Question

The radius of a cylinder is increasing at the rate 2 cm/sec and its height is decreasing at the rate 3 cm/sec, then find the rate of change of volume when the radius is 3cm and the height is 5 cm.

Answer 1

Given:

• r is the radius of the cylinder, and h is its height.
• The volume V of the cylinder is given by:
  \( V = \pi r^2 h \).
• The rate of change of the radius, \( \frac{dr}{dt} = 2 \, \text{cm/sec} \) (radius is increasing).
• The rate of change of the height, \( \frac{dh}{dt} = -3 \, \text{cm/sec} \) (height is decreasing).
• The values to be substituted: r = 3 cm, h = 5 cm.

We need to find:
\( \frac{dV}{dt} \) when \( r = 3 \, \text{cm} \) and \( h = 5 \, \text{cm}. \)

Step 1: Differentiate the volume formula with respect to time \( t \):
The formula for the volume of the cylinder is: \[ V = \pi r^2 h \] Differentiating both sides with respect to \( t \), we get: \[ \frac{dV}{dt} = \frac{d}{dt} \left( \pi r^2 h \right) \]

Step 2: Apply the product rule:
Since the volume formula involves a product of two functions \( r^2 \) and \( h \), we apply the product rule for differentiation: \[ \frac{dV}{dt} = \pi \left( r^2 \frac{dh}{dt} + h \frac{d}{dt}(r^2) \right) \] Next, differentiate \( r^2 \) with respect to \( t \): \[ \frac{d}{dt}(r^2) = 2r \frac{dr}{dt} \] Therefore, we get: \[ \frac{dV}{dt} = \pi \left( r^2 \frac{dh}{dt} + h \cdot 2r \frac{dr}{dt} \right) \]

Step 3: Substitute the given values:
Now, substitute the given values: \( r = 3 \), \( h = 5 \), \( \frac{dr}{dt} = 2 \), and \( \frac{dh}{dt} = -3 \): \[ \frac{dV}{dt} = \pi \left( (3)^2 (-3) + 2(3)(5)(2) \right) \] Simplifying: \[ \frac{dV}{dt} = \pi \left( 9(-3) + 60 \right) \] \[ \frac{dV}{dt} = \pi \left( -27 + 60 \right) \] \[ \frac{dV}{dt} = 33\pi \]

Final Answer:
Therefore, the rate of change of volume is \( \frac{dV}{dt} = 33\pi \, \text{cm}^3/\text{sec}. \)

Answer 2

Given:

• The volume \( V \) of a cylinder: \( V = \pi r^2 h \)
• \( \frac{dr}{dt} = 2 \)
• \( \frac{dh}{dt} = -3 \)

Find \( \frac{dV}{dt} \):

1. Differentiate \( V = \pi r^2 h \):
   \[ \frac{dV}{dt} = \pi \left( 2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} \right) \]
2. Substitute \( \frac{dr}{dt} = 2 \) and \( \frac{dh}{dt} = -3 \):
   \[ \frac{dV}{dt} = \pi \left( 2rh \cdot 2 + r^2 \cdot (-3) \right) \]
   \[ \frac{dV}{dt} = \pi (4rh - 3r^2) \]
3. Substitute \( r = 3 \) and \( h = 5 \):
   \[ \frac{dV}{dt} = \pi (4 \cdot 3 \cdot 5 - 3 \cdot 3^2) \]
   Simplify:
   \[ \frac{dV}{dt} = \pi (60 - 27) \]
   \[ \frac{dV}{dt} = 33\pi \, \text{cm}^3/\text{sec} \]