The radius of a right circular cone is increasing at a rate of 1.3 in/s while its height is decreasing at a rate of 2.9 in/s. At what rate is the volume of the cone changing when the radius is 120 in. and the height is 156 in.?

The volume \( V \) of a right circular cone is given by the formula:

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

where \( r \) is the radius of the base of the cone, and \( h \) is the height of the cone.

### Step 1: Differentiate the volume formula with respect to time \( t \)

We need to find the rate of change of volume \( \frac{dV}{dt} \) with respect to time. Since both \( r \) and \( h \) are changing with time, we will use implicit differentiation:

\[
\frac{dV}{dt} = \frac{d}{dt} \left( \frac{1}{3} \pi r^2 h \right)
\]

Applying the product rule to differentiate:

\[
\frac{dV}{dt} = \frac{1}{3} \pi \left( 2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt} \right)
\]

### Step 2: Plug in the given values

We are given:

- \( r = 120 \, \text{in} \)
- \( h = 156 \, \text{in} \)
- \( \frac{dr}{dt} = 1.3 \, \text{in/s} \) (rate of change of the radius)
- \( \frac{dh}{dt} = -2.9 \, \text{in/s} \) (rate of change of the height, negative because the height is decreasing)

Substitute these values into the differentiated equation:

\[
\frac{dV}{dt} = \frac{1}{3} \pi \left( 2(120)(1.3) \cdot 156 + (120)^2 \cdot (-2.9) \right)
\]

### Step 3: Simplify the expression

First, calculate each part:

- \( 2(120)(1.3) \cdot 156 = 48384 \)
- \( (120)^2 = 14400 \)
- \( 14400 \cdot (-2.9) = -41760 \)

Now, substitute these results back into the equation:

\[
\frac{dV}{dt} = \frac{1}{3} \pi (48384 - 41760)
\]
\[
\frac{dV}{dt} = \frac{1}{3} \pi (6624)
\]
\[
\frac{dV}{dt} = 2208 \pi \, \text{in}^3/\text{s}
\]

### Final Answer:
The volume of the cone is increasing at a rate of:

\[
\frac{dV}{dt} = 2208 \pi \, \text{in}^3/\text{s} \approx 6935.47 \, \text{in}^3/\text{s}
\]

### Do you have any questions, or would you like more details?

Here are 5 related questions to further your understanding:

1. How does the rate of change of height affect the rate of change of volume in a cone?
2. What would happen if both the height and radius were increasing?
3. Can you derive the volume formula for a cone from the geometry of a cone?
4. How would the problem change if the radius were decreasing instead?
5. What happens to the cone's volume when only the height changes but the radius is constant?

**Tip:** In related rates problems, always identify which quantities are changing with time and use implicit differentiation accordingly.

The radius of a right circular cone is increasing at a rate of 1.3 in/s while its height is decreasing at a rate of 2.9 in/s. At what rate is the volume of the cone changing when the radius is 120 in. and the height is 156 in.?

Understand how to find the rate of change of the volume of a cone when the radius and height are changing. Given a radius of 120 in and height of 156 in, this problem demonstrates using calculus concepts such as related rates and differentiation. Suitable for high school students in Grades 11-12.

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