Question

The radius of a cylinder is increasing at the rate of 5 cm/min so that its volume is constant. When its radius is 5 cm and height is 3 cm the rate of decreasing of its height is:

• A. \( 6 \text{ cm/min} \)
• B. \( 3 \text{ cm/min} \)
• C. \( 4 \text{ cm/min} \)
• D. \( 5 \text{ cm/min} \)
• E. \( 2 \text{ cm/min} \)

Hint:

In related rates, a negative derivative signifies a decrease. When the question asks for the "rate of decreasing," provide the absolute value.

Answer 1

The Correct Option is A

Concept: The volume of a cylinder is \( V = \pi r^2 h \). If the volume is constant, its derivative with respect to time \( t \) is zero (\( \frac{dV}{dt} = 0 \)). This is a related rates problem.

Step 1: Differentiate the volume formula with respect to \( t \).
\[ V = \pi r^2 h \] Using the product rule: \[ \frac{dV}{dt} = \pi \left( 2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} \right) \]

Step 2: Substitute the known values.
Given: \( \frac{dV}{dt} = 0 \), \( \frac{dr}{dt} = 5 \), \( r = 5 \), and \( h = 3 \). \[ 0 = \pi \left( 2(5)(3)(5) + (5)^2 \frac{dh}{dt} \right) \] \[ 0 = 150 + 25 \frac{dh}{dt} \]

Step 3: Solve for the rate of change of height.
\[ 25 \frac{dh}{dt} = -150 \] \[ \frac{dh}{dt} = -6 \text{ cm/min} \] The rate of decreasing of height is \( 6 \text{ cm/min} \).