A calculus question involving finding a derivative relating to rate of change in a certain volume. Please help

Question

Here is the question: "Sand is falling onto a conical pile at a rate of 10 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 15 feet high?"

Now of course, I know the answer from the back of the book, but it does me no good if I can't do it myself. The volume of a cone is as follows, since you'll probably need it for something: V=(pi/3)(r^2)(h)
r is the radius of the top of the cone, and h is the height (altitude). pi is pi (3.14...) but I don't know how to write it on the computer.

I know it also has something to do with writing the derivative like this (dh/dt) where this is h'. Remember, the time it takes to go into the pit or whatever is 10 ft^3/m. I am assuming this is the variable 't' for time.

So first person to get the correct answer and with due explanation through their method gets the best answer. Thanks in advance!

Answer 1

You know that dV/dt = 10 ft^3/min. Also, the diameter, 2r, satisfies 2r = 3h. You need to find dh/dt, when h=15.

Start with V = (pi/3)r^2*h, and substitute r = 3h/2 to get

V = (pi/3)(3h/2)^2*h = (3pi/4)(h^3).

Differentiate the equation with respect to t, and get

dV/dt = (3pi/4)(3h^2)dh/dt.

Now plug in h=15 and dV/dt=10, and solve for dh/dt. You have

10 = (3pi/4)675*dh/dt

so dh/dt = 40/(2025*pi) = 0.00628760269005 ft/min

Answer 2

properly... merely think of of it like this: they're asking you for the cost of replace of h, so which you're unlikely to might desire to locate h. you're searching for dV/dh. So, on account which you be attentive to that the DIAMETER is comparable to 3r, then you definately be attentive to that the RADIUS is 3r/2, and that's somewhat information might desire to get dV/dh (technically dV/dh evaluated whilst h=15).