3.1: Related Rates
Learning Objectives
A student will be able to:
 Solve problems that involve related rates.
Introduction
In this lesson we will discuss how to solve problems that involve related rates. Related rate problems involve equations where there is some relationship between two or more derivatives. We solved examples of such equations when we studied implicit differentiation in Lesson 2.6. In this lesson we will discuss some reallife applications of these equations and illustrate the strategies one uses for solving such problems.
Let’s start our discussion with some familiar geometric relationships.
Example 1: Pythagorean Theorem
\begin{align*}x^2 + y^2 = z^2\end{align*}
We could easily attach some reallife situation to this geometric figure. Say for instance that \begin{align*}x\end{align*}
\begin{align*}x^2 + y^2 & = z^2\\ 2x\frac{dx}{dt}+2y \frac{dy}{dt}& =2z\frac{dz}{dt}.\end{align*}
Simplifying, we have
Equation 1. \begin{align*}x\frac{dx}{dt}+y\frac{dy}{dt}=z\frac{dz}{dt}\end{align*}
So we have relationships between the derivatives, and since the derivatives are rates, this is an example of related rates. Let’s say that person \begin{align*}x\end{align*} is walking at \begin{align*}5\;\mathrm{mph}\end{align*} and that person \begin{align*}y\end{align*} is walking at \begin{align*}3\;\mathrm{mph}\end{align*}. The rate at which the distance between the two walkers is changing at any time is dependent on the rates at which the two people are walking. Can you think of any problems you could pose based on this information?
One problem that we could pose is at what rate is the distance between \begin{align*}x\end{align*} and \begin{align*}y\end{align*} increasing after one hour. That is, find \begin{align*}dz/dt.\end{align*}
Solution:
Assume that they have walked for one hour. So \begin{align*}x=5 \;\mathrm{mi}\end{align*} and \begin{align*}y=3.\end{align*} Using the Pythagorean Theorem, we find the distance between them after one hour is \begin{align*}z=\sqrt{34}= 5.83 \;\mathrm{miles}\end{align*}.
If we substitute these values into Equation 1 along with the individual rates we get
\begin{align*}5(5)+3(3)& ={\sqrt{34}}\frac{dz}{dt}\\ {34}& ={\sqrt{34}}\frac{dz}{dt}\\ \frac{34}{\sqrt{34}}& =\frac{dz}{dt}.\end{align*}
Hence after one hour the distance between the two people is increasing at a rate of \begin{align*}\frac{dz}{dt}=\frac{34}{\sqrt{34}}\approx 5.83\;\mathrm{mph}\end{align*}.
Our second example lists various formulas that are found in geometry.
As with the Pythagorean Theorem, we know of other formulas that relate various quantities associated with geometric shapes. These present opportunities to pose and solve some interesting problems
Example 2: Perimeter and Area of a Rectangle
We are familiar with the formulas for Perimeter and Area:
\begin{align*}P & = 2 * l + 2 * w,\\ A & = l * w.\end{align*}
Suppose we know that at an instant of time, the length is changing at the rate of \begin{align*}8\;\mathrm{ft/hour}\end{align*} and the perimeter is changing at a rate of \begin{align*}24\;\mathrm{ft/hour}\end{align*}. At what rate is the width changing at that instant?
Solution:
If we differentiate the original equation, we have
Equation 2: \begin{align*}\frac {dp}{dt}=2 * \frac {dl}{dt}+2* \frac {dw} {dt}.\end{align*}
Substituting our known information into Equation II, we have
\begin{align*}24 & =(2*8)+2 * \frac {dw}{dt}\\ 8 & =2*\frac {dw}{dt}\\ 4 & =\frac {dw}{dt}.\end{align*}
The width is changing at a rate of \begin{align*}4\;\mathrm{ft/hour}\end{align*}.
Okay, rather than providing a related rates problem involving the area of a rectangle, we will leave it to you to make up and solve such a problem as part of the homework (HW #1).
Let’s look at one more geometric measurement formula.
Example 3: Volume of a Right Circular Cone
\begin{align*}V=\frac {1}{3}\pi r^2 {h}\end{align*}
We have a water tank shaped as an inverted right circular cone. Suppose that water flows into the tank at the rate of \begin{align*}5 \;\mathrm{ft}^3/\mathrm{min}.\end{align*} At what rate is the water level rising when the height of the water in the tank is \begin{align*}6 \;\mathrm{feet}\end{align*}?
