The Volume V=(4/3)(pi)r^3 of a spherical balloon changes with the radius. At what rate does the volume change with respect to the radius when r=2ft? I know I have to take the derivative of V to get the answer, but why is the derivative of V 4(pi)r^2? I thought that the derivative of a constant was 0, so shouldn't it be 0(0)r^2, which is 0?

Question

anonymous · Answer

Does that make sense at all?

anonymous · Answer

$$\frac{ dv }{ dr }=3*\frac{ 4 }{ 3 } \pi  r^2=4 \pi r^2$$
when r=2
$$\frac{ dv }{ dr }=4 \pi *2^2=16 \pi$$
here neither v nor r is constant

anonymous · Answer

But 4/3 and pi are both constants, and the derivative of both of those are 0?

anonymous · Answer

Oh wait, is that only when they are being added and not multiplied?

phi · Answer

example:   d/dx (3x)= 3 d/dx (x)  = 3
the constant is "factored out"

anonymous · Answer

Oh, so the (4/3)pi is treated as the coefficient?

anonymous · Answer

So its the same as if it were (4pi)/3

anonymous · Answer

$$\frac{ d }{ dx }\left(  a x^n \right)=a~n x ^{n-1},where ~a~ is ~a ~constant.$$

anonymous · Answer

And because you multiply the coefficient by the power, the 3 becomes 1. Ooooooh, that makes more sense now

phi · Answer

yes.

If we went back to the definition of "derivative",  (which is a pain) you can see that constant coefficients are factored out.

anonymous · Answer

Oh, ok! Thanks a lot guys!!!

phi · Answer

btw, Paul's Online Notes lists almost all the variations on "related rates" problems http://tutorial.math.lamar.edu/Classes/CalcI/RelatedRates.aspx