Question

water is draining from the bottom of a cone - shaped funnel at the rate of 0.8 ft³/sec. the height of the funnel is 6 ft and the radius at the top of the funnel is 5 ft, as shown in the image below. at what rate is the height of the water in the funnel changing when the height of the water is 4 ft? recall that the formula for the volume of a cone is $v = \frac{1}{3}\pi r^{2}h$.

Explanation:

Step1: Establish ratio of radius to height

The cone has a height of $6$ ft and radius of $5$ ft at the top. For similar - triangles (the cross - sections of the cone of water at any time and the whole cone), the ratio of radius $r$ to height $h$ is constant. So, $\frac{r}{h}=\frac{5}{6}$, which gives $r = \frac{5}{6}h$.

Step2: Substitute $r$ into volume formula

The volume formula of a cone is $V=\frac{1}{3}\pi r^{2}h$. Substitute $r=\frac{5}{6}h$ into it: $V=\frac{1}{3}\pi(\frac{5}{6}h)^{2}h=\frac{1}{3}\pi\frac{25}{36}h^{2}\cdot h=\frac{25\pi}{108}h^{3}$.

Step3: Differentiate with respect to time $t$

Differentiate both sides of the equation $V = \frac{25\pi}{108}h^{3}$ with respect to $t$ using the chain rule. $\frac{dV}{dt}=\frac{25\pi}{108}\times3h^{2}\frac{dh}{dt}=\frac{25\pi}{36}h^{2}\frac{dh}{dt}$.

Step4: Solve for $\frac{dh}{dt}$

We know that $\frac{dV}{dt}=- 0.8$ (negative because the volume is decreasing) and $h = 4$. Substitute these values into the equation $\frac{dV}{dt}=\frac{25\pi}{36}h^{2}\frac{dh}{dt}$:
\[

$$\begin{align*} -0.8&=\frac{25\pi}{36}\times4^{2}\times\frac{dh}{dt}\\ -0.8&=\frac{25\pi}{36}\times16\times\frac{dh}{dt}\\ -0.8&=\frac{400\pi}{36}\times\frac{dh}{dt}\\ \frac{dh}{dt}&=\frac{-0.8\times36}{400\pi}\\ \frac{dh}{dt}&=\frac{-28.8}{400\pi}\\ \frac{dh}{dt}&=-\frac{9}{125\pi}\text{ ft/sec} \end{align*}$$

\]

Answer:

$-\frac{9}{125\pi}\text{ ft/sec}$