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(A) Given that, the rate of change of volume is $\frac{dV}{dt} = 30 \ ft^3 / min$ and the radius is $r = 15 \ ft$. The volume of a sphere is given by

Vedclass · Accepted Answer

$\frac{1}{30 \pi} \ ft / min$

Vedclass · Answer

(A) Given that, the rate of change of volume is $\frac{dV}{dt} = 30 \ ft^3 / min$ and the radius is $r = 15 \ ft$. The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$. Differentiating both sides with respect to time $t$, we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$. Substituting the given values: $30 = 4 \pi (15)^2 \frac{dr}{dt}$. $30 = 4 \pi (225) \frac{dr}{dt} = 900 \pi \frac{dr}{dt}$. Therefore, $\frac{dr}{dt} = \frac{30}{900 \pi} = \frac{1}{30 \pi} \ ft / min$.

Gas is being pumped into a spherical balloon at the rate of $30 \ ft^3 / min$. The rate at which the radius increases when it reaches the value $15 \ ft$, is:

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