, so if we multiply these values by $2$ we get their respective distances.

knowing the distance between them horizontally and vertically helps to draw two traingles which we can use to find the distance diagonally between the two. To do so, we use the pythagoreon theorem.

A triangle is helpful in getting us the information we want, which is the rate of change of side c on the triangle, which is the distance between the cars.

we can express that rate in calculus terms as: $\frac{dc}{dt}$

to find $\frac{dc}{dt}$, we can find the derivative of the pythagoreon theorem and solve.

note that though the triangles are the same (albiet mirrored), the rate of change in the context is going to be different for side "a" on the opposite triangle. Knowing this we can use the same distance (40mi) but use $\frac{dB}{dt}$ instead.

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

$2a\frac{dA}{dt}+2b\frac{dB}{dt}=2c\frac{dc}{dt}$

$a\frac{dA}{dt}+b\frac{dB}{dt}=c\frac{dc}{dt}$

$(40mi)(50mi/hr)+(40mi)(70mi/hr)=(50mi)\frac{dc}{dt}$

$(2000mi^2/hr)+(2800mi^2/hr)=50mi\frac{dc}{dt}$

$4800mi^2/hr=50mi\frac{dc}{dt}$

$96mi/hr=\frac{dc}{dt}$

now we can say:

At 2:00pm the distance between the cars is changing at a rate of 96 miles per hour.

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