# Calculus/Related Rates
## Introduction

One useful application of derivatives is as an aid in the calculation of related rates. What is a related rate? In each case in the following examples the related rate we are calculating is a derivative with respect to some value. We compute this derivative from a rate at which some other known quantity is changing. Given the rate at which something is changing, we are asked to find the rate at which a value related to the rate we are given is changing.

## How to Solve

These general steps should be taken in order to complete a related rates problem.
* Write out any relevant formulas and information about the problem.
** The problem should have a variable you "control" (i.e. have knowledge of the value and rate of) and a variable that you want to find the related rate.
** Usually, related rates problem ask for a rate in respect to time. Do not panic if your equations do not appear to have any relationship to time! This will be handled later.
* Combine the formulas together so that the variable you want to find the related rate of is on one side of the equation and everything else is on the other side.
* Differentiate the formula with respect to time. Any other variable not a simple constant (such as $\pi$) should be differentiated as well. Be wary! Chain Rule usually should be used.
* The other variables that you have differentiated should have been given in the question or should be calculated separately. Nevertheless, plug-in known information and simplify.
* The value you get here is your answer.
The steps to solve a related rates problem is strikingly similar to an optimization problem, except that the main variable to find is not assigned to be 0 (it is supposed to be found) and that the extra variables in the optimization problem algorithm are actual variables in this case and are treated as variables instead of constants when differentiating.

## Notation

Newton's dot notation is used to show the derivative of a variable with respect to time. That is, if $f$ is a quantity that depends on time, then ${\dot {f}}={\frac {df}{dt}}$ , where $t$ represents the time. This notation is a useful abbreviation in situations where time derivatives are often used, as is the case with related rates.

## Examples

Example 1:
* Write out any relevant formulas or pieces of information.
    $V={\frac {1}{3}}\pi r^{2}h$
* Take the derivative of the equation above with respect to time. Remember to use the Chain Rule and the Product Rule.
    $V={\frac {1}{3}}\pi r^{2}h$
    ${\dot {V}}={\frac {\pi }{3}}\left(r^{2}{\dot {h}}+2rh{\dot {r}}\right)$
${\dot {V}}={\frac {\pi }{3}}\left(r^{2}{\dot {h}}+2rh{\dot {r}}\right)$

### Example 2

* Write out any relevant formulas and pieces of information.
    $V_{sphere}={\frac {4}{3}}\pi r^{3}$
    ${\dot {V}}=2$
    $r=2$
* Take the derivative of both sides of the volume equation with respect to time.
    $V={\frac {4}{3}}\pi r^{3}$
    ${\dot {V}}$ $={\frac {4}{3}}3\pi r^{2}{\dot {r}}$ $=4\pi r^{2}{\dot {r}}$
* Solve for ${\dot {r}}$ .
    ${\dot {r}}={\frac {\dot {V}}{4\pi r^{2}}}$
* Plug-in known information.
    ${\dot {r}}={\frac {2}{16\pi }}$
${\dot {r}}={\frac {1}{8\pi }}$

### Example 3

Note: Because the vertical distance is downward in nature, the rate of change of y is negative. Similarly, the horizontal distance is decreasing, therefore it is negative (it is getting closer and closer).
The easiest way to describe the horizontal and vertical relationships of the plane's motion is the Pythagorean Theorem.
* Write out any relevant formulas and pieces of information.
    $x^{2}+y^{2}=s^{2}$ (where s is the distance between the plane and the house)
    $x=300$
    $y=400$
    $s={\sqrt {x^{2}+y^{2}}}={\sqrt {300^{2}+400^{2}}}=500$
    ${\dot {x}}={\dot {y}}=-50$
* Take the derivative of both sides of the distance formula with respect to time.
    $x^{2}+y^{2}=s^{2}$
    $2x{\dot {x}}+2y{\dot {y}}=2s{\dot {s}}$
* Solve for ${\dot {s}}$.
    ${\dot {s}}$ $={\frac {2x{\dot {x}}+2y{\dot {y}}}{2s}}$ $={\frac {x{\dot {x}}+y{\dot {y}}}{s}}$
* Plug-in known information
    ${\dot {s}}$ $={\frac {(300)(-50)+(400)(-50)}{(500)}}$ $={\frac {-35000}{500}}$ $=-70$ ft/s
${\dot {s}}=-70$

### Example 4

* Write down any relevant formulas and information.
    $V={\frac {\pi }{3}}r^{2}h$
    ${\dot {V}}=10$
    $r={\frac {h}{2}}$
    $h=5$
Substitute $r={\frac {h}{2}}$ into the volume equation.
    $V$ $={\frac {\pi }{3}}r^{2}h$ $={\frac {\pi }{3}}h\left({\frac {h^{2}}{4}}\right)$ $={\frac {\pi }{12}}h^{3}$
* Take the derivative of the volume equation with respect to time.
    $V={\frac {\pi }{12}}h^{3}$
    ${\dot {V}}={\frac {\pi }{4}}h^{2}{\dot {h}}$
* Solve for ${\dot {h}}$ .
    ${\dot {h}}={\frac {4}{\pi h^{2}}}{\dot {V}}$
* Plug-in known information and simplify.
    ${\dot {h}}$ $={\frac {4}{\pi (5)^{2}}}10$ $={\frac {8}{5\pi }}$ ft/min
${\dot {h}}={\frac {8}{5\pi }}$
Example 5:
* Write out any relevant formulas and information.
Use the Pythagorean Theorem to describe the motion of the ladder.
    $x^{2}+y^{2}=l^{2}$ (where l is the length of the ladder)
    $l=10$
    ${\dot {x}}=2$
    $x=8$
    $y={\sqrt {l^{2}-x^{2}}}={\sqrt {100-64}}={\sqrt {36}}=6$
* Take the derivative of the equation with respect to time.
    $2x{\dot {x}}+2y{\dot {y}}=0$ ($l$ is constant so ${\frac {dl^{2}}{dt}}=0$ .)
* Solve for ${\dot {y}}$ .
    $2x{\dot {x}}+2y{\dot {y}}=0$
    $2y{\dot {y}}=-2x{\dot {x}}$
    ${\dot {y}}=-{\frac {x}{y}}{\dot {x}}$
* Plug-in known information and simplify.
    ${\dot {y}}$ $=\left(-{\frac {8}{6}}\right)(2)$ $=-{\frac {8}{3}}$ ft/sec
${\dot {y}}=-{\frac {8}{3}}$

## Exercises

$100ft^{3}/min$
$4ft$
${\frac {25}{16\pi }}{\frac {ft}{min}}$
${\frac {25}{16\pi }}{\frac {ft}{min}}$
$10ft$
$10ft$
$3ft^{3}/min$
$6ft$
${\frac {1}{3\pi }}{\frac {ft}{min}}$
${\frac {1}{3\pi }}{\frac {ft}{min}}$
$2ft$
$2rpm$
$4\pi {\frac {ft}{min}}$
$4\pi {\frac {ft}{min}}$
$t=0$
$e^{-t}$
$t=-\ln(0.001\pi )$
$t=-\ln(0.001\pi )$
Solutions
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