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 article 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.

Process for solving related rates problems  :

• Write out any relevant formulas and information.
• Take the derivative of the primary equation.
• Solve for the desired variable.
• Plug-in known information and simplify.

Related Rates

As stated in the introduction, when doing related rates, you generate a function which compares the rate of change of one value with respect to change in time. For example, velocity is the rate of change of distance over time. Likewise, acceleration is the rate of change of velocity over time. Therefore, for the variables for distance, velocity, and acceleration, respectively x, v, and a, and time, t:

$v=\frac{dx}{dt}$

$a=\frac{dv}{dt}$

Using derivatives, you can find the functions for velocity and acceleration from the distance function. This is the basic idea behind related rates: the rate of change of a function is the derivative of that function with respect to time.

Common Applications

Filling Tank

This is the easiest variant of the most common textbook related rates problem: the filling water tank.

• The tank is a cube, with volume 1000L.
• You have to fill the tank in ten minutes or you die.
• You want to escape with your life and as much money as possible, so you want to find the smallest pump that can finish the task.

We need a pump that will fill the tank 1000L in ten minutes. So, for pump rate p, volume of water pumped v, and minutes t:

$p=\frac{dv}{dt}$

Newton's Second Law

Using calculus, Sir Issac Newton was able to develop his Second Law of Motion. Let's play around with them. First, we know that velocity is the derivative of the location function, and acceleration is the derivative of velocity. Newton derived the following equations for mass m, velocity v, force F and momentum p:

$p=mv$

therefore, if Force is a change in momentum,

$F=\frac{dp}{dt}$

Examples

Related rates can get complicated very easily.

Example 1:

A cone with a circular base is being filled with water. You need to find a formula which will find the rate with which water is pumped.

• 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.

$V=\frac{1}{3}\pi r^{2}h$

$\frac{dV}{dt}=\frac{\pi }{3}(r^2\cdot \frac{dh}{dt}+2rh\cdot \frac{dr}{dt})$

Answer: $\frac{dV}{dt}=\frac{\pi }{3}(r^2\cdot \frac{dh}{dt}+2rh\cdot \frac{dr}{dt})$

Example 2:

A spherical hot air balloon is being filled with air. The volume is changing at a rate of 2 cubic feet per minute. 
How is the radius changing with respect to time when the radius is equal to 2 feet?

• Write out any relevant formulas and pieces of information.

$V_{sphere}=\frac{4}{3}\pi r^3$

$\frac{dV}{dt}=2$

$r=2$

• Take the derivative of both sides of the volume equation with respect to time.

$V=\frac{4}{3}\pi r^3$

$\frac{dV}{dt}$	=	$\frac{4}{3}\cdot 3\cdot \pi r^2\cdot \frac{dr}{dt}$
	=	$4\pi r^2\cdot \frac{dr}{dt}$

• Solve for $\frac{dr}{dt}.$

$\frac{dr}{dt}=\frac{1}{4\pi r^2}\cdot \frac{dV}{dt}$

• Plug-in known information.

$\frac{dr}{dt}=\frac{1}{16\pi }\cdot 2$

Answer: $\frac{dr}{dt}=\frac{1}{8\pi }$ ft/min.

Example 3:

An airplane is attempting to drop a box onto a house. The house is 300 feet away in horizontal distance and 
400 feet in vertical distance. The rate of change of the horizontal distance with respect to time is the same 
as the rate of change of the vertical distance with respect to time. How is the distance between the box and 
the house changing with respect to time at the moment? The rate of change in the horizontal direction with 
respect to time is -50 feet per second. 

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)

$x=300$

$y=400$

$s=\sqrt{x^2+y^2}=\sqrt{300^2+400^2}=500$

$\frac{dx}{dt}=\frac{dy}{dt}=-50$

• Take the derivative of both sides of the distance formula with respect to time.

$x^2+y^2=s^2$

$2x\cdot \frac{dx}{dt}+2y\cdot \frac{dy}{dt}=2s\cdot \frac{ds}{dt}$

• Solve for $\frac{ds}{dt}$.

