MHB 2214.16 Related Rates With Draining Cone

[karush]

Water is being drained from a container which has the shape of an inverted right circular cone The container has a radius of $6 in$ at the top and a length of $8 in$ at the bottom. when the water in the container is $6 in$ deep, the surface level is falling at a rate of $0.9 in/sec$ find the rate at which water is being drained from the container.

 

[karush]

$$
\begin{align*}\displaystyle
V&=\frac{1}{3}\pi r^2 h\\
\frac{6}{8}&=\frac{r}{h}\\
\therefore \, r&=\frac{3}{4}h\\
V&=\frac{9\pi}{16} h^3\\
\frac{dV}{dt}&=\frac{27\pi}{16} h^2 \frac{dh}{dt}\\
\frac{dV}{dt}&=\frac{27\pi}{16}(6^2)(0.9)\\
&=171.76 \frac{in^3}{sec} \\
0r&= 54.675 \pi \frac{in^3}{sec}
\end{align*}
$$

hopefully

 

[MarkFL]

Water is being drained from a container which has the shape of an inverted right circular cone The container has a radius of $6 in$ at the top and a length of $8 in$ at the bottom. when the water in the container is $6 in$ deep, the surface level is falling at a rate of $0.9 in/sec$ find the rate at which water is being drained from the container.

Let $R$ be the radius of the conical container, $H$ be its height, $r$ be the radius of the conical volume of water, and $h$ be the height of the volume of water.

By similarity, we know:

$$\frac{R}{H}=\frac{r}{h}\implies r=\frac{hR}{H}$$

The formula for the volume of the water is:

$$V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi \left(\frac{hR}{H}\right)^2h=\frac{\pi R^2}{3H^2}h^3$$

Differentiate w.r.t time $t$:

$$\d{V}{t}=\frac{\pi R^2}{H^2}h^2\d{h}{t}$$

We are given:

$$R=6\text{ in},\,H=8\text{ in},\,h=6\text{ in},\,\d{h}{t}=-\frac{9}{10}\,\frac{\text{in}}{\text{s}}$$

And so, for this problem, we find:

$$\d{V}{t}=\frac{\pi\left(6\text{ in}\right)^2}{\left(8\text{ in}\right)^2}\left(6\text{ in}\right)^2\left(-\frac{9}{10}\,\frac{\text{in}}{\text{s}}\right)=-\frac{729\pi}{40}\,\frac{\text{in}^3}{\text{s}}\approx -57.3\,\frac{\text{in}^3}{\text{s}}$$

You used the correct formula for a cone, but when you wrote:

$$V=\frac{9\pi}{16}h^3$$

You neglected the 1/3 in the formula...you should have written:

$$V=\frac{3\pi}{16}h^3$$

Other than that minor error, the method you used itself was correct. :)

 

[HallsofIvy]

$$
\begin{align*}\displaystyle
V&=\frac{1}{3}\pi r^2 h\\
\frac{6}{8}&=\frac{r}{h}\\
\therefore \, r&=\frac{3}{4}h\\
V&=\frac{9\pi}{16} h^3\\$$

According to your first line [math]V= \frac{1}{3}\pi r^2h[/math]
So this should be
[math]V= \frac{1}{3}\pi\left(\frac{9}{16}h^2\right)h= \frac{3}{16}h^3[/math]

[math]\frac{dV}{dt}&=\frac{27\pi}{16} h^2 \frac{dh}{dt}\\
\frac{dV}{dt}&=\frac{27\pi}{16}(6^2)(0.9)\\
&=171.76 \frac{in^3}{sec} \\
0r&= 54.675 \pi \frac{in^3}{sec}
\end{align*}
[/math]

hopefully