A coffeepot has the shape of a cylinder with radius 5 inches, as shown in the figure above. Let $h$ be the depth of the coffee in the pot, measured in inches, where $h$ is a function of time $t$, measured in seconds. The volume $V$ of coffee in the pot is changing at the rate of $-5\pi\sqrt{h}$ cubic inches per second. (The volume $V$ of a cylinder with radius $r$ and height $h$ is $V = \pi r^2 h$.) (a) Show that $\dfrac{dh}{dt} = -\dfrac{\sqrt{h}}{5}$. (b) Given that $h = 17$ at time $t = 0$, solve the differential equation $\dfrac{dh}{dt} = -\dfrac{\sqrt{h}}{5}$ for $h$ as a function of $t$. (c) At what time $t$ is the coffeepot empty?
A coffeepot has the shape of a cylinder with radius 5 inches, as shown in the figure above. Let $h$ be the depth of the coffee in the pot, measured in inches, where $h$ is a function of time $t$, measured in seconds. The volume $V$ of coffee in the pot is changing at the rate of $-5\pi\sqrt{h}$ cubic inches per second. (The volume $V$ of a cylinder with radius $r$ and height $h$ is $V = \pi r^2 h$.)\\
(a) Show that $\dfrac{dh}{dt} = -\dfrac{\sqrt{h}}{5}$.\\
(b) Given that $h = 17$ at time $t = 0$, solve the differential equation $\dfrac{dh}{dt} = -\dfrac{\sqrt{h}}{5}$ for $h$ as a function of $t$.\\
(c) At what time $t$ is the coffeepot empty?