sphere Archives | Jake's Math Lessons

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?

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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 80,424.772 $\mathbf{\frac{mm^3}{s}}$ when its diameter is 80 mm.

If you’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 can’t find the topic or question you’re looking for just let me know by emailing me at jakesmathlessons@gmail.com!

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If a snowball melts so that its surface area decreases at a rate of 1 $\frac{cm^2}{min}$ , find the rate at which the diameter decreases when the diameter is 10 cm.

By looking at the given statement, we can gather a few important fact quickly.

The object in question is a snowball. This means that we will be dealing with a sphere here, so we will likely be needing some formula relating different dimensions and measurements of a sphere.
The problem gives information about the rate of change of a specific measurement of the snowball. When a problem gives information like this, it’s a strong hint that we have a related rates problem.

So we know that we’re dealing with a related rates problem. Therefore, we are going to follow the four steps that these will all follow. If you want to look back at these steps, I discussed them in my related rates lesson.

1. Draw a sketch

Looking back at the problem, you can see that there isn’t a lot of information that has been described to us. This will end up being a very simple sketch, but that’s all it takes sometimes.

So we have a sphere whose surface area is decreasing at a rate of 1 $\frac{cm^2}{min}$ and we are looking at the instant when its diameter is 10 cm.

2. Come up with your equation

Now that we have our situation drawn out, we need our equation. To create this equation, we need to incorporate any relevant information that we were given and we need to consider what the question asks us to find.

What are we looking for?

The question is asking us to find “the rate at which the diameter decreases” at the instant when the diameter is 10 cm. This is the instant that we captured in our drawing. So the important thing here is that we are looking for the rate the diameter is decreasing. Another way to put this is the rate of change of the diameter.

Since we will eventually need to take the derivative, which will provide the rate of change part, we just need to make sure that our equation contains the diameter.

What do we know about?

This question didn’t provide a lot of information to us. We really only know two things:

The rate of change of the sphere’s surface area.
The diameter of the sphere at this instant.

Putting it into an equation

Up to this point we have figured out that we need to include the sphere’s diameter in our equation. We also know that we have some information relating to the snowball’s surface area. So we need to come up with an equation that relates a sphere’s surface area and diameter.

A good place to start may be the equation for the surface area of a sphere.

$$A=4 \pi r^2$$

In this equation A represents the surface area of the sphere and r is the radius.

But we don’t know about the radius

Since we need an equation relating the surface area and the diameter, we will need to make an adjustment. The only thing we need to consider here is that a sphere’s diameter is always double the radius. Or in other words

$$d=2r.$$

But our equation contains the radius. So we’ll want to solve for r so we can plug that into our equation. Dividing both sides by 2 will accomplish this.

$$r=\frac{d}{2}$$

Now we can plug this in for r in our equation for the surface area of a sphere.

$$A=4 \pi \bigg( \frac{d}{2} \bigg)^2$$

So we know have an equation that contains only the surface area and the diameter of our sphere, exactly what we needed. To make things a little easier later, we will simplify this equation a bit.

$$A=4 \pi \bigg( \frac{d^2}{4} \bigg)$$

$$A=\pi d^2$$

3. Implicit differentiation

Now that we have created our equation we need to take its derivative. This will bring in the rates of change we discussed earlier. The most important one being the rate of change of the snowball’s diameter which is what we need to find.

Keep in mind that we will be taking the derivative with respect to time. This means we will need to treat A and d as functions of time, not variables. Therefore, we will need to apply the chain rule here.

$$\frac{d}{dt} \big[ A \big]= \frac{d}{dt} \big[ \pi d^2 \big]$$

$$\frac{dA}{dt} = 2 \pi d \cdot \frac{dd}{dt}$$

4. Solve for the desired rate of change

The fourth and final step of a problem like this is to isolate the rate of change we need and find its value. The question wants us to find the rate at which the diameter is decreasing when the diameter is 10 cm. This tells us we need to solve for the rate of change of the diameter, which is represented by $\frac{dd}{dt}$ .

$$\frac{dA}{dt} = 2 \pi d \cdot \frac{dd}{dt}$$

$$\frac{1}{2 \pi d} \frac{dA}{dt} = \frac{dd}{dt}$$

So now that we have isolated the variable that we need to find we can simply plug in all of the values we have for the other variables.

What values do we know?

There are only two variables we need to plug in a value for: d and $\frac{dA}{dt}$ . The question specifically asked us to find $\frac{dd}{dt}$ when the diameter of the snowball is 10 cm. Therefore, we know that we will have

$$d=10.$$

The question also told us that the surface area of the snowball is decreasing at a rate of 1 $\mathbf{\frac{cm^2}{min}}$ . Since the surface area is decreasing, this tells us that

$$\frac{dA}{dt}=-1.$$

Plugging it all in

Now we just need to go back to our equation and plug in all of these other values.

$$\frac{dd}{dt} = \frac{1}{2 \pi d} \frac{dA}{dt}$$

$$\frac{dd}{dt} = \frac{1}{2 \pi (10)} \cdot (-1)$$

$$\frac{dd}{dt} = \frac{-1}{20 \pi}$$

So this tells us that the diameter of the snowball is changing at a rate of $\frac{-1}{20 \pi} \ \frac{cm}{min}$ . Therefore, we can say that the diameter of the snowball is decreasing at a rate of $\mathbf{\frac{1}{20 \pi} \ \frac{cm}{min}}$ or about 0.0159 $\mathbf{\frac{cm}{min}}$ .

Note that the question asked for the rate at which the diameter is decreasing. As a result of this, our answer will actually be a positive number despite the fact that $\mathbf{\frac{dd}{dt}}$ was negative.

If you’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 these types of 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 can’t find the topic or question you’re looking for just let me know by emailing me at jakesmathlessons@gmail.com!

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