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A tank of water has base a circle of radius 2 meters and vertical sides. If water leaves the tank at a rate of 5 liters per minute, how fast is the water level falling in centimeters per hour? (Remember that 1 liter is 1000 cubic centimeters.)

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lublana

Answer:

0.0398cm/min

Step-by-step explanation:

We are given that

Radius of base circle=r=2 m=[tex]2\times 100=200 cm[/tex]

1 m=100 cm

[tex]\frac{dV}{dt}=5L/minute=5000cm^3/min[/tex]

1 L=1000 cubic cm

We know that

Volume of tank=[tex]\pi r^2 h[/tex]

Differentiate w.r.t t

[tex]\frac{dV}{dt}=\pi r^2\frac{dh}{dt}[/tex]

Substitute the values

[tex]5000=\pi(200)^2\frac{dh}{dt}[/tex]

[tex]\frac{dh}{dt}=\frac{5000}{\pi(200)^2}=0.0398cm/min[/tex]

abidemiokin

The rate at which the water is falling is [tex]\frac{5000}{4 \pi} cm/hr [/tex]

The volume of a cylinder

The formula for calculating the volume of a cylinder is expressed as:

  • [tex]V=\pi r^2h[/tex]

Take the differential

  • [tex]\frac{dV}{dt} =\pi r^2 \frac{dr}{dt} [/tex]

Given the following parameters

  • dV/dt = 5L/min
  • r = 2 meters

Substitute the given parameters into the formula

[tex]5000 = \pi (2)^2 \frac{dr}{dt} \\ \frac{dr}{dt} =\frac{5000}{4 \pi} cm/hr[/tex]

Hence the rate at which the water is falling is [tex]\frac{5000}{4 \pi} cm/hr [/tex]

learn more on differentials here: https://brainly.com/question/18760518

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