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svetlana [45]
3 years ago
11

Water is withdrawn from a conical reservoir, 8 feet in diameter and 10 feet deep (vertex down) at the constant rate of 5 ft³/min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).

Mathematics
1 answer:
denis23 [38]3 years ago
3 0

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.

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