Water flows into a right cylindrical shaped swimming pool with a circular base at a rate of 4 m33/min. The radius of the base is 3 m. How fast is the water level rising inside the swimming pool? The volume of a right cylinder with a circular base is V=πr2hV=πr2h, where rr is the radius of the base and hh is the height of the cylinder.

The water level is rising at a rate of approximately 0.1415 meters per minute.

Step-by-step explanation:

Water is flowing into a right cylindrical-shaped swimming pool at a rate of 4 cubic meters per minute. The radius of the base is 3 meters.

And we want to determine the rate at which the water level of the pool is rising.

Recall that the volume of a cylinder is given by:

Since the radius is a constant 3 meters:

Water is flowing at a rate of 4 cubic meters per minute. In other words, dV/dt = 4 m³ / min.

Take the derivative of both sides with respect to t:

Implicitly differentiate:

The rate at which the water level is rising is represented by dh/dt. Substitute and solve:

Therefore:

In conclusion, the water level is rising at a rate of approximately 0.1415 meters per minute.

step-by-step explanation:

i found the koficent of the linear function and i foun c acordind to the strict formula y=kx+c and then i took 2 as an x and replacet it on the formula i found and then foun the y