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(B) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$. Given, the rate of discharge of air is $\frac{dV}{dt}

Vedclass · Accepted Answer

$1 \,m^2 / min$

Vedclass · Answer

(B) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$. Given, the rate of discharge of air is $\frac{dV}{dt} = -4 \,m^3 / min$ (since air is discharging, volume is decreasing). 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 at $r = 8 \,m$: $-4 = 4 \pi (8)^2 \frac{dr}{dt} \Rightarrow -4 = 256 \pi \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = -\frac{1}{64 \pi} \,m / min$. 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 $r = 8 \,m$ and $\frac{dr}{dt} = -\frac{1}{64 \pi} \,m / min$: $\frac{dS}{dt} = 8 \pi (8) \left( -\frac{1}{64 \pi} \right) = -1 \,m^2 / min$. Since the surface area is shrinking at a rate of $1 \,m^2 / min$, the rate of shrinking is $1 \,m^2 / min$.

Air is discharging from a large spherical balloon at the rate of $4 \,m^3 / min$. The rate at which the surface area is shrinking when the radius of the balloon is $8 \,m$, is

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