Question

The radius r of a right circular cylinder is increasing uniformly at the rate of 0·3 cm/s and its height h is decreasing at the rate of 0·4 cm/s. When r = 3·5 cm and h = 7 cm, find the rate of change of the curved surface area of the cylinder. \[\left[ \text{ Use } \pi = \frac{22}{7} \right]\]

Answer 1

It is given that, \[\frac{dr}{dt} = 0 . 3 cm/s \text { and } \frac{dh}{dt} = - 0 . 4 cm/s\] Curved surface area of a cylinder \[\left( A \right) = 2\pi rh\].

Change in curved surface area of a cylinder is as follows:

\[\frac{dA}{dt} = 2\pi\frac{d\left( rh \right)}{dt}\]

\[ \Rightarrow \frac{dA}{dt} = 2\pi\left( r\frac{dh}{dt} + h\frac{dr}{dt} \right) \left[ \text { By product rule } \right]\]

\[ \Rightarrow \left[ \frac{dA}{dt} \right]_{r = 3 . 5 cm, h = 7 cm} = 2\pi\left[ 3 . 5 \times \left( - 0 . 4 \right) + 7 \times \left( 0 . 3 \right) \right]\] 

\[\Rightarrow \frac{dA}{dt} = 2 \times \frac{22}{7}\left[ - 1 . 4 + 2 . 1 \right]\]

\[ \Rightarrow \frac{dA}{dt} = 2 \times \frac{22}{7}\left[ 0 . 7 \right]\]

\[ \Rightarrow \frac{dA}{dt} = 4 . 4 {cm}^2 /s\]