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  4. The rate of change of the surface area of a sphere of radius r when the radius is increasing at the rate of 2 cm / sec is proportional to

Q. The rate of change of the surface area of a sphere of radius $r$ when the radius is increasing at the rate of $2\, cm / \sec$ is proportional to

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Solution:

Surface area of the sphere is $s=4 \pi r^{2}\, \&\, \frac{d r}{d t}=2$
$\frac{d s}{d t}=8 \pi r \frac{d r}{d t}$
$=8 \pi r(2)$
$=16 \pi r$
$\therefore \frac{d s}{d t} \propto r$

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