# Rates of change: surface area and volume of a sphere

• AN630078
In summary: Is it true there?Yes, for a cube, the surface area is the derivative of the volume as well. The formula for the volume of a cube is V = s^3, where s is the length of one side. Taking the derivative with respect to s gives dV/ds = 3s^2, which is equal to the surface area of the cube. So, for a cube, the rate of change of volume with respect to its side length is equal to its surface area.
AN630078
Homework Statement
Hello, I have been practising some related rates problems and have found the following problem which I am a little wary of. I feel a little uncertain of my solution and would greatly appreciate any advice.

The radius of a sphere is increasing at a constant rate of 3cms^-1. Given that the radius of the sphere is 5cm find in terms of π the rates at which its surface area and volume are increasing.
Relevant Equations
Volume of sphere= 4/3πr^3
Surface area of sphere = 4πr^2
The surface area of the sphere is 4πr^2.

dr/dt is given as 3cm^-1.
dS/dt=dS/dr*dr/dt
Differentiating 4πr^2 is dS/dr= 8πr
dS/dt=8πr*3
dS/dt=24πr

Given that r=5 dS/dt=24π*5=120 π

The volume of the sphere is 4/3πr^3, differentiating which is dV/dr=4πr^2
dV/dt=dV/dr*dr/dt
dV/dt= 4πr^2*3
dV/dt=12πr^2

Given that r=5, dV/dt=12π*(5^2)=300π

I honestly do not know whether this is correct and have only been introduced to solving similar problems through reading their mark schemes which is why I am very unsure of my method. I found a similar problem, here at https://www.toppr.com/ask/question/the-volume-of-a-sphere-is-increasing-at-the-rate-of-3-cubic-centimetre-per/

but following this approach of finding the rate at which the volume is increasing first and using this to find dr/dt I got a different answer and have now confused myself.

Just to check, would dr/dt=3cm^-1? I think this value is what is confusing me most of all. I would very much appreciate if anyone could tidy up my workings or suggest an alternative method with greater clarity

AN630078 said:
Given that r=5 dS/dt=24π*5=120 π

Given that r=5, dV/dt=12π*(5^2)=300π
You work is okay, but you need units.

As as aside, did you notice that: $$\frac{dV}{dr} = 4\pi r^2 = S$$

AN630078
PeroK said:
You work is okay, but you need units.

As as aside, did you notice that: $$\frac{dV}{dr} = 4\pi r^2 = S$$
Thank you very much for your reply. Would the correct units be dS/dt=120 π cm^2 s^-1 and dV/dt=300π cm^3 s^-1? Yes, indeed I did notice that the surface area is the derivative of the volume. Is this always true?

AN630078 said:
Thank you very much for your reply. Would the correct units be dS/dt=120 π cm^2 s^-1 and dV/dt=300π cm^3 s^-1? Yes, indeed I did notice that the surface area is the derivative of the volume. Is this always true?
Yes, those are the correct units.

It's always true for a sphere, as ##\frac{dV(r)}{dr} = S(r)##. What about a cube?

hutchphd

## 1. What is the formula for finding the surface area of a sphere?

The formula for finding the surface area of a sphere is 4πr2, where r is the radius of the sphere.

## 2. How do you calculate the volume of a sphere?

The formula for calculating the volume of a sphere is (4/3)πr3, where r is the radius of the sphere.

## 3. Can you explain the concept of rate of change in relation to the surface area and volume of a sphere?

The rate of change in relation to the surface area and volume of a sphere refers to how the surface area and volume of a sphere change as the radius of the sphere changes. As the radius increases, the surface area and volume also increase, and vice versa.

## 4. How can I use rates of change to solve problems involving spheres?

To solve problems involving spheres, you can use the formulas for surface area and volume, along with the concept of rate of change. This can help you calculate how much the surface area and volume change when the radius changes by a certain amount, or vice versa.

## 5. Are there any real-life applications of rates of change for spheres?

Yes, rates of change for spheres have many real-life applications, such as in engineering and architecture, where calculations for surface area and volume are necessary for designing structures like domes and spherical tanks. They are also used in physics and astronomy to calculate the surface area and volume of planets and other celestial bodies.

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