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(A) The volume of a sphere $(V)$ with radius $(r)$ is given by $V = \frac{4}{3} \pi r^3$. Differentiating with respect to time $(t)$,we get the rate o

Vedclass · Accepted Answer

$\frac{1}{\pi} \, cm/s$

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(A) The volume of a sphere $(V)$ with radius $(r)$ is given by $V = \frac{4}{3} \pi r^3$. Differentiating with respect to time $(t)$,we get the rate of change of volume: $\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \frac{dr}{dt}$. Given that $\frac{dV}{dt} = 900 \, cm^3/s$,we substitute this into the equation: $900 = 4 \pi r^2 \frac{dr}{dt}$. Solving for $\frac{dr}{dt}$: $\frac{dr}{dt} = \frac{900}{4 \pi r^2} = \frac{225}{\pi r^2}$. When the radius $r = 15 \, cm$: $\frac{dr}{dt} = \frac{225}{\pi (15)^2} = \frac{225}{225 \pi} = \frac{1}{\pi} \, cm/s$. Thus,the rate at which the radius increases is $\frac{1}{\pi} \, cm/s$.

$A$ balloon,which always remains spherical on inflation,is being inflated by pumping in $900 \, cm^3$ of gas per second. Find the rate at which the radius of the balloon increases when the radius is $15 \, cm$.

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