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(A) The volume of a cylinder is given by $V = \pi r^2 h$. Since the volume is constant,its derivative with respect to time $t$ is zero: $\frac{dV}{dt}

Vedclass · Accepted Answer

$6$

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(A) The volume of a cylinder is given by $V = \pi r^2 h$. Since the volume is constant,its derivative with respect to time $t$ is zero: $\frac{dV}{dt} = 0$.Applying the product rule to $V = \pi r^2 h$,we get: $\frac{dV}{dt} = \pi (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}) = 0$.Dividing by $\pi r$,we have $2h \frac{dr}{dt} + r \frac{dh}{dt} = 0$.Given $\frac{dr}{dt} = 5 \ cm/min$,$r = 5 \ cm$,and $h = 3 \ cm$,we substitute these values:$2(3)(5) + 5 \frac{dh}{dt} = 0$.$30 + 5 \frac{dh}{dt} = 0$.$5 \frac{dh}{dt} = -30$.$\frac{dh}{dt} = -6 \ cm/min$.The negative sign indicates a decrease. Therefore,the rate of decrease of the height is $6 \ cm/min$.

The radius of a cylinder is increasing at the rate of $5 \ cm/min$,so that its volume remains constant. When its radius is $5 \ cm$ and height is $3 \ cm$,the rate of decrease of its height is .......... $cm/min$.

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