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?
1 Answer
Using
#(dr)/(dt) = 4 "mm"/"s"#
or
#r = r(t) = 4t#
The formula for a solid sphere's volume is:
#V = V(r) = 4/3pir^3#
When you take the derivative of both sides with respect to time...
#(dV)/(dt) = 4/3pi(3r^2)((dr)/(dt))#
...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
Now what you can do is simply plug in what
#(dV)/(dt) = 4/3pi(3(20 "mm")^2)(4 "mm"/"s")#
#= 6400pi "mm"^3/"s"#
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.