$\frac{ds}{dt}=\frac{1}{2s}(2x\cdot \frac{dx}{dt}+2y\cdot \frac{dy}{dt})$

• Plug-in known information

$\frac{ds}{dt}$	=	$\frac{1}{2(500)}[2(300)\cdot (-50)+2(400)\cdot (-50)]$
	=	$\frac{1}{1000}(-70000)$
	=	$-70$ ft/s

Answer: $\frac{ds}{dt}=-70$ ft/sec.

Example 4:

Sand falls onto a cone shaped pile at a rate of 10 cubic feet per minute.  The radius of the pile's base
is always 1/2 of its altitude.  When the pile is 5 ft deep, how fast is the altitude of the pile increasing?

• Write down any relevant formulas and information.

$V=\frac{1}{3}\pi r^{2}h$

$\frac{dV}{dt}=10$

$r=\frac{1}{2}h$

$h=5$

Substitute $r=\frac{1}{2}h$ into the volume equation.

$V$	=	$\frac{1}{3}\pi r^{2}h$
	=	$\frac{1}{3}\pi h\cdot \left(\frac{h^2}{4}\right)$
	=	$\frac{1}{12}\pi h^3$

• Take the derivative of the volume equation with respect to time.

$V=\frac{1}{12}\pi h^3$

$\frac{dV}{dt}=\frac{1}{4}\pi h^2\cdot \frac{dh}{dt}$

• Solve for $\frac{dh}{dt}$.

$\frac{dh}{dt}=\frac{4}{\pi h^2}\cdot \frac{dV}{dt}$

• Plug-in known information and simplify.

$\frac{dh}{dt}$	=	$\frac{4}{\pi (5{)}^2}\cdot 10$
	=	$\frac{8}{5\pi }$ ft/min

Answer: $\frac{dh}{dt}=\frac{8}{5\pi }$ ft/min.

Example 5:

A 10 ft long ladder is leaning against a vertical wall.  The foot of the ladder is being pulled away from
the wall at a constant rate of 2 ft/sec.  When the ladder is exactly 8 ft from the wall, how fast is 
the top of the ladder sliding down the wall?

• 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)

$l=10$

$\frac{dx}{dt}=2$

$x=8$

$y=\sqrt{l^2-x^2}=\sqrt{100-64}=\sqrt{36}=6$

• Take the derivative of the equation with respect to time.

$2x\cdot \frac{dx}{dt}+2y\cdot \frac{dy}{dt}=0$ ($l^2$ so $\frac{dl}{dt}=0$.)

• Solve for $\frac{dy}{dt}$.

$2x\cdot \frac{dx}{dt}+2y\cdot \frac{dy}{dt}=0$

$2y\cdot \frac{dy}{dt}=-2x\cdot \frac{dx}{dt}$

$\frac{dy}{dt}=-\frac{x}{y}\cdot \frac{dx}{dt}$

• Plug-in known information and simplify.

$\frac{dy}{dt}$	=	$(-\frac{8}{6})(-2)$
	=	$\frac{8}{3}$ ft/sec

Answer: $\frac{dy}{dt}=\frac{8}{3}$ ft/sec.

Exercises

Problem Set

Here's a few problems for you to try:

1. A spherical balloon is inflated at a rate of 100 $f{t}^3$/min. Assuming the rate of inflation remains constant, how fast is the radius of the balloon increasing at the instant the radius is 4 ft?
2. Water is pumped from a cone shaped reservoir (the vertex is pointed down) 10 ft in diameter and 10 ft deep at a constant rate of 3 $f{t}^3$/min. How fast is the water level falling when the depth of the water is 6 ft?
3. A boat is pulled into a dock via a rope with one end attached to the bow of a boat and the other end held by a man standing 6 ft above the bow of the boat. If the man pulls the rope at a constant rate of 2 ft/sec, how fast is the boat moving toward the dock when 10 ft of rope is out?

Solution Set

1. $\frac{25}{16\pi }\frac{ft}{min}$
2. $\frac{1}{3\pi }\frac{ft}{min}$
3. $\frac{5}{2}\frac{ft}{sec}$