Question

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)

Answer 1

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;

$V = \frac 13\pi r^2h$

This gives

$V = \frac 13\pi (\frac{3h}{7})^2h$

Expand

$V = \frac{3h^3}{49}\pi$

Differentiate

$V' = \frac{9h^2}{49}\pi h'$

Make h' the subject

$h' = \frac{49}{9\pi h^2}V'$

When the water level is 3.5.

We have:

$h' = \frac{49}{9\pi * 3.5^2}V'$

Also, we have:

V' = 4.3

So, the equation becomes

$h' = \frac{49}{9\pi * 3.5^2} * 4.3$

Evaluate the products

$h' = \frac{210.7}{346.36}$

Evaluate the quotient

h' = 0.608

Hence, the water level increases by 0.608 meters per minute

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