Math Problem Statement

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.

Solution

To solve this, we’ll use the formula for the volume of a cylinder: V=πr2hV = \pi r^2 h Given:

  • The volume VV is constant at 889 cubic inches.
  • The height hh is increasing at a constant rate of dhdt=7\frac{dh}{dt} = 7 inches per minute.
  • We need to find the rate of change of the radius rr, drdt\frac{dr}{dt}, when r=99r = 99 inches.

Step 1: Differentiate the Volume Equation with Respect to Time

Since VV is constant, we can set the derivative of VV with respect to time to zero: dVdt=0\frac{dV}{dt} = 0 Differentiating both sides of V=πr2hV = \pi r^2 h with respect to tt: ddt(πr2h)=0\frac{d}{dt}(\pi r^2 h) = 0 Applying the product rule: πddt(r2)h+πr2dhdt=0\pi \cdot \frac{d}{dt}(r^2) \cdot h + \pi r^2 \cdot \frac{dh}{dt} = 0

Step 2: Substitute the Derivatives

We know: ddt(r2)=2rdrdtanddhdt=7\frac{d}{dt}(r^2) = 2r \frac{dr}{dt} \quad \text{and} \quad \frac{dh}{dt} = 7 Substitute these into the equation: π(2rdrdt)h+πr27=0\pi (2r \frac{dr}{dt}) h + \pi r^2 \cdot 7 = 0 Simplify by dividing through by π\pi: 2rhdrdt+7r2=02r h \frac{dr}{dt} + 7r^2 = 0

Step 3: Solve for drdt\frac{dr}{dt}

Rearrange to isolate drdt\frac{dr}{dt}: 2rhdrdt=7r22r h \frac{dr}{dt} = -7r^2 drdt=7r22rh\frac{dr}{dt} = -\frac{7r^2}{2r h} Simplify by canceling an rr term: drdt=7r2h\frac{dr}{dt} = -\frac{7r}{2h}

Step 4: Substitute Known Values

We know r=99r = 99 inches and need to find hh when r=99r = 99. Using the original volume equation: V=πr2hV = \pi r^2 h 889=π(99)2h889 = \pi (99)^2 h Solve for hh: h=889π×992h = \frac{889}{\pi \times 99^2}

Calculating this value:

  1. 992=980199^2 = 9801
  2. π×980130787.608\pi \times 9801 \approx 30787.608
  3. h88930787.6080.0289h \approx \frac{889}{30787.608} \approx 0.0289 inches.

Step 5: Substitute hh and Solve for drdt\frac{dr}{dt}

drdt=7×992×0.0289\frac{dr}{dt} = -\frac{7 \times 99}{2 \times 0.0289}

Calculate the value:

  1. 7×99=6937 \times 99 = 693
  2. 2×0.0289=0.05782 \times 0.0289 = 0.0578
  3. drdt6930.057811991.348\frac{dr}{dt} \approx -\frac{693}{0.0578} \approx -11991.348

Final Answer:

The rate of change of the radius is approximately: drdt11991.348 inches per minute\frac{dr}{dt} \approx -11991.348 \text{ inches per minute}

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


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 drdt\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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Related Rates
Differentiation

Formulas

Volume of a cylinder: V = πr^2h
Product rule in differentiation

Theorems

Chain Rule
Product Rule

Suitable Grade Level

Grades 11-12 or College