Ok. I have a problem, I thought I was doing this problem correctly but my answers aren't making since logically.
A tank of water in the shape of a cone is leaking water at a constant rate of 2 ft^3/hour. The base of the radius of the tank is 5 ft. and the height of the tank is 14 ft.
a) At what rate is the depth of the water in the tank changing when the depth of the water is 6 ft?
b) At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 ft?
Now I set up the first problem using the Volume formula V=pi(r^2)(h)/3. And then multiplied each side by d/dt. Since the problem told us dV/dt=2 ft^3/hr and the radius is 5 I could solve for dh/dt and I got .0764 ft/hr for a). I then took the original height of the cone 14 ft minus the new depth of the water 6 ft which gives me a change of 8 ft and divided that by .0764 to get the number of hours that it took for the depth to change that much. Which gave me 104.7120 hrs I multiplied that times 2 ft^3/hr to find the amount of volume that had drained out in this time and got 209.4241 ft^3. I subtracted this from the original volume 366.5191 and got 157.0950 which is the volume when the height is 6. I then plugged that volume in to find the radius of what is left of the water and the answer was slightly larger than the original radius which shouldn't happen since it gets smaller as you get closer to the tip of the cone. Where did I go wrong?
Thanks. Any help would be greatly appreciated.