my circular swimming pool has a 29 foot diameter and is 8 feet tall. at the beginning of the summer i need to fill the pool with water. the flow rate from my spigot into the pool is about 0.75 cubic feet per minute. how fast is the water level, h, rising at the moment when the water level is at a height is 3.7 feet. Plus find the equation relating the two changing quantities.
The pool has Volume = π * 29^2 * 8/4 = 1682π ft^3
However, the water h feet deep has volume π * 29^2 * 1/4 * h ft^3 = 210.25πh ft^3
V = 841πh/4
dV = 841π/4 dh
0.75 = 841π/4 dh
dh = 0.00113544 ft/min
This rate does not change, since the pool has a circular cross-section, and the radius is constant.
Not only that, it will take 8/0.00028386 = 7045 min to fill the pool. That's 4.9 days!
To find the rate at which the water level is rising when it is at a height of 3.7 feet, we can use similar triangles and the concept of related rates.
Let's start by setting up the equation that relates the changing quantities. The volume of the pool, V, can be expressed as the product of its height, h, and the area of its circular base, A:
V = A * h
The area of a circle is given by the formula:
A = π * r^2
where r is the radius. Since the diameter of your pool is 29 feet, the radius can be calculated by dividing the diameter by 2:
r = 29 ft / 2 = 14.5 ft
So, the equation relating the volume and height of the pool is:
V = π * (14.5 ft)^2 * h
Now, let's differentiate both sides of this equation with respect to time, t, and apply the chain rule to find the rate at which the water level is rising, dh/dt:
dV/dt = π * (14.5 ft)^2 * dh/dt
The rate at which the volume is changing can be given by the flow rate from the spigot, which is 0.75 cubic feet per minute. So, dV/dt can be substituted:
0.75 ft^3/min = π * (14.5 ft)^2 * dh/dt
Now, we can solve for dh/dt:
dh/dt = (0.75 ft^3/min) / (π * (14.5 ft)^2)
To find the value of dh/dt when the water level is at a height of 3.7 feet, we substitute h = 3.7 ft into the equation:
dh/dt = (0.75 ft^3/min) / (π * (14.5 ft)^2) ≈ 0.0011 ft/min
Therefore, the water level is rising at a rate of approximately 0.0011 feet per minute when the water level is at a height of 3.7 feet.
The equation relating the changing quantities is:
dh/dt = (0.75 ft^3/min) / (π * (14.5 ft)^2)