Rates of change: surface area and volume of a sphere

[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

 

[PeroK]

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]

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?

 

[PeroK]

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?