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(C) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$.Given that $\frac{dV}{dt} = 40 \ cm^3/\min$.The volume

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$.Given that $\frac{dV}{dt} = 40 \ cm^3/\min$.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 the given values: $40 = 4\pi (8)^2 \frac{dr}{dt} \implies 40 = 256\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{40}{256\pi} = \frac{5}{32\pi} \ cm/\min$.The surface area of a sphere is $S = 4\pi r^2$.Differentiating with respect to $t$,we get $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.Substituting $r = 8$ and $\frac{dr}{dt} = \frac{5}{32\pi}$:$\frac{dS}{dt} = 8\pi 	imes 8 	imes \frac{5}{32\pi} = 64\pi 	imes \frac{5}{32\pi} = 2 	imes 5 = 10 \ cm^2/\min$.

The volume of a spherical balloon is increasing at the rate of $40 \ cm^3/\min$. The rate of change of the surface area of the balloon at the instant when its radius is $8 \ cm$ is ........ $cm^2/\min$.

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