QUESTION: A sphere with radius $r$ has volume $\dfrac{4\pi}{3}r^3$ and surface area $4\pi r^2$.
A spherical snowball is melting in such a way that its volume is decreasing at a rate equal to twice the surface area. How quickly is the radius of the snowball decreasing?
So I did:
$V' = 2SA'$
or $4\pi r^2 = 16\pi r$
or $r = 4$.
This turned out to be wrong. What did I do wrong?
Best Answer
Hint
We have that $$V'(t)=-2A(t).$$ In other words
$$4\pi r^2(t)r'(t)=-8\pi r^2(t).$$
Now, it should be easy to get $r'(t).$