A water tank is in the the shape of cone with its vertex pointing towards the
ground. The tank has a height of 20 meters and its radius at the top of the tank is 5 meters.
Water is being pumped into the tank at a rate of 3m3/min, but due to a hole at the bottom
of tank, water is also leaking out at a rate proportional the water’s depth.
(a) Derive an expression for the rate of change (with respect to time) of the depth of water
in the tank.
(b) Assuming that the water tank is able to be completely filled under these conditions, what
can be determined about the proportionality constant?
(c) The hole in the bottom of the tank has widened and water is now flowing out of the tank
at a rate proportional to the cube of the depth of the water. Find a function for the
height of the water in the tank at a given time. Assume that the tank was initially empty.
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Let's tackle this problem step by step.

### Part (a) - Derive an expression for the rate of change of the depth of water in the tank:

Given:

- The tank is conical with height \( H = 20 \) meters and radius \( R = 5 \) meters.
- The volume of the cone when filled with water up to height \( h \) is given by:

\[
V = \frac{1}{3} \pi r^2 h
\]

To express \( r \) (the radius at height \( h \)) in terms of \( h \), use the similarity of triangles:

\[
\frac{r}{h} = \frac{R}{H} = \frac{5}{20} = \frac{1}{4}
\]

Thus,

\[
r = \frac{h}{4}
\]

Substituting \( r = \frac{h}{4} \) into the volume formula:

\[
V = \frac{1}{3} \pi \left(\frac{h}{4}\right)^2 h = \frac{1}{3} \pi \frac{h^3}{16} = \frac{\pi}{48} h^3
\]

Now, differentiate \( V \) with respect to time \( t \) to get the rate of change of volume:

\[
\frac{dV}{dt} = \frac{\pi}{16} h^2 \frac{dh}{dt}
\]

Given that water is being pumped into the tank at a rate of \( 3 \, \text{m}^3/\text{min} \), and water is leaking out at a rate proportional to the depth, say \( kh \), the rate of change of volume is:

\[
\frac{dV}{dt} = 3 - kh
\]

Equating the two expressions for \( \frac{dV}{dt} \):

\[
\frac{\pi}{16} h^2 \frac{dh}{dt} = 3 - kh
\]

Thus, the expression for the rate of change of the depth of water \( h(t) \) is:

\[
\frac{dh}{dt} = \frac{16}{\pi h^2} (3 - kh)
\]

### Part (b) - Determine the proportionality constant \( k \):

To determine \( k \), we need to consider the scenario when the tank is full. When the tank is full, \( h = 20 \) meters, and \( \frac{dh}{dt} \) should be 0 because the tank can hold no more water:

\[
\frac{16}{\pi (20)^2} (3 - 20k) = 0
\]

Simplifying:

\[
3 - 20k = 0
\]

\[
k = \frac{3}{20} = 0.15 \, \text{m/min}
\]

So, the proportionality constant \( k \) must be \( 0.15 \, \text{m/min} \).

### Part (c) - Find a function for the height of the water in the tank at a given time:

Now, the rate of water leaking out is proportional to the cube of the depth, so the differential equation changes to:

\[
\frac{dV}{dt} = 3 - k h^3
\]

Following the same steps as in part (a), equate the rates:

\[
\frac{\pi}{16} h^2 \frac{dh}{dt} = 3 - k h^3
\]

Thus, the differential equation governing the height \( h(t) \) becomes:

\[
\frac{dh}{dt} = \frac{16}{\pi h^2} (3 - k h^3)
\]

This equation is separable:

\[
\int \frac{h^2}{3 - k h^3} dh = \frac{16}{\pi} \int dt
\]

Solving this integral gives us the function for \( h(t) \), which may involve an implicit or explicit solution depending on the complexity of the integral. If \( k \) and initial conditions are given, this equation could be further simplified or solved numerically.

Would you like more details or have any questions?

### Further questions:
1. What methods could be used to solve the differential equation for \( h(t) \) numerically?
2. How does the proportionality constant \( k \) affect the time it takes to fill the tank?
3. What assumptions are made in deriving the equations, and how could these affect the model's accuracy?
4. Could the problem be approached differently if the rate of water entering the tank was also variable?
5. How would the problem change if the tank had a different shape, such as a cylinder or sphere?

**Tip:** When dealing with rates of change, always ensure units are consistent throughout the problem to avoid errors in your calculations.

Explore the solution to a conical water tank problem involving rates of change and proportional leakage. Derive expressions for depth change rate, determine a proportionality constant, and solve for height of water with a cubic leakage rate. Suitable for advanced high school or college-level students.

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