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(A) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$. Given: $\frac{dV}{dt} = 2 \ cm^3/sec$. We know that th

Vedclass · Accepted Answer

$1$

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(A) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$. Given: $\frac{dV}{dt} = 2 \ cm^3/sec$. We know that 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: $2 = 4 \pi r^2 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{2 \pi r^2}$. 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 $\frac{dr}{dt} = \frac{1}{2 \pi r^2}$ into the equation for $\frac{dS}{dt}$: $\frac{dS}{dt} = 8 \pi r \left( \frac{1}{2 \pi r^2} \right) = \frac{4}{r}$. At $r = 4 \ cm$, the rate of change of surface area is $\frac{dS}{dt} = \frac{4}{4} = 1 \ cm^2/sec$. Thus, the correct option is $A$.

The volume of a spherical balloon is increasing at the rate of $2 \ cm^3/sec$. When its radius is $4 \ cm$, the rate of change of its surface area (in $cm^2/sec$) is

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