Question

Water is withdrawn from a conical reservoir, 8 feet in diameter and 10 feet deep (vertex down) at the constant rate of 5 ft³/min. How fast is the water level falling when the depth of the water in the reservoir is 5 ft? ($V = \frac{1}{3} \pi r^2h$).

Answer 1

Answer:

Water level in the reservoir is falling at the rate of 0.398 ft per minute.

Step-by-step explanation:

From the figure attached,

Water level in the reservoir has been given as 10 feet and radius of the reservoir is 4 feet.

Let the level of water in the reservoir after time t is h where radius of the water level becomes r.

ΔABE and ΔCDE are similar.

Therefore, their corresponding sides will be in the same ratio.

$\frac{r}{h}=\frac{4}{10}$

$r=\frac{2}{5}h$ --------(1)

Now volume of the water V = $\frac{1}{3}\pi r^{2}h$

From equation (1),

V = $\frac{1}{3}\pi (\frac{2}{5}h)^{2} h$

V = $\frac{4\pi h^{2}\times h}{75}$

$\frac{dV}{dt}=\frac{4\pi }{75}\times \frac{d}{dt}(h^{3})$

$\frac{dV}{dt}=\frac{4\pi }{75}\times (3h^{2})\frac{dh}{dt}$

$\frac{dV}{dt}=\frac{12\pi h^{2}}{75}\times \frac{dh}{dt}$

Since $\frac{dV}{dt}=5$ feet³ per minute and h = 5 feet

$5=\frac{12\pi (5)^{2}}{75}\times \frac{dh}{dt}$

$5=4\pi \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{5}{4\pi}$

$\frac{dh}{dt}=0.398$ feet per minute

Therefore, water level in the reservoir is falling at the rate of 0.398 feet per minute.