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A water tank in the shape of an inverted circular cone has a base radius of 3 m and height of 7 m . If water is being pumped into the tank at a rate of 4.3 m^3 / min , find the rate at which the water level is rising when the water is 3.5 m deep. (Round your answer to three decimal places if required)

Sagot :

MrRoyal

The water level increases by 0.608 meters per minute when the water is 3.5 m deep

How to determine the rate?

The given parameters are:

  • Radius, r = 3
  • Height, h = 7
  • Rate in, V' = 4.3m^3/min

The relationship between the radius and height is:

r/h = 3/7

Make r the subject

r = 3h/7

The volume of a cone is;

[tex]V = \frac 13\pi r^2h[/tex]

This gives

[tex]V = \frac 13\pi (\frac{3h}{7})^2h[/tex]

Expand

[tex]V = \frac{3h^3}{49}\pi[/tex]

Differentiate

[tex]V' = \frac{9h^2}{49}\pi h'[/tex]

Make h' the subject

[tex]h' = \frac{49}{9\pi h^2}V'[/tex]

When the water level is 3.5.

We have:

[tex]h' = \frac{49}{9\pi * 3.5^2}V'[/tex]

Also, we have:

V' = 4.3

So, the equation becomes

[tex]h' = \frac{49}{9\pi * 3.5^2} * 4.3[/tex]

Evaluate the products

[tex]h' = \frac{210.7}{346.36}[/tex]

Evaluate the quotient

h' = 0.608

Hence, the water level increases by 0.608 meters per minute

Read more about volumes at:

https://brainly.com/question/10373132

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