https://file.aiquickdraw.com/mathbot/file/1729455172828y0j0rny3.jpg

Let's break down and solve the problem based on the image provided. The situation describes water draining from a conical tank, and we need to calculate how fast the water level is changing when \( h = 2 \, \text{ft} \).

### (a) **GIVEN:**
- The rate at which water is draining (rate of change of volume) is \( \frac{dV}{dt} = -3 \, \text{ft}^3/\text{min} \).
  - (It's negative because the volume is decreasing.)
- The cone has a height of \( 10 \, \text{ft} \) and a top radius of \( 4 \, \text{ft} \).
- We are asked to find the rate of change of the water height \( h \) when \( h = 2 \, \text{ft} \).

### (b) **FIND:**
- The rate of change of the water level, \( \frac{dh}{dt} \), when \( h = 2 \, \text{ft} \).

### (c) **Equation that Relates Volume and Height:**
The volume \( V \) of a cone is given by the formula:

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

However, the radius \( r \) of the water surface in the tank changes as the height \( h \) changes. From the geometry of the cone (similar triangles), the ratio of the radius to the height of the cone is constant:

\[
\frac{r}{h} = \frac{4}{10} \quad \Rightarrow \quad r = \frac{2}{5}h
\]

Substitute this expression for \( r \) into the volume formula:

\[
V = \frac{1}{3} \pi \left( \frac{2}{5}h \right)^2 h = \frac{1}{3} \pi \frac{4}{25} h^3
\]

Simplifying:

\[
V = \frac{4\pi}{75} h^3
\]

### (d) **Differentiate with Respect to Time:**

To find how fast the water level is changing, we need to differentiate the volume equation with respect to time:

\[
\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4\pi}{75} h^3 \right)
\]

Using the chain rule:

\[
\frac{dV}{dt} = \frac{4\pi}{75} \cdot 3h^2 \cdot \frac{dh}{dt}
\]

Simplifying:

\[
\frac{dV}{dt} = \frac{4\pi}{25} h^2 \cdot \frac{dh}{dt}
\]

Now, substitute \( \frac{dV}{dt} = -3 \, \text{ft}^3/\text{min} \) and \( h = 2 \, \text{ft} \):

\[
-3 = \frac{4\pi}{25} \cdot (2)^2 \cdot \frac{dh}{dt}
\]

\[
-3 = \frac{4\pi}{25} \cdot 4 \cdot \frac{dh}{dt}
\]

\[
-3 = \frac{16\pi}{25} \cdot \frac{dh}{dt}
\]

Solve for \( \frac{dh}{dt} \):

\[
\frac{dh}{dt} = \frac{-3 \cdot 25}{16\pi}
\]

\[
\frac{dh}{dt} = \frac{-75}{16\pi} \, \text{ft/min}
\]

Thus, the water level is changing at a rate of \( \frac{dh}{dt} = \frac{-75}{16\pi} \, \text{ft/min} \approx -1.49 \, \text{ft/min} \) when \( h = 2 \, \text{ft} \).

### Would you like more details or have any further questions?

Here are 5 related questions:
1. How does the rate of water level change vary for different heights \( h \)?
2. What happens to \( \frac{dh}{dt} \) if the rate of draining \( \frac{dV}{dt} \) is increased?
3. How does the relationship between \( r \) and \( h \) affect the rate at which the water level decreases?
4. Can this method be applied to other cone-like tanks with different dimensions?
5. What would be the height of the water if the volume was halved?

**Tip:** When solving related rates problems, always check for geometric relationships that simplify the variables before differentiation.

Water drains from the conical tank shown at a rate of 3 cubic feet per minute. Use r for radius, h for height, and t for time. Find how fast the water level is changing when h = 2 feet.

Solve the related rates problem where water drains from a conical tank at 3 cubic feet per minute. Find how fast the water level changes when h = 2 feet. This problem covers the volume of a cone, similarity of triangles, and calculus differentiation.

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