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(A) Let $r$ be the radius and $V$ be the volume of the spherical bubble. The surface area $A$ is given by $A = 4\pi r^2$.Differentiating with respect

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Let $r$ be the radius and $V$ be the volume of the spherical bubble. The surface area $A$ is given by $A = 4\pi r^2$.Differentiating with respect to time $t$,we get $\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$.Given $\frac{dA}{dt} = 2 	ext{ cm}^2/	ext{s}$ and $r = 6 	ext{ cm}$,we substitute these values:$2 = 8\pi(6) \frac{dr}{dt} \Rightarrow 2 = 48\pi \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{24\pi} 	ext{ cm/s}$.The volume $V$ of the 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 $\frac{dr}{dt} = \frac{1}{24\pi}$:$\frac{dV}{dt} = 4\pi(6)^2 \left(\frac{1}{24\pi}ight) = 4\pi(36) \left(\frac{1}{24\pi}ight) = \frac{144\pi}{24\pi} = 6 	ext{ cm}^3/	ext{s}$.

The surface area of a spherical bubble increases at a rate of $2 \text{ cm}^2/\text{s}$. The rate at which the volume of the bubble increases when the radius is $6 \text{ cm}$ is . . . . . . $\text{cm}^3/\text{s}$.

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