Water is leaking out of an inverted conical tank at a rate of cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height centimeters and the diameter at the top is centimeters. If the water level is rising at a rate of centimeters per minute when the height of the water is centimeters, find the rate at which water is being pumped into the tank in cubic centimeters per minute

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To solve this problem, we need to apply the principles of related rates. Let's label the given information:

- The rate at which water is leaking out of the tank is given as \( \frac{dV}{dt} \) cubic centimeters per minute.
- The rate at which water is being pumped into the tank is what we need to find, let's call it \( \frac{dV}{dt}_{in} \) cubic centimeters per minute.
- The height of the tank is labeled as \( h \) centimeters.
- The diameter at the top of the tank is given as \( D \) centimeters.
- The rate at which the water level is rising is given as \( \frac{dh}{dt} \) centimeters per minute.
- The height of the water is \( h \) centimeters.

Now, we need to use the formulas of volume and related rates for a cone to connect these variables:

Volume of a cone, \( V \), is given by \( V = \frac{1}{3} \pi r^2 h \), where \( r \) denotes the radius:

1. To find the radius \( r \) in terms of \( D \), we divide the diameter by 2: \( r = \frac{D}{2} \).
2. We need to express \( V \) in terms of \( h \).

Since the height of the tank is constant, we assume the radius is also changing over time.

To differentiate the volume function with respect to time, we use the chain rule:

\( \frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} \)

We know that \( \frac{dV}{dt} \) is the rate at which the water is leaking out of the tank, so \( \frac{dV}{dt} = - \frac{dV}{dt}_{out} \).

Now, we differentiate the volume formula:

\( \frac{dV}{dh} = \frac{d}{dh} \left(\frac{1}{3} \pi r^2 h \right) \)

To differentiate \( r \) with respect to \( h \), we use similar triangles:

\( \frac{r}{h} = \frac{D/2}{h} \)

Simplifying, we find \( r = \frac{D}{2h} \).

Differentiating \( r \) with respect to \( h \):

\( \frac{dr}{dh} = -\frac{D}{2h^2} \)

Now we can substitute the values we obtained:

\( \frac{dV}{dh} = \frac{d}{dh} \left(\frac{1}{3} \pi \left(\frac{D}{2h}\right)^2 h \right) \)

\( \frac{dV}{dh} = \frac{d}{dh} \left(\frac{\pi D^2 h}{12} \right) \)

\( \frac{dV}{dh} = \frac{\pi D^2}{12} \)

Substituting the values we know:

\( - \frac{dV}{dt}_{out} = \frac{\pi D^2}{12} \cdot \frac{dh}{dt} \)

Now we can solve for the rate at which water is being pumped into the tank:

\( \frac{dV}{dt}_{in} = \frac{dV}{dt}_{out} + \frac{\pi D^2}{12} \cdot \frac{dh}{dt} \)

Since we know that the water level is rising at a constant rate of \( \frac{dh}{dt} \) centimeters per minute when the height is \( h \) centimeters, we can substitute those values into the equation and calculate \( \frac{dV}{dt}_{in} \), the rate of water being pumped into the tank in cubic centimeters per minute.