1.
Create your variable table. Use \(x\) for the side length. (DR)
| Variable | Variable Description | Unit |
| \(\mathstrut\) | ||
| \(\mathstrut\) | ||
| \(\mathstrut\) | ||
| \(\mathstrut\) | ||
| \(\mathstrut\) |
Activity 8.1 Introduction to Related Rates
Consider the following problem.
A perfectly spherical snowball melts at a constant rate of 0.3 cm3⁄min. At what rate is the radius decreasing when the radius of the snowball is exactly 1.4 cm?
Before we actually solve the problem, let’s take a moment to think about the situation. If the snowball were really large, there would be a lot of surface area to account for the loss of water. When the snowball gets small, however, there’s less surface area to contribute water and, consequently, the radius will decrease at a faster rate than it does when the snowball is large. Mathematically, this increase in the rate at which the radius changes over time is a result of the manner in which the radius changes with respect to volume.
Figure 8.1.1 shows a graph of \(r = \fe{f}{V}\) where \(r\) is the radius of a sphere of volume \(V\text{.}\) Note that the tangent lines to the curve have greater slope the closer \(V\) gets to zero. This tells us that a small change in volume has a greater effect on the radius the closer \(V\) is to zero.
Since the volume is decreasing at a constant rate, and a change in volume has a greater and greater effect on the radius as the volume decreases, the rate at which the radius decreases must increase as time goes by. This relationship between the rates at which the volume and radius change is an example of what is called related rates.
An acronym sometimes associated with the solution process for related rates problems is DREDS. The first four letters of the acronym stand for Diagram, Rates, Equation, and Differentiate. The final letter stands for Substitute. The reason the substitution step comes last is that you need to have a rates equation before you have a place to substitute your values.
The first thing we need to do is clearly identify our variables. In related rates problems there are three broad types of variables.
-
Variables that measure physical quantities. These include, among other things, length variables, angle measurement variables, area variables, and volume variables. We will call these quantity-variables.
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Variables that measure rates (with respect to time). These variables always emerge by taking the derivatives (with respect to time) of the quantity-variables. We will call these rate-variables.
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A time variable. Depending on context, time may be measured in seconds, minutes, hours, days, weeks, years, etc.
The first three letters of DREDS (Diagram, Rates, Equation) are all interrelated and are not always to be taken literally. Basically they are meant to suggest that you need to identify the rates in the problem (either a given rate or the unknown rate you seek to find), identify quantity-variables that differentiate to each rate, and find an equation that relates those quantity-variables. A diagram is frequently useful during this part of the problem solving process.
Let’s make a variable table for the snowball problem. Please note that we will need a volume variable. We know this because of the presence of cm3 in the given rate unit; cubic-centimeters always measure volume (just as square-centimeters always measure area and centimeters always measure length). We will also need a radius variable (as the rate of change in the radius is stated in the problem).
| Variable | Variable Description | Unit |
| \(V\) |
The volume of the snowball \(t\) minutes after the snowball begins to melt.
|
cm3 |
| \(r\) |
The radius of the snowball \(t\) minutes after the snowball begins to melt.
|
cm |
| \(\lz{V}{t}\) |
The rate of change in the volume of the snowball \(t\) minutes after the snowball begins to melt.
|
cm3⁄min |
| \(\lz{r}{t}\) |
The rate of change in the radius of the snowball \(t\) minutes after the snowball begins to melt.
|
cm⁄min |
| \(t\) |
The amount of time that elapses after the snowball begins to melt.
|
min |
Since we are asked to determine a rate at which the radius changes, we are ultimately going to need to determine a value of \(\lz{r}{t}\text{.}\) We cannot do that until we have an equation that includes the symbol \(\lz{r}{t}\text{.}\) There is a two-step process that we follow to find that equation.
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Find an equation that relates the quantity-variables. This step is one of the places where a diagram sometimes comes into play. This step corresponds to E in the DREDS acronym.
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Differentiate both sides of the first equation with respect to time and simplify each side. The resultant equation is called the related rates equation. This step corresponds to the second D in the DREDS acronym.
