The volume of a sphere increases at the rate | vedclass

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(C) Let $V$ be the volume and $r$ be the radius of the sphere.Given that the rate of change of volume is $\frac{dV}{dt} = \pi 	ext{ cm}^3/	ext{s}$.T

Vedclass · Accepted Answer

$\frac{1}{16}$

Vedclass · Answer

(C) Let $V$ be the volume and $r$ be the radius of the sphere.Given that the rate of change of volume is $\frac{dV}{dt} = \pi 	ext{ cm}^3/	ext{s}$.The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.Differentiating both sides with respect to $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Substituting the given values $\frac{dV}{dt} = \pi$ and $r = 2 	ext{ cm}$:$\pi = 4 \pi (2)^2 \frac{dr}{dt}$$\pi = 16 \pi \frac{dr}{dt}$$\frac{dr}{dt} = \frac{\pi}{16 \pi} = \frac{1}{16} 	ext{ cm/s}$.Thus,the rate at which the radius increases is $\frac{1}{16} 	ext{ cm/s}$.

The volume of a sphere increases at the rate of $\pi \text{ cm}^3/\text{s}$. When the radius is $2 \text{ cm}$,the rate at which the radius increases is . . . . . . $\text{cm/s}$.

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