Water flows from a tap into cylindrical container at the rate 5πcm\(^3\) per second. If the radius of the container is 3cm, calculate the level of water in ...

Water flows from a tap into cylindrical container at the rate 5πcm\(^3\) per second. If the radius of the container is 3cm, calculate the level of water in the container at the end of 9 seconds.

Answer Details

To solve the problem, we can use the formula for the volume of a cylinder: V = πr\(^2\)h where V is the volume of the water in the cylinder, r is the radius of the cylinder, and h is the height of the water in the cylinder. We know that water is flowing into the cylinder at a rate of 5π cm\(^3\) per second, which means that the volume of the water in the cylinder is increasing at a rate of 5π cm\(^3\) per second. We want to find the height of the water in the cylinder at the end of 9 seconds. Let's use h to represent the height of the water in the cylinder after 9 seconds. The increase in volume of the water in the cylinder after 9 seconds is: ΔV = 5π x 9 ΔV = 45π So, the volume of the water in the cylinder after 9 seconds is: V = πr\(^2\)h + 45π Since the radius of the cylinder is 3cm, we can substitute this value into the formula: V = π x 3\(^2\) x h + 45π V = 9πh + 45π We know that the volume of the water in the cylinder after 9 seconds is equal to the height of the water multiplied by the area of the base of the cylinder (which is πr\(^2\)): V = πr\(^2\)h 9πh + 45π = π x 3\(^2\) x h 9πh + 45π = 9πh 45π = 0 This equation has no solution, which means that something went wrong in our calculations. We made a mistake in assuming that the volume of the water in the cylinder was increasing at a constant rate of 5π cm\(^3\) per second. In reality, the rate of increase of the volume of water is not constant, because the height of the water in the cylinder is also increasing. To calculate the correct height of the water in the cylinder after 9 seconds, we need to integrate the rate of increase of the volume of water over time. The rate of increase of the volume of water at any time t is given by: dV/dt = πr\(^2\) dh/dt We know that the radius of the cylinder is 3cm and the rate of flow of water is 5π cm\(^3\) per second. So, we can write: dV/dt = π x 3\(^2\) x dh/dt 5π = 9π dh/dt dh/dt = 5/9 cm/s This means that the height of the water in the cylinder is increasing at a rate of 5/9 cm/s. We want to find the height of the water in the cylinder after 9 seconds. Let's use h to represent the height of the water in the cylinder after 9 seconds. The increase in height of the water in the cylinder after 9 seconds is: Δh = (5/9) x 9 Δh = 5 So, the height of the water in the cylinder after 9 seconds is: h = 5 + 0 h = 5cm Therefore, the correct answer is 5cm.