Water is draining from the bottom of a cone-shaped funnel at the rate of 0.4ft3/sec. The height of the funnel is 8ft and the radius at the top of the funnel is 5ft, as shown in the image below. At what rate is the height of the water in the funnel changing when the height of the water is 2ft?

Submit an exact answer in terms of π.

We are given that water is draining from a cone-shaped funnel, and we need to find the rate at which the height of the water in the funnel is changing when the water's height is 2 ft. The volume of water is decreasing at a rate of \( \frac{dV}{dt} = -0.4 \, \text{ft}^3/\text{sec} \).

### Step 1: Express the volume of water in terms of the height.
The volume \( V \) of a cone is given by the formula:
\[
V = \frac{1}{3} \pi r^2 h
\]
where \( r \) is the radius of the water surface, and \( h \) is the height of the water.

### Step 2: Relate the radius \( r \) to the height \( h \).
Since the funnel is cone-shaped, the radius and height are proportional. At the top of the funnel, the radius is 5 ft and the height is 8 ft. So, the relationship between \( r \) and \( h \) is:
\[
\frac{r}{h} = \frac{5}{8} \quad \Rightarrow \quad r = \frac{5}{8} h
\]

### Step 3: Substitute \( r = \frac{5}{8} h \) into the volume formula.
\[
V = \frac{1}{3} \pi \left( \frac{5}{8} h \right)^2 h = \frac{1}{3} \pi \frac{25}{64} h^3 = \frac{25 \pi}{192} h^3
\]

### Step 4: Differentiate the volume with respect to time.
To find the rate at which the height is changing, we need to differentiate the volume \( V \) with respect to time \( t \):
\[
\frac{dV}{dt} = \frac{25 \pi}{192} \cdot 3h^2 \frac{dh}{dt}
\]
\[
\frac{dV}{dt} = \frac{25 \pi}{64} h^2 \frac{dh}{dt}
\]

### Step 5: Solve for \( \frac{dh}{dt} \).
We are given that \( \frac{dV}{dt} = -0.4 \, \text{ft}^3/\text{sec} \) and we are asked to find \( \frac{dh}{dt} \) when \( h = 2 \, \text{ft} \). Plugging in these values:
\[
-0.4 = \frac{25 \pi}{64} (2)^2 \frac{dh}{dt}
\]
\[
-0.4 = \frac{25 \pi}{64} \cdot 4 \frac{dh}{dt}
\]
\[
-0.4 = \frac{100 \pi}{64} \frac{dh}{dt}
\]
\[
\frac{dh}{dt} = \frac{-0.4 \cdot 64}{100 \pi}
\]
\[
\frac{dh}{dt} = \frac{-25.6}{100 \pi}
\]
\[
\frac{dh}{dt} = \frac{-0.256}{\pi} \, \text{ft/sec}
\]

### Final Answer:
The rate at which the height of the water is changing when the height is 2 ft is:
\[
\frac{dh}{dt} = \frac{-0.256}{\pi} \, \text{ft/sec}
\]

Would you like further details on the steps or clarifications?

Here are 5 related questions to expand on this topic:
1. How does the rate of change of the height vary as the height increases?
2. What would happen if the rate of draining \( \frac{dV}{dt} \) changed?
3. Can you derive the volume formula for other shapes, such as a cylinder?
4. How does the proportionality between the radius and height affect the flow rate?
5. What happens to the rate of height change if the funnel has a larger or smaller radius?

**Tip:** Always relate variables using known proportions when dealing with cones or similar shapes. This simplifies complex rate problems.

Water is draining from the bottom of a cone-shaped funnel at the rate of 0.4ft3/sec. The height of the funnel is 8ft and the radius at the top of the funnel is 5ft. At what rate is the height of the water in the funnel changing when the height of the water is 2ft?

Learn how to solve a related rates problem involving a cone-shaped funnel. Water is draining at a rate of 0.4ft³/sec, and this solution calculates the rate of change of the water's height when the water is 2ft high. The solution uses key calculus concepts like differentiation and proportional relationships between the funnel's height and radius.

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