A sphere with radius 10 cm is being filled at a rate of 1 cm3 per minute. What rate is the depth of the water increasing when it is 5 cm?
So what I did was:
$$r=10$$
$$\frac{dV}{dt} = 1$$
$$d = 5$$
$$V= \frac{4}{3}\pi r^2$$
$$\frac{dV}{dt} = \frac{8}{3}\pi r \frac{dr}{dt}$$
However, what I'm really looking for isn't $\frac{dr}{dt}$ but instead $\frac{dD}{dt}$, and I can't think of any way to relate volume to depth or radius. How can I do this?
Best Answer
It seems that you're using $V$ for two different things -- both the volume of the entire sphere, and the volume of water yet in the sphere.
What you need to do is choose a variable for for the volume of water in the sphere, and then use the formula for a spherical cap to relate it to the depth of water. Only after you have this relation will it make sense to start doing calculus on the situation.