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?

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:

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.

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.

50 inches should be the resultant vector