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.