A conical water tank with vertex down has a radius of 13 feet at the top and is 28 feet high. If water flows into the tank at a rate of 10 {\rm ft}^3{\rm /min}, how fast is the depth of the water increasing when the water is 17 feet deep?

Answer 1

make a sketch.
let the radius of the water level be r ft
let the height of the water level by h ft

by ratios:
r/h = 13/28
r = 13h/28

V = (1/3)π r^2 h
= (1/3)π (13h/28)^2 h
= 169/84 π h^3

dV/dt = 169/28 π h^2 dh/dt
for the given data ...
10 = 169/28 π (289) dh/dt
dh/dt = 280/(169π*289)
= ... ft/min

check my arithmetic, and finish the calculation

Answer 2

A conical water tank with vertex down has a radius of 13 feet at the top and is 28 feet high. If water flows into the tank at a rate of 10 {\rm ft}^3{\rm /min}, how fast is the depth of the water increasing when the water is 17 feet deep? I need the right answer that shows all the work so i understand

Answer 3

I did all the work, except the final calculation
Surely you can do that.
Don't you have a calculator?
Mine is not handy

Answer 4

I put it in my calculator, but it is the wrong answer though.

Answer 5

just noticed that I did not square the 28 in
= 169/84 π h^3

should have been
= 169/2352 π h^3
...

dVdt = 169/984 π h^2 dh/dt

try it now

Answer 6

I put it in my calculator and I keep getting 0.0641299171, so when I put that in my homework (that is online), it says that it is wrong.

Answer 7

I don't know what is wrong, but thank you for helping me out.