Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 6cm and a height of 12cm, at the rate of 3 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 9 cm deep? Give an exact answer showing all work and include units in your answer.

Question

I am not sure if I have the right answer I got -0.141 cm^3/min I think I did something wrong

Answer 1

Explain how you got this answer so i can help :)

Answer 2

I and also Steve have answered a million of these. For example:

http://www.jiskha.com/display.cgi?id=1415757944

Answer 3

The answer should be -0.140cm^4/min
depending on what way their teaching you.

Answer 4

No. The answer is in cm/min not cm^4/min. I did not check the arithmetic.

Answer 5

Oh well

h = 9 cm
r = (6/12)9 = 4.5 cm

dV = pi r^2 dh
so
dV/dt = pi r^2 dh/dt

-3 cm^3/min = pi (4.5)^2 dh/dt
so
dh/dt = -3/[ pi(4.5)^2]
about .047 cm/min

Answer 6

Let h = height of water
Let r = radius of surface of water
By similar triangles, r/h = 6/12 = 0.5, so r = 0.5h
The volume of the water is:
v = (1/3) * pi * r^2 * h
v = (1/3) * pi * (0.5h)^2 * h
v = (1/12) * pi * h^3
dv/dh = 0.25*pi*h^2
The question tells us that dv/dt = -3
By the Chain Rule:
dv/dt = dv/dh * dh/dt
-3 = 0.25*pi*h^2 * dh/dt
dh/dt = -3 / (0.25*pi*h^2)
dh/dt = -12 / (pi*h^2)
When h = 9 we have:
dh/dt = -12 / (pi*9^2)
dh/dt = -12 / (81*pi)
dh/dt = -4 / (9*pi) cm^3/min
dh/dt =~ -0.14147106052612918735011890077557 cm^3/min
this is how I did it thank you!!!

Answer 7

I agree with Damon's answer

I followed your steps and found and error around the third last step

ok at :
dh/dt = -12 / (81*pi) , but then ....
dh/dt = -4/(27π) ----> you had -4/(9π)
= -.04716

you can also simplify your steps:
from v = (1/12) * pi * h^3
why not just differentiate with respect to t to get
dv/dt = (1/4)π h^2 dh/dt ?

Answer 8

Thank you!!! :)