Question

A cylinder with a height of 5ft and a base radius of 10in is filled with water. The water is being drained out at a rate of 3 cubic inches per minute. How fast is the water level decreasing?

Answer 1

To find the water level decrease rate in a cylinder, convert the radius to feet, use the given rate of volume decrease, apply the formula for the volume of a cylinder, and solve for the change in height over time. The water level is decreasing at approximately 0.000796 feet per minute.

Step-by-step explanation:

To find out how fast the water level in a cylinder is decreasing, we can use the formula for the volume of a cylinder, which is V = πr^2h, where V is the volume, r is the radius, and h is the height. We're given the rate at which the volume is decreasing, which is 3 cubic inches per minute, and we want to find the rate of the height decrease in feet per minute.

First, convert the base radius from inches to feet. There are 12 inches in a foot: 10 inches ÷ 12 inches/foot = 0.833 feet.

With a volume decrease rate of 3 cubic inches per minute: 3 cubic inches/minute × (1/1728 cubic feet/cubic inch) = 0.001736 cubic feet/minute. This is because there are 12^3, or 1728, cubic inches in one cubic foot.

Now, using the volume formula and solving for dh/dt, the rate of height decrease:

dV/dt = πr^2 × dh/dt

0.001736 = π(0.833)^2 × dh/dt

/dh/dt = 0.001736 ÷ (0.833)^2π feet per minute

After calculating, the rate at which the water level is decreasing is approximately 0.000796 feet per minute.

Answer 2

The water level in the cylinder is decreasing at a rate of approximately 3/100π inches per minute.

Calculating the cross-sectional area: You correctly calculated the area of the cylinder's base using the formula πr². Plugging in the radius of 10 inches, we get:

Area = π * (10 inches)^2

Area = 100π square inches

Relating volume drained and area to water level decrease: While it's true that the water level decrease is proportional to the volume drained divided by the base area, the actual relation involves the rate of change (derivative) of these quantities.

Let:

• V(t) be the volume of water in the cylinder at time t
• h(t) be the height of the water level at time t
• r be the base radius of the cylinder (10 inches)
• Q be the rate of water draining (3 cubic inches per minute)

The volume of the water can be expressed as:

V(t) = πr² * h(t)

Taking the derivative of both sides with respect to time t, we get:

dV/dt = πr² * dh/dt

Since the rate of water draining is Q, we have:

dV/dt = -Q

(Negative sign indicates the water is decreasing)

Now, we can relate the rate of volume change to the base area and water level decrease:

dh/dt = -Q / (πr²)

Substituting values and simplifying:

dh/dt = -3 cubic inches/minute / (100π square inches)

dh/dt ≈ 3 / (100π) inches/minute

Therefore, the rate of water level decrease is approximately 3/100π inches per minute.