Question

Water is being drained from a container which has the shape of an inverted right circular cone. The container has a radius of 7.00 inches at the top and a height of 9.00 inches. At the instant when the water in the container is 8.00 inches deep, the surface level is falling at a rate of 1.4 in./sec. Find the rate at which water is being drained from the container.

Answer 1

Answer:

Explanation:

Given

radius of cone is $R=7 in.$

height of cone $H=9 in.$

rate of change of height is $\frac{\mathrm{d} h}{\mathrm{d} t}=1.4 in./s$

From similar triangle Property at any instant where radius is r and height is h

$\frac{r}{7}=\frac{h}{9}$

differentiating we get

$9\times \frac{\mathrm{d} r}{\mathrm{d} t}=7\times \frac{\mathrm{d} h}{\mathrm{d} t}$

$\frac{\mathrm{d} h}{\mathrm{d} t}=\frac{7}{9}\times 1.4=1.08 in./s$

Volume is given by

$V=\frac{1}{3}\pi r^2h$

Differentiate

$\frac{\mathrm{d} V}{\mathrm{d} t}=\frac{\pi }{3}\left [ 2r\frac{\mathrm{d} r}{\mathrm{d} t}\cdot h+\frac{\mathrm{d} h}{\mathrm{d} t}\cdot r^2\right ]$

$\frac{\mathrm{d} V}{\mathrm{d} t}=\frac{\pi }{3}\left [ 2\times 6.22\times 1.08+1.4\times 6.22^2\right ]$

$\frac{\mathrm{d} V}{\mathrm{d} t}=70.803 in.^3/s$