The height of a cylinder is increasing at a constant rate of 7 inches per minute. The volume remains a constant 889 cubic inches. At the instant when the radius of the cylinder is 99 inches, what is the rate of change of the radius? The volume of a cylinder can be found with the equation V, equals, pi, r, squared, h, .V=πr
2
h. Round your answer to three decimal places.

To solve this, we’ll use the formula for the volume of a cylinder:
\[
V = \pi r^2 h
\]
Given:
- The volume \( V \) is constant at 889 cubic inches.
- The height \( h \) is increasing at a constant rate of \( \frac{dh}{dt} = 7 \) inches per minute.
- We need to find the rate of change of the radius \( r \), \( \frac{dr}{dt} \), when \( r = 99 \) inches.

### Step 1: Differentiate the Volume Equation with Respect to Time
Since \( V \) is constant, we can set the derivative of \( V \) with respect to time to zero:
\[
\frac{dV}{dt} = 0
\]
Differentiating both sides of \( V = \pi r^2 h \) with respect to \( t \):
\[
\frac{d}{dt}(\pi r^2 h) = 0
\]
Applying the product rule:
\[
\pi \cdot \frac{d}{dt}(r^2) \cdot h + \pi r^2 \cdot \frac{dh}{dt} = 0
\]
### Step 2: Substitute the Derivatives
We know:
\[
\frac{d}{dt}(r^2) = 2r \frac{dr}{dt} \quad \text{and} \quad \frac{dh}{dt} = 7
\]
Substitute these into the equation:
\[
\pi (2r \frac{dr}{dt}) h + \pi r^2 \cdot 7 = 0
\]
Simplify by dividing through by \( \pi \):
\[
2r h \frac{dr}{dt} + 7r^2 = 0
\]

### Step 3: Solve for \( \frac{dr}{dt} \)
Rearrange to isolate \( \frac{dr}{dt} \):
\[
2r h \frac{dr}{dt} = -7r^2
\]
\[
\frac{dr}{dt} = -\frac{7r^2}{2r h}
\]
Simplify by canceling an \( r \) term:
\[
\frac{dr}{dt} = -\frac{7r}{2h}
\]

### Step 4: Substitute Known Values
We know \( r = 99 \) inches and need to find \( h \) when \( r = 99 \). Using the original volume equation:
\[
V = \pi r^2 h
\]
\[
889 = \pi (99)^2 h
\]
Solve for \( h \):
\[
h = \frac{889}{\pi \times 99^2}
\]

Calculating this value:
1. \( 99^2 = 9801 \)
2. \( \pi \times 9801 \approx 30787.608 \)
3. \( h \approx \frac{889}{30787.608} \approx 0.0289 \) inches.

### Step 5: Substitute \( h \) and Solve for \( \frac{dr}{dt} \)
\[
\frac{dr}{dt} = -\frac{7 \times 99}{2 \times 0.0289}
\]

Calculate the value:
1. \( 7 \times 99 = 693 \)
2. \( 2 \times 0.0289 = 0.0578 \)
3. \( \frac{dr}{dt} \approx -\frac{693}{0.0578} \approx -11991.348 \)

### Final Answer:
The rate of change of the radius is approximately:
\[
\frac{dr}{dt} \approx -11991.348 \text{ inches per minute}
\]

Would you like more details on any step or have questions?

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**Related Questions:**
1. How does the height change affect the volume if the radius also changes?
2. What happens to the radius if the volume were to decrease instead?
3. How would the problem differ if the volume were not constant?
4. What is the impact on \( \frac{dr}{dt} \) if the height's rate of increase changes?
5. How does using different units (e.g., cm) affect the results?

**Tip:** When rates of change are involved, identifying constant parameters can simplify differentiation.

The height of a cylinder is increasing at a constant rate of 7 inches per minute. The volume remains a constant 889 cubic inches. At the instant when the radius of the cylinder is 99 inches, what is the rate of change of the radius? The volume of a cylinder can be found with the equation V = πr^2h. Round your answer to three decimal places.

Explore how to find the rate of change of the radius of a cylinder given a constant volume of 889 cubic inches and a height increasing at 7 inches per minute. This calculus-based related rates problem involves differentiation and the product rule. Ideal for advanced high school or college students.

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