The quantity-variables in the snowball problem are \(V\) and \(r\text{.}\) We don’t need a diagram in this problem as it is “standard knowledge” that the equation that relates the volume and radius of a sphere is \(V = \frac{4}{3}\pi r^3\text{.}\) Now that we have our (E)quation, we need to (D)ifferentiate both sides of this equation with respect to \(t\text{,}\) keeping in mind that both \(V\) and \(r\) are variables dependent upon \(t\text{.}\)
\begin{align*}
V \amp= \frac{4}{3}\pi r^3\\
\lzoo{t}{V} \amp= \lzoo{t}{\frac{4}{3}\pi r^3}\\
\lz{V}{t} \amp= 4\pi r^2\lz{r}{t}\amp\text{This is our related rates equation.}
\end{align*}
Now that we have our related rates equation (and only now), we can finally (S)ubstitute the given values into the appropriate variables. Before doing so, we need to alert the reader with an introductory sentence or phrase. We also need to clearly state the values of each of the variables for which we know the value. In most math texts this part of the problem is worked without units. This is because some problems contain constants with “hidden” units, and it can be confusing when the units do not work out in the end.
Getting back to the actual solution…
| Variable | What we know |
| \(V\) |
This variable is not in the related rates equation so it is no longer relevant in the problem solving process.
|
| \(r\) |
The value of this variable is \(1.4\text{.}\)
|
| \(\lz{V}{t}\) |
The value of this variable is \(-0.3\text{.}\) The value is negative because the volume decreases over time.
|
| \(\lz{r}{t}\) |
This is the variable for which we need to solve.
|
From our related rates equation and the values in Table 8.1.3:
\begin{align*}
\lz{r}{t} \amp= \frac{\lz{V}{t}}{4\pi r^2}\\
\\
\lzoa{r}{t}{r = 1.4,\lz{V}{t} = -0.3} \amp= \frac{-0.3}{4\pi (1.4)^2}\\
\amp\approx -0.01218\ldots
\end{align*}
To finish up we need to write a conclusion.
At the moment the radius of the snowball is 1.4 cm, the radius is decreasing at the rate of about 0.01218 cm⁄min.
Exercises Exercises
Exercise Group.
Consider the following problem.
A square is getting larger. The sides are each growing at a constant rate of 3 in⁄min. At what rate is the area of the square growing at the moment its area is 25 in2?
1.
2.
Write an equation that relates the two quantity-variables. (E)
3.
Differentiate both sides of this equation with respect to \(t\text{,}\) and simplify. This equation is called the related rates equation. (D)
4.
Write an orientation sentence or phrase and create a summary table for your quantity and rate variables. (preparing for S)
| Variable | What we know |
| \(\mathstrut\) | |
| \(\mathstrut\) | |
| \(\mathstrut\) | |
| \(\mathstrut\) |
5.
Substitute the known values into the related rates equation and solve for the unknown rate. (S)
6.
Write a conclusion.
7.
The radius of a circle is decreasing at the rate of 0.2 cm⁄min. At what rate is the area of the circle decreasing when the area of the circle is \(4\pi\) cm2?
Use the problem solving strategy:
-
Create your variable table. (DR)
-
Write an equation that relates the two quantity-variables. (E)
-
Differentiate both sides of the equation you just wrote with respect to \(t\text{.}\) Simplify the result. This equation is called the related rates equation. (D)
-
Write an orientation sentence or phrase and create a summary table for your quantity and rate variables. (preparing for S)
-
Substitute the known values into the related rates equation and solve for the unknown rate. (S)
-
Write a conclusion.
8.
As salt water boils, the only thing that evaporates is the water—the amount of salt in the mixture remains constant. As a consequence, the concentration of the salt in the mixture increases over time.
Suppose that the concentration of salt in a pot of boiling water is given by \(C = \frac{200}{V}\) where \(C\) is the concentration of salt in the mixture (mg⁄L) and \(V\) is the total amount of mixture in the pot (L). Determine the rate at which the salt concentration is changing at the moment there are 5.2 L of liquid in the pot if at that moment water is evaporating at the rate of 0.13 L⁄min.
