We are given the following information:

• The radius $r$ is decreasing at a constant rate: $\frac{dr}{dt} = -5$ inches per minute.
• The volume $V$ is decreasing at a rate: $\frac{dV}{dt} = -249$ cubic inches per minute.
• At the instant when the radius is $r = 22$ inches and the volume is $V = 51$ cubic inches, we need to find the rate of change of the height $h$, i.e., $\frac{dh}{dt}$.

The volume of a cone is given by the formula: $V = \frac{1}{3} \pi r^2 h$

Step 1: Differentiate the volume formula

To find the rate of change of the height, we differentiate both sides of the equation with respect to time $t$, using the product and chain rule:

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

Applying the product rule, we get:

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

Step 2: Substitute known values

Now, we can substitute the known values:

• $\frac{dV}{dt} = -249$ cubic inches per minute.
• $r = 22$ inches.
• $\frac{dr}{dt} = -5$ inches per minute.

Substitute these values into the differentiated equation:

$-249 = \frac{1}{3} \pi \left( 2(22)(h)(-5) + (22)^2 \frac{dh}{dt} \right)$

Simplify the expression:

$-249 = \frac{1}{3} \pi \left( -220h + 484 \frac{dh}{dt} \right)$

Step 3: Solve for $\frac{dh}{dt}$

We are given the volume at the instant is $V = 51$ cubic inches. Using the volume formula, solve for $h$:

$51 = \frac{1}{3} \pi (22)^2 h$ $51 = \frac{1}{3} \pi (484) h$ $51 = \frac{484 \pi h}{3}$ $h = \frac{51 \times 3}{484 \pi} = \frac{153}{484 \pi}$ $h \approx 0.100 \text{ inches}$

Now substitute $h = 0.100$ into the equation we derived for $\frac{dV}{dt}$:

$-249 = \frac{1}{3} \pi \left( -220(0.100) + 484 \frac{dh}{dt} \right)$

Simplify:

$-249 = \frac{1}{3} \pi \left( -22 + 484 \frac{dh}{dt} \right)$ Multiply both sides by 3:

$-747 = \pi \left( -22 + 484 \frac{dh}{dt} \right)$ Divide both sides by $\pi$:

$-747 / \pi = -22 + 484 \frac{dh}{dt}$ $-237.35 = -22 + 484 \frac{dh}{dt}$ Solve for $\frac{dh}{dt}$:

$-237.35 + 22 = 484 \frac{dh}{dt}$ $-215.35 = 484 \frac{dh}{dt}$ $\frac{dh}{dt} = \frac{-215.35}{484} \approx -0.445 \text{ inches per minute}$

Final Answer:

The rate of change of the height is approximately $\frac{dh}{dt} \approx -0.445$ inches per minute.

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Would you like further details or have any questions?

Here are some related questions:

1. How does the rate of change of the radius affect the volume of the cone over time?
2. What happens to the height of the cone if the volume were to decrease at a slower rate?
3. How would the rate of change of height change if the radius were increasing instead of decreasing?
4. What is the relationship between the radius, height, and volume in a cone?
5. How would you solve a similar problem with a different shape, like a cylinder?

Tip: When differentiating a product of two functions, remember the product rule: $\frac{d}{dt}(f(t)g(t)) = f'(t)g(t) + f(t)g'(t)$.