Solution:
We first note that this problem presents some challenges that the other examples did not. Both \begin{align*}r\end{align*} and \begin{align*}h\end{align*} are functions of \begin{align*}t\end{align*}, and so implicit differentiation of the \begin{align*}V(t)\end{align*} function is going to produce several variables in the form \begin{align*}r(t)\end{align*}, \begin{align*}\frac{dr}{dt}\end{align*}, \begin{align*}h(t)\end{align*}, and \begin{align*}\frac{dh}{dt}\end{align*}, and that will give us too many variables to solve for \begin{align*}\frac{dh}{dt}\end{align*}. So we need to find a way to eliminate \begin{align*}r\end{align*} and \begin{align*}\frac{dr}{dt}\end{align*}.
When we differentiate the original equation, \begin{align*}V=(1/3) \pi r^2 h,\end{align*} we get
\begin{align*}\frac{dV}{dt} = \frac {1}{3} {\pi}({h})({2r})\frac {dr}{dt} + \frac{1}{3}\pi r^2 \frac {dh}{dt}.\end{align*}
The difficulty here is that we have no information about the radius when the water level is at \begin{align*}6\;\mathrm{feet}\end{align*}. So we need to relate the radius a quantity that we do know something about. Starting with the original equation, let’s find a relationship between \begin{align*}h\end{align*} and \begin{align*}r.\end{align*} Let \begin{align*} r_1 \end{align*} be the radius of the surface of the water as it flows out of the tank.
Note that the two triangles are similar and thus corresponding parts are proportional. In particular,
\begin{align*}\frac {r_1}{h} & = \frac{8}{20}\\ r_1 & = \frac{8h}{20}=\frac{2h}{5}.\end{align*}
Now we can solve the problem by substituting \begin{align*}r_1 = (2h/5)\end{align*} into the original equation:
\begin{align*}V=\frac {1}{3} \pi \left (\frac{2h}{5}\right )^2 {h}=\frac {4\pi}{75}h^3.\end{align*}
Hence \begin{align*}\frac{dV}{dt}=\frac{12 \pi}{75} h^2 \frac {dh}{dt}\end{align*}, and by substitution,
\begin{align*}5 &=\frac{12 \pi}{75}({36}) \frac {dh}{dt}\\ \frac{dh}{dt}& =\frac{375} {43 2\pi} \approx 0.28 \frac {\text{ft}}{\text{min}}.\end{align*}
Lesson Summary
 We learned to solve problems that involved related rates.
Multimedia Links
For a video presentation of related rates (12.0), see Math Video Tutorials by James Sousa, Related Rates (10:34).
In the following applet you can explore a problem about a melting snowball where the radius is decreasing at a constant rate. Calculus Applets Snowball Problem. Experiment with changing the time to see how the volume does not change at a constant rate in this problem. If you'd like to see a video of another example of a related rate problem worked out (12.0), see Khan Academy Rates of Change (Part 2) (5:37).
Review Questions

 Make up a related rates problem about the area of a rectangle.
 Illustrate the solution to your problem.
 Suppose that a particle is moving along the curve \begin{align*}4x^2 + 16y^2 = 32.\end{align*} When it reaches the point \begin{align*}(2,1),\end{align*} the \begin{align*}x\end{align*}coordinate is increasing at a rate of \begin{align*}3\;\mathrm{ft/sec}\end{align*}. At what rate is the \begin{align*}y\end{align*}coordinate changing at that instant?
 A regulation softball diamond is a square with each side of length \begin{align*}60\;\mathrm{ft}\end{align*}. Suppose a player is running from first base to second base at a speed of \begin{align*}18\;\mathrm{ft/sec}\end{align*}. At what rate is the distance between the runner and home plate changing when the runner is \begin{align*}2/3\end{align*} of the way from first to second base?
 At a recent Hot Air Balloon festival, a hot air balloon was released. Upon reaching a height of \begin{align*}300\;\mathrm{ft}\end{align*}, it was rising at a rate of \begin{align*}20 \;\mathrm{ft/sec}\end{align*}. Mr. Smith was \begin{align*}100 \;\mathrm{ft}\end{align*} away from the launch site watching the balloon. At what rate was the distance between Mr. Smith and the balloon changing at that instant?
