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(C) Let $v$ be the volume and $s$ be the surface area of the spherical balloon with radius $r$. Given: $\frac{dv}{dt} = 30 \ cm^3/min$. The volume of

Vedclass · Accepted Answer

$10 \ cm^2/min$

Vedclass · Answer

(C) Let $v$ be the volume and $s$ be the surface area of the spherical balloon with radius $r$. Given: $\frac{dv}{dt} = 30 \ cm^3/min$. The volume of a sphere is $v = \frac{4}{3} \pi r^3$. Differentiating with respect to $t$: $\frac{dv}{dt} = 4 \pi r^2 \frac{dr}{dt}$. Substituting the given values: $30 = 4 \pi r^2 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{30}{4 \pi r^2} = \frac{15}{2 \pi r^2}$. The surface area of a sphere is $s = 4 \pi r^2$. Differentiating with respect to $t$: $\frac{ds}{dt} = 8 \pi r \frac{dr}{dt}$. Substituting $\frac{dr}{dt}$: $\frac{ds}{dt} = 8 \pi r \left( \frac{15}{2 \pi r^2} \right) = \frac{60}{r}$. For $r = 6 \ cm$: $\frac{ds}{dt} = \frac{60}{6} = 10 \ cm^2/min$.

The volume of a spherical balloon is increasing at the rate of $30 \ cm^3/min$. Find the rate of change of the surface area of the balloon when its radius is $6 \ cm$.

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