Related Rates Problem: Solving for Volume Increase with Changing Sphere Radius

[RJLiberator]

Hey guys,

I want to make sure I am on the right track with this problem:

The radius of a sphere is increasing at a rate of 4 cm/s. How fast is the volume increasing when the radius is 40 cm? (Recall the formula relating the area A and radius r of a sphere: A = 4πr^2.)

So, I use the equation A=4πr^2
I take the derivate with respect to time.

dA/dt = 4π*2r*dr/dt
Simplifying : dA/dt = 8π*r*dr/dt

Inputing radius of 40cm for variable r and inputting rate of 4cm/s for variable "dr/dt" The answer becomes

dA/dt = 1280π cm^2/sec

The answer seems to make sense (units). This just seems too... easy for me. In class we were doing a bit more difficult problems.

Does everything check out?

Thanks.

 

[LCKurtz]

Hey guys,

I want to make sure I am on the right track with this problem:

The radius of a sphere is increasing at a rate of 4 cm/s. How fast is the volume increasing when the radius is 40 cm? (Recall the formula relating the area A and radius r of a sphere: A = 4πr^2.)

So, I use the equation A=4πr^2
I take the derivate with respect to time.

dA/dt = 4π*2r*dr/dt
Simplifying : dA/dt = 8π*r*dr/dt

Inputing radius of 40cm for variable r and inputting rate of 4cm/s for variable "dr/dt" The answer becomes

dA/dt = 1280π cm^2/sec

The answer seems to make sense (units). This just seems too... easy for me. In class we were doing a bit more difficult problems.

Does everything check out?

Thanks.

Everything except you either stated or worked the wrong problem. You asked how fast the volume is changing, not the surface area. But your work is correct for the area and you would work the volume problem similarly.

 

[RJLiberator]

Oh, I see! That would be incredibly important.

I am not sure if the instructor meant to do this, or misused an equation. Either way, I will discuss it with him.

Anyway, V=4/3π*r^3

dV/dt = 4/3π*3r^2*dr/dt

plugging in 40 for r and 4 for dr/dt

The answer becomes 25600π cm^2/sec

Quite a large answer.

Thank you for verifying my previous work/efforts.

 

[LCKurtz]

Oh, I see! That would be incredibly important.

I am not sure if the instructor meant to do this, or misused an equation. Either way, I will discuss it with him.

Anyway, V=4/3π*r^3

dV/dt = 4/3π*3r^2*dr/dt

plugging in 40 for r and 4 for dr/dt

The answer becomes 25600π cm^2/sec

Quite a large answer.

Thank you for verifying my previous work/efforts.

Are you sure about those units?

 

[RJLiberator]

Well,

The radius is increasing at 4cm/s
The radius is 40 cm
Aha.

The radius is squared so it becomes cm^3.

Volume should be in cubed units.

Ah. I love it, such a tiny error could throw off the problem. One of the beauties of mathematics :D.

Thank you... once again. =)

 

[Orodruin]

dV/dt = 4/3π*3r^2*dr/dt

You can still "use" the area to find the more beautiful relation
$$
\frac{dV}{dt} = \frac{4\pi}{3} 3r^2 \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt} = A \frac{dr}{dt}
$$

Depending on what was being considered in the particular part of the book, this may even have been the intended way of solving the problem (in particular if the problem is in a part treating rotational volumes and computing the volume of a sphere by summing infinitesimal shells, i.e., integrating):
The radius increases by dr in the infinitesimal time dt. Thus, a spherical shell with radius r and thickness dr is added to the volume in time dt. The volume of said shell is A dr, thus resulting in dV/dt = A dr/dt.