The radius $r$ of a right circular cylinder increases at $0.2\text{ cm/s}$ and its height $h$ decreases at $0.1\text{ cm/s}$. At the instant when $r = 5\text{ cm}$ and $h = 10\text{ cm}$, the rate of change of volume $\frac{dV}{dt}$ in $\text{cm}^3/\text{s}$ is $k \pi$. The value of $k$ is ________ (round off to one decimal place).
Answer Key:
17.5
Step-by-Step Solution & Explanation
The volume of a cylinder is $V = \pi r^2 h$.
Using the total derivative with respect to time $t$:
$\frac{dV}{dt} = \frac{\partial V}{\partial r} \frac{dr}{dt} + \frac{\partial V}{\partial h} \frac{dh}{dt} = 2\pi r h \frac{dr}{dt} + \pi r^2 \frac{dh}{dt}$
Given: $r = 5$, $h = 10$, $\frac{dr}{dt} = 0.2$, and $\frac{dh}{dt} = -0.1$:
$\frac{dV}{dt} = 2\pi(5)(10)(0.2) + \pi(5^2)(-0.1) = \pi [20 - 2.5] = 17.5 \pi$
Therefore, $k = 17.5$.