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

The volume of a cylinder is given by the formula V = πr^2h, 

where r is the radius and h is the height. 

We are given that the radius is increasing at the rate of 2 cm/sec, which means dr/dt = 2 cm/sec, and that the height is decreasing at the rate of 3 cm/sec, which means dh/dt = -3 cm/sec. 

We want to find the rate of change of volume when the radius is 3 cm and the height is 5 cm. 

So, we need to find dV/dt when r = 3 cm and h = 5 cm. 

Using the product rule of differentiation, we can write: dV/dt = π(2rh)(dr/dt) + π(r^2)(dh/dt) 

Substituting the given values, we get: dV/dt = π(2 x 3 x 5)(2) + π(3^2)(-3) dV/dt = 30π - 27π dV/dt = 3π 

Therefore, the rate of change of volume when the radius is 3cm and the height is 5 cm is 3π cubic cm/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:
   \[ \frac{dV}{dt} = \pi (60 - 27) \]
   \[ \frac{dV}{dt} = 33\pi \, \text{cm}^3/\text{sec} \]