Question

The radius of the base of a cone is increasing at the rate of 3 cm/minute and the altitude is decreasing at the rate of 4 cm/minute. The rate of change of lateral surface when the radius = 7 cm and altitude 24 cm is

Options

• 54π cm²/min

• 7π cm²/min

• 27 cm²/min

• none of these

Answer 1

\[\text { Let r be the radius, h be the height and S be the lateral surface area of the cone at any time t } .\]

\[\text { Given }: \frac{dr}{dt} = 3 cm/\min\text {  and } \frac{dh}{dt} = - 4 cm/\min\]

\[\text { Here }, \]

\[ l^2 = h^2 + r^2 \]

\[ \Rightarrow l = \sqrt{\left( 24 \right)^2 + \left( 7 \right)^2}\]

\[ \Rightarrow l = \sqrt{625}\]

\[ \Rightarrow l = 25\]

\[S=\pi rl\]

\[ \Rightarrow S^2 = \left( \pi rl \right)^2 \]

\[ \Rightarrow S^2 = \pi^2 r^2 \left( h^2 + r^2 \right)\]

\[ \Rightarrow S^2 = \pi^2 r^4 + \pi^2 h^2 r^2 \]

\[ \Rightarrow 2S\frac{dS}{dt} = 4 \pi^2 r^3 \frac{dr}{dt} + 2 \pi^2 r^2 h\frac{dh}{dt} + 2 \pi^2 h^2 r\frac{dr}{dt}\]

\[ \Rightarrow 2\pi rl\frac{dS}{dt} = 2 \pi^2 rh\left[ \frac{2 r^2}{h}\frac{dr}{dt} + r\frac{dh}{dt} + h\frac{dr}{dt} \right]\]

\[ \Rightarrow 25\frac{dS}{dt} = 24\pi\left[ \frac{2 \left( 7 \right)^2}{24} \times 3 - 7 \times 4 + 24 \times 3 \right] \left[ \text { Given }: r = 7, h = 24 \right]\]

\[ \Rightarrow 25\frac{dS}{dt} = 24\pi\left[ \frac{49}{4} - 28 + 72 \right]\]

\[ \Rightarrow 25\frac{dS}{dt} = 24\pi\left[ \frac{49 + 288 - 112}{4} \right]\]

\[ \Rightarrow \frac{dS}{dt} = 24\pi\left[ \frac{225}{100} \right]\]

\[ \Rightarrow \frac{dS}{dt} = 24\pi\left( 2 . 25 \right)\]

\[ \Rightarrow \frac{dS}{dt} = 54\pi \text{cm}^2 /\sec\]