Question

The radius of the base of a cone is increasing at the rate of

3cm/minute and the altitude is decreasing at the rate of

4cm/minute. The rate of change of lateral surface when the radius = 7 cm and altitude = 24 cm, is –

• A. 54π cm² / min
• B. 7π cm² / min
• C. 27π cm² / min
• D. None of these

Answer 1

Answer: (A)

54π cm² / min

Answer 2

Let r, l and h denotes respectively the radius, slant height and height of the cone at any time t. Then,

l² = r² + h²

⇒  2l $\frac{dl}{dt}$ = 2r $\frac{dr}{dt}$ + 2h$\frac{dh}{dt}$

⇒ l $\frac{dl}{dt}$= r $\frac{dr}{dt}$ + h$\frac{dh}{dt}$

⇒ l$\frac{dl}{dt}$ = 7 × 3 + 24 × (–4) $\left\lbrack \because\frac{dh}{dt} = - 4and\frac{dr}{dt} = 3 \right\rbrack$

⇒  l$\frac{dl}{dt}$ = –75

Where r = 7 and h = 24, we have

l² = 7² + 24²

[Q t² = r² + h²]

⇒ l = 25

∴ l $\frac{dl}{dt}$ = –75 ⇒ $\frac{dl}{dt}$ = – 3

Let S denote the lateral surface area. Then,

⇒ $\frac{dS}{dt}$ = π$\left\{ \frac{drl}{dt} \right\}$ = π$\left\{ \frac{dr}{dt}l + r\frac{dl}{dt} \right\}$ = π {3 × 25 + 7 × (–3)}

= 54 π cm²/min.

Hence (1) is the correct answer.