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Q.

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

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a

$3 π {cm}^{2} / \min$

b

$54 π {cm}^{2} / \min$

c

$27 π {cm}^{2} / \min$

d

$9 π {cm}^{2} / min$

answer is A.

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Detailed Solution

$\begin{array}{l} l = \sqrt{r^{2} + h^{2}} \\ \frac{dl}{dt} = \frac{1}{2 \sqrt{r^{2} + h^{2}}} (2 r \cdot \frac{dr}{dt} + 2 h \cdot \frac{dh}{dt}) \\ = \frac{1}{25} [7 \times 3 + (24) \cdot (- 4)] = - 3 \\ S = πrl \Rightarrow \frac{ds}{dt} = π [r \cdot \frac{dl}{dt} + l \cdot \frac{dr}{dt}] \\ = π [(7) (- 3) + 25 (3)] = 54 π \end{array}$

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