Question

water is being pumped into an inverted conical tank at some constant rate. the tank has a height of 600 cm and the diameter of the tank across the top is 400 cm. if the water level is rising at a rate of 20 cm/min when the height of the water is 200 cm, find the rate at which water is being pumped into the tank.

Answer 1

It is found that water is being pumped at a rate of 838095.24 cm³/min using implicit differentiation.

A conical tank's volume (V) is calculated as follows:

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

where the height is h and the radius is r.

When we use implicit differentiation, we get the following:

$\frac{dV}{dt} = \frac{\pi }{3} (2rh \frac{dr}{dt} + r^{2} \frac{dh}{dt} )$

We have that:

Height of 600 cm, thus h = 600 cm

Diameter of 400 cm, thus r = $\frac{400}{2}$ = 200 cm

Water level rising at a rate of 20 cm/min, thus $\frac{dh}{dt} = 20 cm/min$

Since the radius is constant, $\frac{dr}{dt} = 0$

Pumping water into the tank happens at a rate $\frac{dV}{dt}$ of

$\frac{dV}{dt} = \frac{\pi }{3} (2rh \frac{dr}{dt} + r^{2} \frac{dh}{dt} )$

$\frac{dV}{dt} = \frac{\pi }{3} ((2 * 200 *600 * 0) + (200)^{2} * 20)$

$\frac{dV}{dt} = \frac{\pi }{3} ( 0 + 40000 * 20)$

$\frac{dV}{dt} = \frac{\pi }{3} (800000)$

$\frac{dV}{dt} = \frac{22 }{7*3} (800000)$

$\frac{dV}{dt} = \frac{22 }{21} (800000)$

$\frac{dV}{dt} = 838095.24\; cm^{3}/min$

Therefore, the rate of water pumping into the tank is 838095.24 cm³/min.

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