I am going through James Steward Calculus 8th edition and I'm completely stuck on this problem. I've looked everywhere and can't find any help.
The question is: Water is leaking out of an inverted conical tank at a rate of 10,000cm^3/min at the same time water is being pumped into the tank at a constant rate. The tank has height 6m and the diameter at the top is 4m. If the water level is rising at a rate of 20cm/min when the height of the water is 2m, find the rate at which water is being pumped into the tank.
The back of the book has the solution as 2.89 x 10^5 which is far from what I got. My incorrect solution:
First I found the angle at the bottom of the triangle made by the height and radius, which i will call A. tan(A) = 2/6 = 1/3 This means r = (1/3)h
I know a cone has volume = pi(r^2)(h/3) And I can rewrite it with my new r so that V = pi (h/3)^2 (h/3) = (pi h^3) / 27
Then I took the derivative with respect to t. I will write dv/dt as v' and dh/dt as h'. So,
v' = (pi (3h^2) h') / 27 = (3pi h^2 h') / 27 = (pi h^2 h') / 9 Then I plugged the 10cm for h, and 20cm/s for h'
v' = (pi (10)^2 (20)) / 9 v' = 2000pi / 9 which is about 222.22... pi And finialy, total change(v') = incoming rate - outgoing rate (10,000)
therefore, incoming rate = 222.22pi + 10,000 = 10,222.22 pi
This is very far from what the answer should be, but I have no idea why! If you could let me know where I went wrong It would be greatly appreciated, Thanks!
