Question

The radius of a sphere is increasing at a rate of 4 mm/s. how fast is the volume increasing when the diameter is 40 mm?

Answer 1

Using r to represent the radius and t for time, you can write the first rate as:

drdt=4mms

or

r=r(t)=4t

The formula for a solid sphere's volume is:

V=V(r)=43πr3

When you take the derivative of both sides with respect to time...

dVdt=43π(3r2)(drdt)

...remember the Chain Rule for implicit differentiation. The general format for this is:

dV(r)dt=dV(r)dr(t)⋅dr(t)dt with V=V(r) and r=r(t) .

So, when you take the derivative of the volume, it is with respect to its variable r (dV(r)dr(t)) , but we want to do it with respect to t (dV(r)dt) . Since r=r(t) and r(t) is implicitly a function of t , to make the equality work, you have to multiply by the derivative of the function r(t) with respect to t (dr(t)dt) as well. That way, you're taking a derivative along a chain of functions, so to speak (V→r→t ).

Now what you can do is simply plug in what r is (note you were given diameter) and what drdt is, because dVdt describes the rate of change of the volume over time, of a sphere.

dVdt=43π(3(20mm)2)(4mms) =6400πmm3s

Since time just increases, and the radius increases as a function of time, and the volume increases as a function of a constant times the radius cubed, the volume increases faster than the radius increases, so we can't just say the two rates are the same.