Water is leaking out of an inverted conical tank at a rate of 10,000 cm^3/min at the same time that 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.

Question

Water is leaking out of an inverted conical tank at a rate of 10,000 cm^3/min at the same time that 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.

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First, I know we'll be using the formula for volume of a cone, V=(1/3)pi*r^2*h. I know dV/dt = -10000, h=600cm, d=400cm (or 200cm for radius), and dh/dt=20cm/min.
I also know when h=200cm, r=150cm (I did a proportion with the given heights and diameter/radius).
I believe the question is asking for another dV/dt, but I don't know how to go about solving for that with all this given information. Help please?

Thank you!

Answer 1

V=1/2 PI r^2 h

dv/dt= 1/3 PI (r^2 dh/dt + 2hr dr/dt)

now relate dr/dt to dh/dt (proportion)
then solve for dv/dt given dh/dt

Answer 2

To find the rate at which water is being pumped into the tank, we need to find the rate of change of the volume of the water in the tank with respect to time, which is denoted as dV/dt.

Given information:
dV/dt = -10000 cm^3/min (water is leaking out of the tank at this rate)
h = 600 cm (height of the tank)
d = 400 cm (diameter at the top of the tank, or 2 times the radius)
dh/dt = 20 cm/min (rate at which the height of the water level is rising)

We want to find dV/dt when h = 200 cm.

First, let's find the radius of the water level when h = 200 cm. We can use a proportion:

h1/d1 = h2/d2

Solving for d2 (radius when h = 200 cm):

200/600 = d2/400
d2 = (200/600) * 400
d2 = 133.33 cm

Now that we have the radius when h = 200 cm, we can use the volume formula for a cone to relate the variables:

V = (1/3) * π * r^2 * h

Differentiating both sides of the equation with respect to time (t), we have:

dV/dt = (1/3) * π * [(2 * r * dr/dt * h) + (r^2 * dh/dt)]

We have values for r, dh/dt, and h:

r = 133.33 cm
dh/dt = 20 cm/min (given)
h = 200 cm (given)

The only unknown is dr/dt, the rate of change of radius with respect to time. We can solve for it using the information that dh/dt = 20 cm/min when h = 200 cm:

dV/dt = (1/3) * π * [(2 * r * dr/dt * h) + (r^2 * dh/dt)]
-10000 = (1/3) * π * [(2 * 133.33 * dr/dt * 200) + (133.33^2 * 20)]

Now we can solve this equation for dr/dt.