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(D) Let $\frac{dr}{dt} = c$ and $h = ar + b$.Given $\frac{dh}{dt} = 3 \frac{dr}{dt}$,so $a \frac{dr}{dt} = 3 \frac{dr}{dt}$,which implies $a = 3$.Thus

Vedclass · Accepted Answer

$33$

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(D) Let $\frac{dr}{dt} = c$ and $h = ar + b$.Given $\frac{dh}{dt} = 3 \frac{dr}{dt}$,so $a \frac{dr}{dt} = 3 \frac{dr}{dt}$,which implies $a = 3$.Thus,$h = 3r + b$.When $r = 1, h = 6$,so $6 = 3(1) + b$,which gives $b = 3$.Therefore,$h = 3r + 3 = 3(r + 1)$.The volume $V = \pi r^2 h = \pi r^2 (3r + 3) = 3\pi (r^3 + r^2)$.Differentiating with respect to $t$,$\frac{dV}{dt} = 3\pi (3r^2 + 2r) \frac{dr}{dt}$.When $r = 6$,$\frac{dV}{dt} = 1$,so $1 = 3\pi (3(36) + 2(6)) \frac{dr}{dt} = 3\pi (108 + 12) \frac{dr}{dt} = 360\pi \frac{dr}{dt}$.Thus,$\frac{dr}{dt} = \frac{1}{360\pi}$.When $r = 36$,$\frac{dV}{dt} = n = 3\pi (3(36)^2 + 2(36)) \frac{dr}{dt}$.$n = 3\pi (3888 + 72) \frac{1}{360\pi} = 3\pi (3960) \frac{1}{360\pi} = 3 	imes 11 = 33$.

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