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.
I am not sure if I have the right answer I got -0.141 cm^3/min I think I did something wrong
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
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 ?
Explain how you got this answer so i can help :)
I and also Steve have answered a million of these. For example:
http://www.jiskha.com/display.cgi?id=1415757944
Thank you!!! :)
To find the rate at which the depth of the water is changing, we can use related rates. Let's denote the radius of the water at a particular depth as r and the depth as h.
Given:
The radius of the tank is 6 cm.
The height of the tank is 12 cm.
The water is drained at a rate of 3 cm³/min.
We need to find dh/dt, the rate at which the depth of the water is changing at the instant when the water is 9 cm deep.
To solve this problem, we'll use similar triangles. We'll consider a small change in time Δt. During this time, the water level decreases by Δh, and the corresponding decrease in radius is Δr.
Since the tank is shaped like an inverted right circular cone, the ratio of the change in radius to the change in height remains constant. So we have:
Δr/Δh = r/h
We can rewrite this as:
(Δr/Δt) / (Δh/Δt) = r/h
Now, we can take the derivative with respect to time (dt) on both sides:
(dr/dt) / (dh/dt) = (dr/dt) / (dh/dt)
At the instant when the water is 9 cm deep, the radius can be calculated using similar triangles:
r/h = 6/12
r = (1/2)h
Now, let's differentiate this equation with respect to time:
dr/dt = (1/2)(dh/dt)
We are given that dr/dt = -3 cm³/min because the water is being drained. To find dh/dt, we substitute the known values into our derivative equation:
-3 = (1/2)(dh/dt)
Now we can solve for dh/dt:
dh/dt = -6 cm³/min
Therefore, the depth of the water is changing at a rate of -6 cm³/min when the water in the tank is 9 cm deep.
Note: The negative sign indicates that the depth is decreasing.
The answer should be -0.140cm^4/min
depending on what way their teaching you.
No. The answer is in cm/min not cm^4/min. I did not check the arithmetic.
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!!!