A vertical cylinder is leaking water at a rate of 4m3/sec. If the cylinder has a height of 10m and a radius of 2m, at what rate is the height of the water changing when the height is 3m?

The height of the water is changing at a rate of approximately 0.318 meters per second.

The volume formula of the cylinder (), in cubic meters, is described below:

(1)

Where:

According to the statement, the leakage is represented by a decrease in the height of the water. By differential calculus and under the assumption that no rate of change exists for the radius, we have the expression for the rate of change in the height of the water within the cylinder (), in meters per second:

(2)

Where is the rate change in the volume occupied by the water, in cubic meters per second.

If we know that and , then the rate of change in the height is:

The height of the water is changing at a rate of approximately 0.318 meters per second.

We kindly invite to check this question on rates of change: link

Therefore the rate change of height is   m/s.

Step-by-step explanation:

Given that a vertical cylinder is leaking water at rate of 4 m³/s.

It means the rate change of volume is 4 m³/s.

The radius of the cylinder remains constant with respect to time, but the height of the water label changes with respect to time.

The height of the cylinder be h(say).

The volume of a cylinder is

Differentiating with respect to t.

Putting the value

The rate change of height does not depend on the height.

Therefore the rate change of height is   m/s.

in problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. to find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

hope it