A conical water tank is 24 feet high and has a radius of 10 feet at the top. If water flows into the tank at the rate of 20 cubic feet per minute, how fast is the depth of the water increasing when the water is 16 feet deep?

We are dealing with a conical water tank, and we need to find how fast the depth of the water is increasing when the water is 16 feet deep. Here's a step-by-step approach to solve this related rates problem:

### Step 1: Set up variables and relationships
- Let \( r \) be the radius of the water's surface at any given time.
- Let \( h \) be the depth of the water (which changes over time).
- We know the cone's dimensions: the height is 24 feet and the radius at the top is 10 feet.
- The volume of the water in the tank is increasing at a rate of \( \frac{dV}{dt} = 20 \) cubic feet per minute.

The formula for the volume of a cone is:
\[
V = \frac{1}{3} \pi r^2 h
\]
Since \( r \) and \( h \) are both changing, we need to relate them to each other.

### Step 2: Relate \( r \) and \( h \)
From the geometry of the cone, the radius and height are proportional. At the top of the cone:
\[
\frac{r}{h} = \frac{10}{24} = \frac{5}{12}
\]
This gives:
\[
r = \frac{5}{12}h
\]

### Step 3: Substitute \( r \) into the volume formula
Substitute \( r = \frac{5}{12}h \) into the volume formula to express \( V \) in terms of \( h \) alone:
\[
V = \frac{1}{3} \pi \left( \frac{5}{12}h \right)^2 h = \frac{1}{3} \pi \cdot \frac{25}{144} h^3 = \frac{25 \pi}{432} h^3
\]

### Step 4: Differentiate with respect to time
We need to find how fast the depth of the water \( h \) is changing, which is \( \frac{dh}{dt} \). To do this, differentiate the volume equation with respect to time \( t \):
\[
\frac{dV}{dt} = \frac{25 \pi}{432} \cdot 3h^2 \cdot \frac{dh}{dt}
\]
Simplifying:
\[
\frac{dV}{dt} = \frac{25 \pi}{144} h^2 \frac{dh}{dt}
\]

### Step 5: Plug in known values
We are given that \( \frac{dV}{dt} = 20 \) cubic feet per minute, and we are asked to find \( \frac{dh}{dt} \) when \( h = 16 \) feet. Substituting these values into the equation:
\[
20 = \frac{25 \pi}{144} (16)^2 \frac{dh}{dt}
\]
Simplify \( 16^2 = 256 \):
\[
20 = \frac{25 \pi}{144} \cdot 256 \cdot \frac{dh}{dt}
\]
Simplify further:
\[
20 = \frac{25 \pi \cdot 256}{144} \cdot \frac{dh}{dt}
\]
\[
20 = \frac{6400 \pi}{144} \cdot \frac{dh}{dt}
\]
\[
20 = \frac{1600 \pi}{36} \cdot \frac{dh}{dt}
\]
\[
20 = \frac{400 \pi}{9} \cdot \frac{dh}{dt}
\]

### Step 6: Solve for \( \frac{dh}{dt} \)
Now, solve for \( \frac{dh}{dt} \):
\[
\frac{dh}{dt} = \frac{20 \cdot 9}{400 \pi} = \frac{180}{400 \pi} = \frac{9}{20 \pi}
\]

Thus, the depth of the water is increasing at a rate of \( \frac{9}{20 \pi} \) feet per minute when the water is 16 feet deep. 

In decimal form, this is approximately:
\[
\frac{dh}{dt} \approx 0.143 \, \text{feet per minute}
\]

### Conclusion:
The depth of the water is increasing at approximately 0.143 feet per minute when the water is 16 feet deep.

---

Do you want more details or have any questions?

Here are 5 related questions you might want to explore:
1. How would the rate change if the radius of the tank were larger?
2. What if the water flow rate was 30 cubic feet per minute instead of 20?
3. How does the rate of depth change vary with different water depths?
4. How would the answer change if the tank had a cylindrical shape instead of a conical one?
5. What if the cone were inverted, with the wide base at the bottom—how would this affect the calculation?

**Tip:** When solving related rates problems, always look for relationships between changing quantities and use implicit differentiation to connect their rates of change.

Solve a related rates problem involving water flow into a conical tank with a height of 24 feet and radius of 10 feet. Learn how to calculate the rate at which the depth of water increases when the water is 16 feet deep, using the formula for the volume of a cone and implicit differentiation. This detailed solution is suitable for students in Grades 10-12 studying calculus.

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