# Related Rates

Consider the following problem: one day it snows, and you and a friend decide to form a giant snowball. The next day, however, temperatures are back up above freezing, and the snowball starts to melt. If the snowball’s radius decreases at a rate of one centimeter per hour, can we determine the rate at which the snowball’s volume is decreasing as well? We can! Let’s discuss how.

This snowball problem belongs to a type of calculus problem we like to call “**related rates**.” These kinds of problems have that name because we are given information about one rate of change and are asked about *another* rate of change, where the two are related to each other in some way. For example, we know from geometry that the radius and volume of a sphere are related by the equation \(V=\frac{4}{3}\pi r^{3}\).

Let’s work through how to solve the problem now. We want to know the rate at which the snowball’s volume is decreasing.

To start, let’s ask: “what do we already know?” We know that the radius of the snowball decreases at a rate of one centimeter per hour, so we can write this as \(r'(t)=\frac{dr}{dt}=-1\frac{cm}{hr}\). I chose the letter \(r\) because we are using this equation to represent the radius. \(r\) is prime because we are describing the radius’s rate of change. This rate is described in units “centimeters per hour,” so mathematically we can express it with \(\frac{dr}{dt}\), the derivative of \(r\) with respect to the time \(t\).

What else do we know? As I mentioned earlier, we know that the volume of a sphere is:

\(V=\frac{4}{3}\pi r^{3}\)

Let’s use this information to write a function describing the snowball’s volume at time \(t\).

\(V(t)=\frac{4}{3}\pi \cdot r(t)^{3}\)

Here, instead of just writing \(r^{3}\), we need to write \(r(t)^{3}\) to remind ourselves that the radius is changing with time just like the volume is. Since we are trying to find the rate at which the snowball’s volume is decreasing, all we have to do from here is calculate \(V'(t)\).

Notice that because \(r\) is a function of \(t\), *within the function V*, we need to use the **chain rule** to compute this derivative.

The derivative of \(V(t)\) is then:

\(V'(t)=\frac{d}{dt}\left [ \frac{4}{3}\pi r(t)^{3} \right ]\)

Taking the derivative with the respect to \(t\), the **power rule** states that we multiply by 3 and drop the exponent to 2. This will gives us:

\(V'(t)=\frac{4}{3}\pi \cdot 3r(t)^{2}\cdot \frac{dr}{dt}\)

The \(dr\) at the end is to help us to remember to use the chain rule. We now need to multiply by the derivative of \(r\), \(r'(t)\), which is -1.

\(V'(t)=\frac{4}{3}\pi \cdot 3[r(t)]^{2}\cdot (-1)\)

Then, all this will simplify to:

\(V'(t)=-4\pi [r(t)]^{2}\)

This means that the rate of change in volume at \(t\) hours is equal to \(-4\pi\) times the radius at that time squared.

Let’s look at another example.

Elizabeth throws a stone into a lake and observes that the ripples grow in diameter at the rate of 2 feet per second. By what speed does the area inside the ripples grow when they are 8 feet across?

Let’s start by asking, “what do we want to know?” We want to know the rate of change in area at a particular time. Let’s call area at time \(t\), \(A(t)\), and let’s call the rate of change in area \(A'(t)\). Once we find \(A'(t)\), we will need to plug in the time when diameter equals 8 feet to get the final solution.

Now, what do we already know? So, we know that the ripples grow in diameter by 2 feet per second. Let’s call this rate of change \(d'(t)=2\). We will let \(d(t)\) represent the diameter at time \(t\).

We also know that a circle’s diameter is related to its area because \(A=\pi r^{2}=\pi (\frac{d}{2})^{2}\).

This is everything we need to solve the problem. Remember, since we want to know \(A'(t)\), we first need to write \(A(t)\). Then we’ll take its derivative.

\(A(t)\), or area at time \(t\), is equal to \(\pi\) times half the diameter at time \(t\), squared.

\(A(t)=\pi \left [ \frac{d(t)}{2} \right ]^{2}\)

Notice that we can go ahead and square this out. That will give us:

\(A(t)=\pi \frac{[d(t)]^{2}}{4}\)

Then we can rewrite this as:

\(A(t)=\frac{\pi}{4}[d(t)]^{2}\)

Now let’s take the derivative. Since \(\frac{\pi}{4}\) is constant, its derivative is 0, so we can focus on taking the derivative of \([d(t)]^{2}\). We need to use the chain rule here because \(d(t)\) is of course a function. So \(A'(t)\) will equal:

\(A'(t)=\frac{\pi}{4}\cdot 2\cdot d(t)\cdot d'(t)\)

Remember, we were told in the problem statement that \(d'(t)=2\), so we can plug this in.

\(A'(t)=\frac{\pi}{4}\cdot 2\cdot d(t)\cdot (2)\)

Then, we can simplify this out and get:

\(A'(t)=\pi d(t)\)

At this point, we have found the rate of change in area at time \(t\). The last step is to find the rate of change in area when the ripple is 8 feet across. In other words, when \(d(t)=8\). So, our final answer is that the rate of change when \(d\) of \(t\) is equal to 8 is:

\(\pi d(t)=\pi (8)=8\pi ft^{2}/sec\)

I have one more problem to go over with you in this video, but for this one, let’s see if you can work it out on your own!

The sides of a cube are growing at the rate of 4 inches per minute. By what speed will the cube’s surface area be growing when the sides of the cube are 24 inches long?

Pause the video now and work it out on your own. Then we will compare our work.

Let’s jump right in. For this problem, we want to know the rate of change in the cube’s surface area at a specific time. This means we need to write a formula for surface area, then take its derivative to get that rate of change.

We are told that the sides of a cube are growing in length at the rate of 4 inches per minute. I’m going to write this fact as \(s'(m)=4\), where \(s\) represents side length at time \(m\).

Now, because cubes have 6 sides of equal size, we can write the surface area with respect to time \(m\), in minutes, as:

\(SA(m)=6[s(m)]^{2}\)

Time to take the derivative, once again using the chain rule. The surface area has rate of change:

\(SA'(m)=6\cdot 2\cdot s(m )\cdot s'(m)\)

Since \(s'(m)=4\), we can rewrite this so that it’s:

\(SA'(m)=6\cdot 2\cdot s(m)\cdot (4)\)

Then, this will simplify down to:

\(SA'(m)=48s(m)\)

So the rate of change in surface area at a particular time equals 48 times the side length at that time.

We are interested in the rate of change in surface area when sides have length 24 inches, so substitute in 24 where you see \(s(m)\). The solution is then:

\(48s(m)=48(24)=1,152\text{ in}^{2}/\text{min}\)

Many students and teachers acknowledge that related rates is typically the most difficult section in Calculus 1. Even so, these problems are certainly doable if you keep these main steps in mind:

- Ask yourself, “What is the problem asking for? What do we want to know?” Typically, you want to find some rate of change. Write down exactly what you’re looking for.
- Next, ask, “What do I already know from the problem?” Write down the information that you’re given in the problem statement, which should include some rate of change. Also make sure to write down the relationship between the two rates of interest (for example, \(V=\frac{4}{3}\pi r^{3}\)).
- Use what you know to write a formula with the respect to time.
- Then take the derivative to acquire the rate of change you’re interested in. Remember that this step almost always incorporates the chain rule.
- Finally, plug in the appropriate condition to get your solution.

Even though that seems like a lot, you’ll get the hang of it with practice.

Thanks for watching, and happy studying!