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(A) Let $r$ be the radius of the spherical balloon. The volume $V$ of the sphere is given by $V = \frac{4}{3}\pi r^3$.Differentiating with respect to

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$10$

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(A) Let $r$ be the radius of the spherical balloon. The volume $V$ of the sphere is given by $V = \frac{4}{3}\pi r^3$.Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.Given $\frac{dV}{dt} = 35 \, cm^3/min$,we have $35 = 4\pi r^2 \frac{dr}{dt}$,which implies $\frac{dr}{dt} = \frac{35}{4\pi r^2} \quad (1)$.The surface area $S$ of the sphere is $S = 4\pi r^2$.Differentiating with respect to $t$,we get $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.Substituting the value of $\frac{dr}{dt}$ from $(1)$ into the expression for $\frac{dS}{dt}$:$\frac{dS}{dt} = 8\pi r \left( \frac{35}{4\pi r^2} \right) = \frac{70}{r}$.Given the diameter is $14 \, cm$,the radius $r = 7 \, cm$.Substituting $r = 7$ into the expression for $\frac{dS}{dt}$:$\frac{dS}{dt} = \frac{70}{7} = 10 \, cm^2/min$.

$A$ spherical balloon is being inflated at the rate of $35 \, cm^3/min$. The rate of increase in the surface area (in $cm^2/min$) of the balloon when its diameter is $14 \, cm$,is

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