The diameter and altitude of a right circular cone at a certain instant, were found to be 10 cm and 20 cm respectively. If the diameter is increasing at a rate of 2 cm/s, then at what rate must its altitude change in order to keep its volume is constant?

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-8 cm/s

4 cm/s

-4 cm/s

6 cm/s

answer is D.

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Detailed Solution

$x = 10, h = 20, \frac{d x}{d t} = 2,$
$V \to constant \Rightarrow \frac{d h}{d t} =$
Radius $r = \frac{x}{2}$
Volume $V = \frac{1}{3} π r^{2} h = \frac{π}{12} x^{2} h$
Diff.
$0 = \frac{π}{12} (x^{2} \frac{d h}{d t} + h (2 x) \frac{d x}{d t})$
$\frac{d h}{d t} = - 8 c m / s$

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