A conical water tank with vertex down has a radius of 10 feet at the top and is 22 feet high. If water flows into the tank at a rate of 30 {\rm ft}^3{\rm /min}, how fast is the depth of the water increasing when the water is 14 feet deep?
I have tried several times but keep getting .38197. Please help!
To find the rate at which the depth of the water is increasing, we need to use related rates. We can start by writing the formula for the volume of a cone:
V = (1/3)πr^2h,
where V is the volume, r is the radius of the base of the cone, h is the height of the cone, and π is a constant.
In this problem, the radius of the cone is changing with time, so we need to use the chain rule to find the rate of change of the volume with respect to time. The chain rule states that:
dV/dt = (∂V/∂r) * (dr/dt) + (∂V/∂h) * (dh/dt),
where dV/dt is the rate of change of the volume, (∂V/∂r) and (∂V/∂h) are the partial derivatives of the volume with respect to the radius and height, respectively, and dr/dt and dh/dt are the rates of change of the radius and height, respectively.
Since the tank is being filled with water, the rate of change of the volume with respect to time is given as 30 ft^3/min.
To find the rate at which the depth of the water is increasing, we need to find dh/dt when h = 14 ft.
First, let's find (∂V/∂r) and (∂V/∂h):
(∂V/∂r) = 2πrh,
(∂V/∂h) = (1/3)πr^2.
Now, we need to find dr/dt. Since the radius of the tank is given as 10 ft at the top, we can assume that the radius is changing linearly with respect to the height, h. This means that:
r = (10/22)h.
Now, we can find dr/dt by differentiating this equation with respect to time:
dr/dt = (10/22) * (dh/dt).
Finally, substituting all the values into the chain rule equation:
30 = (2πrh) * (10/22) * (dh/dt) + ((1/3)πr^2) * (dh/dt).
We now have one equation with one unknown, dh/dt. Let's solve for it:
30 = (2πrh) * (10/22) * (dh/dt) + ((1/3)πr^2) * (dh/dt),
30 = (2π * (10/22) * (h * (10/22))) * (dh/dt) + ((1/3)π * (h/2)^2) * (dh/dt),
30 = ((10π/11) * h^2) * (dh/dt) + ((1/12)π * h^2) * (dh/dt),
30 = (10π/11 + π/12) * h^2 * (dh/dt),
30 = (130π/132) * h^2 * (dh/dt),
1 = (h^2 * (dh/dt)) / ((132/130) * π),
1 = (h^2 * (dh/dt)) / (1.0094 * π),
1 = (h^2 * (dh/dt)) / 3.168,
dh/dt = 3.168.
So, when the water is 14 feet deep, the depth of the water is increasing at a rate of approximately 3.168 ft/min.
at depth y, the radius of the water surface is r = (10/22)x = 5/12 x
So, at depth y, the volume of water is
v = 1/3 π (5/12 y)^2 = 25/432 π y^2
dv/dt = 25/216 π y dy/dt
30 = 25/216 π (14) dy/dt
30 = 5.08 dy/dt
dy/dt = 5.89 ft/min
As always, check my math.