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(D) Given that the rate of change of surface area $\frac{dA}{dt} = 2 	ext{ cm}^2/	ext{sec}$.The surface area of a sphere is $A = 4\pi r^2$.Different

Vedclass · Accepted Answer

$6 	ext{ cm}^3/	ext{sec}$

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(D) Given that the rate of change of surface area $\frac{dA}{dt} = 2 	ext{ cm}^2/	ext{sec}$.The surface area of a sphere is $A = 4\pi r^2$.Differentiating with respect to $t$,we get $\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$.Substituting the given values: $2 = 8\pi(6) \frac{dr}{dt} \implies 2 = 48\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{24\pi} 	ext{ cm/sec}$.The volume of a sphere is $V = \frac{4}{3}\pi r^3$.Differentiating with respect to $t$,we get $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.Substituting $r = 6$ and $\frac{dr}{dt} = \frac{1}{24\pi}$:$\frac{dV}{dt} = 4\pi(6)^2 \left(\frac{1}{24\pi}ight) = 4\pi(36) \left(\frac{1}{24\pi}ight) = \frac{144\pi}{24\pi} = 6 	ext{ cm}^3/	ext{sec}$.

The surface area of a spherical balloon is increasing at the rate of $2 \text{ cm}^2/\text{sec}$. Find the rate of increase in the volume of the balloon when the radius of the balloon is $6 \text{ cm}$.

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