To find the rate at which the volume of the cone is changing, we need to use related rates. The formula for the volume $V$ of a cone is given by:

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

where $r$ is the radius and $h$ is the height.

Step 1: Differentiate the volume formula

We will differentiate the volume formula with respect to time $t$. Applying the product rule:

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

Step 2: Substitute known values

• $r = 105$ in
• $h = 149$ in
• $\frac{dr}{dt} = 1.6$ in/s (rate at which the radius is increasing)
• $\frac{dh}{dt} = -2.4$ in/s (rate at which the height is decreasing)

Now we substitute these values into the equation:

$\frac{dV}{dt} = \frac{1}{3} \pi \left( 2(105)(1.6)(149) + (105)^2(-2.4) \right)$

Let's compute this.The volume of the cone is increasing at a rate of approximately $24,718.05 \, \text{in}^3/\text{s}$ when the radius is 105 inches and the height is 149 inches.

Would you like more details or explanations on any of the steps?

Here are 5 related questions for further practice:

1. How does the rate of change of the height affect the rate of change of volume in this scenario?
2. What would happen to the volume if both the radius and height were decreasing?
3. How would the rate of change of volume differ if the height was increasing at the same rate as the radius?
4. Can you derive the formula for the surface area of the cone and apply related rates to it?
5. What is the physical significance of a negative rate of change in height?

Tip: Always keep track of the units when working with related rates, as it helps prevent errors in interpretation.