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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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What must be found: 
$$\frac{dx}{dt}$$
What is known: x=13 ft, $\frac{dy}{dt}$=-$2$ ft/s
Formula: $x^2+5^2=y^2$
derivative of the formula: $2y$ $\frac{dy}{dt}$=$2x$ $\frac{dy}{dt}$
                Divide the formula by 2x 
                $\frac{2y}{2x}$ $\frac{dy}{dt}$=$\frac{2x}{2x}$ $\frac{dy}{dt}$
                .
                $\frac{dx}{dt}$=$\frac{y}{x}$$\frac{dy}{dt}$
                .
                find y with the pythagorean theorem formula:
                $$5^2+x^2=y^2$$
                $$5^2+13^2=y^2$$
                $$25+169=y^2$$
                $$\sqrt{194}=\sqrt{y^2}$$
                $$\sqrt{194}=y$$
                substitute into $\frac{dx}{dt}$=$\frac{y}{x}$$\frac{dy}{dt}$
                .
             $$\frac{dx}{dt}=\frac{\sqrt{194}}{13}(-2)$$
             $$\frac{dx}{dt}=-2.142828966$$
             The boat is approaching the dock at 2.14 feet per second 
                
![](https://i.imgur.com/REHXtBs.jpg)



:::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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What must be found: 
$$\frac{d\theta}{dt}$$
What is known: x=30, $\frac{dx}{dt}$=-24 ft/s, 
Formula: Tan$\theta$= $\frac{x}{90}$
Derivative of formula:  $\frac{1}{cos^2\theta}$$\frac{d\theta}{dt}$= $\frac{1}{90}$ $\frac{dx}{dt}$
Solve for $\frac{d\theta}{dt}$
$$\frac{d\theta}{dt}=\frac{1}{90}*\frac{dx}{dt}*\cos^2\theta$$

solve for $cos^2\theta$ beforehand
$$30^2+90^2=c^2$$
$$900+8100=c^2$$
 $$\sqrt{9000}=\sqrt{c^2}$$
$$\sqrt{9000}=c$$

$cos\theta$=$(90/\sqrt{9000})$
Substiture back into original formula 

$$\frac{d\theta}{dt}=\frac{1}{90}*(-24)* (90/ \sqrt{9000})^2$$


$$\frac{d\theta}{dt}=\frac{1}{90}*(-24)* (8100/ 9000)$$
$$\frac{d\theta}{dt}=-0.24 rad/sec$$ 
When the runner is 30 ft from third base, he is running at 0.24 radians per second 
![](https://i.imgur.com/taZvO3n.jpg)










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