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

Let r be the radius of the cylinder and h be its height.
The volume V of the cylinder is given by:
V = π r² h
We are given:
dr/dt = 2 cm/sec (radius is increasing)
dh/dt = -3 cm/sec (height is decreasing)
We need to find dV/dt when r = 3 cm and h = 5 cm.
Differentiate V with respect to time t:
dV/dt = d/dt(π r² h)
Using the product rule:
dV/dt = π ( r² dh/dt + h d/dt(r²) )
dV/dt = π ( r² dh/dt + h (2r) dr/dt )
dV/dt = π ( r² dh/dt + 2rh dr/dt )
Now, substitute the given values: r = 3, h = 5, dr/dt = 2, and dh/dt = -3:
dV/dt = π ( (3)² (-3) + 2(3)(5)(2) )
dV/dt = π ( 9(-3) + 60 )
dV/dt = π ( -27 + 60 )
dV/dt = 33π
Therefore, the rate of change of volume is 33π cm³/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:
   \[ \frac{dV}{dt} = \pi (60 - 27) \]
   \[ \frac{dV}{dt} = 33\pi \, \text{cm}^3/\text{sec} \]