 Two trains left the St. Louis train station in the late morning. The first train was traveling East at a constant speed of \begin{align*}65 \;\mathrm{mph}\end{align*}. The second train traveled South at a constant speed of \begin{align*}75 \;\mathrm{mph}\end{align*}. At \begin{align*}3\end{align*} PM, the first train had traveled a distance of \begin{align*}120 \;\mathrm{miles}\end{align*} while the second train had traveled a distance of \begin{align*}130 \;\mathrm{miles}\end{align*}. How fast was the distance between the two trains changing at that time?
 Suppose that a \begin{align*}17 \;\mathrm{ft}\end{align*} ladder is sliding down a wall at a rate of \begin{align*}6\;\mathrm{ft/sec}\end{align*}. At what rate is the distance between the bottom of the ladder and the wall increasing when the top is \begin{align*}8\;\mathrm{ft}\end{align*} from the ground?
 Suppose that the length of a rectangle is increasing at the rate of \begin{align*}6 \;\mathrm{ft/min}\end{align*} and the width is increasing at a rate of \begin{align*}2 \;\mathrm{ft/min}\end{align*}. At what rate is the area of the rectangle changing when its length is \begin{align*}25 \;\mathrm{ft}\end{align*} and its width is \begin{align*}15 \;\mathrm{ft}\end{align*}?
 Suppose that the quantity demand of new \begin{align*}40''\end{align*} plasma TVs is related to its unit price by the formula \begin{align*}p + x^2 = 1200\end{align*}, where \begin{align*}p\end{align*} is measured in dollars and \begin{align*}x\end{align*} is measured in units of one thousand. How is the quantity demand changing when \begin{align*}x = 20,\end{align*} \begin{align*}p = 1500,\end{align*} and the price per TV is decreasing at a rate of \begin{align*}\$10\end{align*} per week?
 The volume of a cube with side \begin{align*}s\end{align*} is changing. At a certain instant, the sides of the cube are \begin{align*}6 \;\mathrm{inches}\end{align*} and increasing at the rate of \begin{align*}1/4 \;\mathrm{in/min}\end{align*}. How fast is the volume of the cube increasing at that time?
 Suppose that the area of a circle is increasing at a rate of \begin{align*}24 \;\mathrm{in}^2/\;\mathrm{min}.\end{align*} How fast is the radius increasing when the area is \begin{align*}36 \pi \;\mathrm{in}^2\end{align*}?
 How fast is the circumference changing at that instant?
Review Answers
 Answers will vary.
 \begin{align*}\frac{dy}{dt}=\frac {3}{2} \frac{\mathrm{ft}}{\mathrm{sec}}\end{align*}
 Using the following diagram, \begin{align*}\frac{dy}{dt}=\frac {720} {\sqrt{5200}} \frac{\mathrm{ft}}{\mathrm{sec}} \approx 9.98 \frac{\mathrm{ft}}{\mathrm{sec}}.\end{align*}
 Using the following diagram, \begin{align*}\frac{dy}{dt}=\frac {6000} {\sqrt{100000}} \frac{\mathrm{ft}}{\mathrm{sec}} \approx 18.97 \frac{\mathrm{ft}}{\mathrm{sec}}. \end{align*}
 Using the following diagram, \begin{align*}\frac{ds}{dt}=\frac {17550} {\sqrt{31300}} \approx 99.20\;\mathrm{mph}.\end{align*}
 Using the following diagram, \begin{align*}\frac {dx}{dt} = \frac {16}{5} \frac{\mathrm{ft}}{\mathrm{sec}}.\end{align*}
 \begin{align*}\frac {dA}{dt}=140 \frac{\mathrm{ft}}{\mathrm{min}}\end{align*}
 The demand is increasing at a rate of \begin{align*}1/4\end{align*} per thousand units, or \begin{align*}250 \;\mathrm{units}\end{align*} per week.

\begin{align*}\frac {dV}{dt}=27 \frac {{\mathrm{in}}^3}{\mathrm{min}}\end{align*}
 \begin{align*}\frac {dr}{dt}=\frac{2}{\pi} \frac {\mathrm{in}}{\mathrm{min}}\end{align*}
 \begin{align*}\frac {dp}{dt}=4 \frac {\mathrm{in}}{\mathrm{min}}\end{align*}