If the semi-vertical angle of a cone is 45°, then the rate of change of volume of the cone is

If the semi-vertical angle of a cone is 45°, then the rate of change of volume of the cone is

A
curved surface area times the rate of change of $r$
B
base area times the rate of change of $I$
C
base area times the rate of change of $r$
D
none of these

Solution:

Let $r$ be the radius of the base, $h$ the height and

$l$ slant height of the cone with semi-vertical angle 45°. Then,

$r = h$ and $l^{2} = r^{2} + h^{2} = 2 r^{2}$

Let V be the volume. Then,

$V = \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{3}$

$\Rightarrow \frac{d V}{d t} = π r^{2} \frac{d r}{d t} =$ Area of the base $\times \frac{d r}{d t}$

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