Question

The radius of a cylinder is increasing at \(2\,m/s\) and height is decreasing at \(3\,m/s\). When \(r=3\,m, h=5\,m\), rate of change of volume is:

• A. \(87\pi\, m^3/s\)
• B. \(33\pi\, m^3/s\)
• C. \(27\pi\, m^3/s\)
• D. \(15\pi\, m^3/s\)

Hint:

Use product rule when differentiating volume with respect to time when both radius and height change.

Answer 1

The Correct Option is B

To find the rate of change of volume of the cylinder, we begin by recalling the formula for the volume of a cylinder:

\(V = \pi r^2 h\)

where \(r\) is the radius and \(h\) is the height of the cylinder.

Given that the radius \((r)\) is increasing at a rate of \(2\, m/s\) and the height \((h)\) is decreasing at a rate of \(3\, m/s\), we need to find the rate of change of the volume when the radius is \(r = 3\, m\) and height is \(h = 5\, m\).

We use the chain rule of calculus to differentiate the volume with respect to time:

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

Substitute the given values:

• \(r = 3\, m\)
• \(h = 5\, m\)
• \(\frac{dr}{dt} = 2\, m/s\)
• \(\frac{dh}{dt} = -3\, m/s\) (negative because the height is decreasing)

Substitute these into the differentiated formula:

\(\frac{dV}{dt} = \pi \left(2 \cdot 3 \cdot 2 \cdot 5 + 3^2 \cdot (-3)\right)\)

This simplifies to:

\(\frac{dV}{dt} = \pi \left(2 \cdot 3 \cdot 2 \cdot 5 - 27\right)\)

\(\frac{dV}{dt} = \pi \left(60 - 27\right)\)

\(\frac{dV}{dt} = \pi \cdot 33\)

Thus, the rate of change of the volume of the cylinder is \(33\pi \, m^3/s\).

Therefore, the correct answer is \(33\pi \, m^3/s\).