A swimming pool is a rectangular solid 15 ft wide, 40 ft long, and at most 4 ft deep. Water is being added to the pool at the rate of 25 cubic ft per minute. How fast is the water rising when there is 1800 cubic ft of water in the pool?

Question

A swimming pool is a rectangular solid 15 ft wide, 40 ft long, and at most 4 ft deep. Water is being added to the pool at the rate of 25 cubic ft per minute. How fast is the water rising when there is 1800 cubic ft of water in the pool?

Answer 1

since the pool is a rectangular solid, it has a cross-section of constant size.

the pool is 15x40 = 600 ft^2

Thus, each ft of depth adds 600ft^3 of volume.

So, at 25ft^3/min, the height rises 25/600 = .0417 ft/min

Now, since this is for calculus class, they must want a solution involving related rates.

v = 40*15*x = 600x when the water is x feet deep.

dv/dt = 600 dx/dt
25 = 600 dx/dt
dx/dt = 25/600 = .0417

Answer 2

To solve this problem, we can use the concept of related rates. First, let's set up the equation for the dimensions of the pool:

Width (w) = 15 ft
Length (l) = 40 ft
Depth (h) = 4 ft (at most)

Since the pool is a rectangular solid, we can use the formula for volume to calculate the amount of water in the pool:

Volume (V) = length × width × depth

Given that the pool is filling up at a rate of 25 cubic ft per minute, we can express the rate of change of the volume with respect to time as:

dV/dt = 25 ft³/min

Now, we need to find the rate at which the water level is rising (dh/dt) when the volume is 1800 cubic ft. To do this, we differentiate the volume equation with respect to time:

dV/dt = d/dt (lwh)

Since we want to find dh/dt, we need to determine how the height (h) changes with time (dt). To do this, we can divide both sides of the volume equation by h:

V/h = l × w

Differentiating both sides of the equation with respect to time (dt), we get:

dV/dt × (1/h) = d/dt (l × w)

Now, let's substitute the given values l = 40 ft, w = 15 ft, and the known rate of change dV/dt = 25 ft³/min into the equation:

25 ft³/min × (1/h) = d/dt (40 ft × 15 ft)

We are interested in finding dh/dt when V = 1800 cubic ft. So, we need to find h when V = 1800. Rearranging the equation, we get:

h = V / (l × w)

Substituting the given values V = 1800 cubic ft, l = 40 ft, and w = 15 ft into the equation:

h = 1800 ft³ / (40 ft × 15 ft)

h ≈ 3 ft

Now that we know the height (h) is approximately 3 ft, we can substitute this value back into the equation:

25 ft³/min × (1/3 ft) = dh/dt

Simplifying the equation:

dh/dt = 25/3 ft/min

Therefore, when there is 1800 cubic ft of water in the pool, the water level is rising at a rate of approximately 25/3 ft/min.