The radius of a sphere is increasing at a rate of *4* \(\frac{mm}{s}\). How fast is the volume increasing when the diameter is *80 mm*?

This question states pretty clearly that we will be working with a sphere here. Since it gives information about how it is changing and asks us to find how quickly another value is changing, we know it’s a related rates problem.

As with all of the other related rates problems I’ve worked through, we are going to be following the same **four step process**. I will go through them one step at a time, but you can also find where I introduce the steps here.

## 1. Draw a sketch

The first thing we need to do is draw a sketch of the scene being described. Obviously we are dealing with a sphere, but we are really only told two things:

- how quickly the
**radius is changing**, and - what the
**diameter**is at the specific moment we’re concerned with.

Since that’s all we know, it will be pretty simple to put that into a drawing. It would look something like this:

So we have a sphere whose radius is increasing at a rate of *4* \(\frac{mm}{s}\) and we need to consider the moment when its diameter is *80 mm*.

## 2. Come up with your equation

As with any related rates problem, we need to create our equation once we have created our drawing. To do this we need to consider the information we have been given and how it relates to the piece of information we’re looking for.

*What are we looking for?*

The question asks us to find **how fast the volume is increasing** when the diameter is* 80 mm*. Asking about how fast something is changing refers to its rate of change. Therefore, we can tell that this question is asking us about the **rate of change of the volume**.

Since we will later be taking the derivative of the equation we are currently building, **we only need to make sure to include the volume**. Once we take the derivative of the equation, this will introduce the rate of change of the volume.

*What do we know about? *

This question didn’t provide a lot of information and it’s fairly straight forward.

**Rate of change of the radius**.- The
**diameter**of the figure at this moment.

*Putting it into an equation*

Up to this point we know that we need to include the sphere’s volume in this equation. We have also figured out that we know some information about the diameter, which can easily be used to find the radius. And we also know about the rate of change of the radius. So we basically know everything about the radius that we might need.

What is the first equation or formula you think of that **relates the volume of a sphere to its radius?**

I think a good place to start is with the formula for the volume of a sphere.

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

We can see that this equation only contains *V* (volume) and *r* (radius). As a result, I’d say this is as good of a place to start as any. Let’s proceed with this equation.

## 3. Implicit differentiation

Now that we have come up with our equation, we need to **take its derivative with respect to time**. This will allow us to introduce and work with the rates of change of our measurements.

Since we will be taking the derivative with respect to time, we will need to treat *V* and *r* as functions of time rather than variables. In order to do this we will need to use the chain rule. So, taking the derivative of our equation gives us:

$$\frac{d}{dt} \big[ V \big] = \frac{d}{dt} \bigg[ \frac{4}{3} \pi r^3 \bigg]$$

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

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

## 4. Solve for the desired rate of change

Finally, all we need to do now is solve for the rate of change the question is asking us to find. The problem asked us to find how fast the volume is changing at this moment. This is exactly what \(\frac{dV}{dt}\) represents. Since this is already isolated, all we need to do is plug in the other values we know about.

The other values that are needed are *r* and \(\frac{dr}{dt}\), which represent the **radius **and **its rate of change** respectively.

The radius of our figure wasn’t given directly, but we do know that its diameter is *80 mm* at this instant. Since the radius of a sphere is always half of the diameter, this tells us that the radius is *40 mm*, or

$$r=40.$$

We were given that the figure’s radius is increasing at a rate of *4* \(\frac{mm}{s}\). Therefore, we know

$$\frac{dr}{dt} = 4.$$

#### Plugging it all in

Now we simply need to plug these values into the differentiated equation we found in step three.

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

$$\frac{dV}{dt} = 4 \pi (40)^2 (4)$$

$$\frac{dV}{dt} = 25,600 \pi$$

$$\frac{dV}{dt} \approx 80,424.772$$

So this tells us that the **volume of the sphere is increasing at a rate of 25,600**\(\mathbf{\pi \ \frac{mm^3}{s}}\), or about

*\(\mathbf{\frac{mm^3}{s}}\) when its diameter is*

**80,424.772***80 mm*.

If youâ€™re still having some trouble with related rates problems or just want some more practice you should check out my related rates lesson. At the bottom of this lesson there is a list of related rates practice problems that I have posted a solution of. I also have several other lessons and problems on the derivatives page you can check out. If you canâ€™t find the topic or question youâ€™re looking for just let me know by emailing me at **jakesmathlessons@gmail.com**!

I also encourage you to join my email list! **Just enter your name and email below and I’ll send you a copy of my calculus 1 study guide to help you get through your homework faster and the exams with better scores!**