The rate of change of the volume is ____ cubic meters per hour.

Question

The radius of a sphere is increasing at a rate of 5 meters per hour. When the radius is 12 meters, then how fast is the volume changing?

Note: Volume of a sphere = (4/3)⋅π⋅(radius)3

The rate of change of the volume is ____ cubic meters per hour.

(Enter your answer as a decimal number rounded to 2 places.)

Answer 1

Hello,

For your question, let's define some things real quick:

Let V = (4pi/3)*r^3 where V is the equation for a sphere with radius 'r'

Our goal is to find: dV/dt (which is the rate of change of the sphere in respect to time, because we are trying to figure out how much it is changing as time progresses)

Moving forward...

We take the 1st derivative in respect to 't - time' and plug in what we are given (r = 12m and dr/dt = 5m)

note: Keep in mind that in this case that r = r(t) which means the radius is a function of 't - time' since it's the only variable that's changing.

dV/dt = (4pi/3)*d/dt(r^3) (I placed the constants in the front since they have no weight to the problem)

Now...

dV/dt = (4pi/3)*(3*r^2)*(dr/dt) where (3*r^2) is just a power rule and respecting the chain rule and implicit differentiation, we also get (dr/dt) out.

Now it's just a matter of plugging things in from our premise.

I will leave that part to you.

Hope this helps!