I have a problem. Volume of the sphere is given as: $$V(r) = \frac{4}{3}\pi r^3$$
I am then told (part A of the question) to find the average rate of change of V with respect to r, when r changes from 5 ft to 8 ft. (exact quote)
I know we're going to form up the equation: $$\frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt}$$
After deriving $V(r)$, we get: $$\frac{dV}{dr} = 4\pi r^2$$ Is this correct so far?
Now, I don't have a RATE of change with respect to the radius. I have a change from 5 feet to 8 feet, but it does not indicate a duration of time over which this occurs. Is this missing or am I misunderstanding what I'm needing to require.
The answer to this question states that part A is $387 ft^3$
The problem is that I have no idea how they got to the answer.
I THINK that I'm supposed to use a time range of 1 minute for part A, meaning $\frac{dr}{dt} = 3 ft/min$. However, $$\frac{dV}{dr}\frac{dr}{dt} = 4\pi (5^2)(3 ft/min) = 942.4778$$ which is DEFINITELY not the answer.
Where am I going wrong?
You're thinking far too hard. A derivative finds instantaneous rate-of-change - which is explicitly not what the question was asking for. You have correctly computed the instantaneous rate of change with respect to time at the moment that $r = 5$; this is more or less irrelevant to the average rate of change.
Average rate of change is much simpler to compute, and takes no calculus whatsoever. For example, if I travel $500$ miles in $8$ hours, the average rate of change of my position is just $500/8 = 62.5 mph$, even though at some points in time my instantaneous rate of change may be much higher.
Secondly, you computed $\frac{dV}{dt}$; the question is asking with respect to r, which means that even if it were asking for instantaneous rate of change you'd want $\frac{dV}{dr}$ instead.
So what you want is just $\frac{\Delta V}{\Delta r}$ - which is just $\frac{V(8) - V(5)}{8 - 5}$.