A conical water tank with vertex down has a radius of 11 feet at the top and is 27 feet high. If water flows into the tank at a rate of 20 ft3/min
, how fast is the depth of the water increasing when the water is 17 feet deep?

We are tasked with finding the rate at which the depth of water is increasing in a conical tank as it is being filled. The cone has a radius of 11 feet at the top and a height of 27 feet. Water flows into the tank at a rate of \( \frac{dV}{dt} = 20 \, \text{ft}^3/\text{min} \), and we need to determine how fast the water depth (\( h \)) is increasing when the depth of water is 17 feet.

### Step 1: Relate Volume of the Cone to its Dimensions
The formula for the volume \( V \) of a cone is:
\[
V = \frac{1}{3} \pi r^2 h
\]
Where \( r \) is the radius of the water surface at a certain depth \( h \), and \( h \) is the height (or depth) of the water.

Since the cone is similar at all levels, we can relate \( r \) and \( h \) proportionally. At the top of the cone (when \( h = 27 \, \text{ft} \)), the radius \( r = 11 \, \text{ft} \). Therefore, by similar triangles:
\[
\frac{r}{h} = \frac{11}{27}
\]
Thus,
\[
r = \frac{11}{27}h
\]

### Step 2: Substitute \( r \) in the Volume Formula
Substitute \( r = \frac{11}{27}h \) into the volume formula:
\[
V = \frac{1}{3} \pi \left( \frac{11}{27}h \right)^2 h
\]
Simplify:
\[
V = \frac{1}{3} \pi \cdot \frac{121}{729} h^3 = \frac{121 \pi}{2187} h^3
\]

### Step 3: Differentiate with Respect to Time
Now, differentiate the volume equation with respect to time \( t \) to find the rate of change of volume with respect to time:
\[
\frac{dV}{dt} = \frac{121 \pi}{2187} \cdot 3h^2 \frac{dh}{dt}
\]
Simplify:
\[
\frac{dV}{dt} = \frac{363 \pi}{2187} h^2 \frac{dh}{dt}
\]

### Step 4: Solve for \( \frac{dh}{dt} \)
We are given that \( \frac{dV}{dt} = 20 \, \text{ft}^3/\text{min} \), and we need to find \( \frac{dh}{dt} \) when \( h = 17 \, \text{ft} \). Plug these values into the equation:
\[
20 = \frac{363 \pi}{2187} \cdot 17^2 \cdot \frac{dh}{dt}
\]
Simplify:
\[
20 = \frac{363 \pi}{2187} \cdot 289 \cdot \frac{dh}{dt}
\]
\[
20 = \frac{105007 \pi}{2187} \cdot \frac{dh}{dt}
\]
Now, solve for \( \frac{dh}{dt} \):
\[
\frac{dh}{dt} = \frac{20 \cdot 2187}{105007 \pi}
\]
\[
\frac{dh}{dt} \approx 0.0132 \, \text{ft}/\text{min}
\]

Thus, the depth of the water is increasing at a rate of approximately \( 0.0132 \, \text{ft}/\text{min} \) when the water is 17 feet deep.

Would you like further clarification on any of the steps or details? Here are some related questions to explore:

1. How would the result change if the flow rate were 30 ft³/min instead of 20 ft³/min?
2. What happens to the rate of depth increase as the water level approaches the top of the cone?
3. How is the rate of change affected if the cone had a different height-to-radius ratio?
4. Could this problem be extended to consider evaporation from the tank’s surface?
5. How would this calculation change if the tank were cylindrical instead of conical?

**Tip:** When dealing with related rates problems, always look for geometric similarities or relationships between variables to simplify the equations.

A conical water tank with vertex down has a radius of 11 feet at the top and is 27 feet high. If water flows into the tank at a rate of 20 ft³/min, how fast is the depth of the water increasing when the water is 17 feet deep?

Learn how to solve a related rates problem involving water flowing into a conical tank. Given a flow rate of 20 ft³/min, find the rate at which the depth of water is increasing when the water is 17 feet deep. Includes step-by-step calculus-based solution using geometry of cones and differentiation. Suitable for advanced calculus students in Grades 11-12.

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