A trough (a place where horses and sheep drink water) has the shape of a half-cylinder with length 4 m and radius 20 cm. Water is poured into the trough at a constant rate of 300 cm3/s. At what rate (in cm/s) is the level of the water in the trough rising at the moment the level of the water is 10 cm below the upper surface of the trough?

Question

A trough (a place where horses and sheep drink water) has the shape of a half-cylinder with length 4 m and radius 20 cm. Water is poured into the trough at a constant rate of 300 cm3/s. At what rate (in cm/s) is the level of the water in the trough rising at the moment the level of the water is 10 cm below the upper surface of the trough?

Answer 1

when the water is y down from the top, the surface subtends an angle θ, so the cross-section has area

1/2 r^2(θ - sinθ)

where cosθ = y/r

So, plugging in our numbers,

v = (400)(1/2)(20^2)(arccos(y/20)-√(20^2-y^2))
v = 800(arccos(y/20)-√(400-y^2))
so,

dv/dt = 800(y-1)/√(400-y^2) dy/dt

Now, when y=10, we have

300 = 7200/√300 dy/dt
dy/dt = √300/7200 = √3/720 = 0.0024 cm/s

seems kinda small, right? Let's see whether is is. When the water is 10 cm deep, the surface has an area of 400*10√3 = 6928 cm^2

So, since the volume is changing at 300cm^3/s, we'd expect the height to be changing at 300/6928 = 0.043 cm/s.

I'll think on it a bit. In the meantime, maybe you can see where my mistake is.

Answer 2

Thank you for the solution! I think the mistake is at the calculations of finding dy/dt, I think it is 0.72. Maybe I am wrong but one more time I thank you in advance.

Answer 3

To find the rate at which the water level is rising, we need to find the derivative of the water level with respect to time.

Let's first find the volume of water in the trough as a function of the water level. Since the trough is shaped like a half-cylinder, its volume is given by:

V = (1/2) * π * r^2 * h

where V represents the volume, r is the radius, and h is the water level.

Given that the length of the trough is 4 m and the radius is 20 cm (or 0.2 m), we know that when the trough is empty, the water level is 0 m and when it is full, the water level is 4 m.

Using these values, we can express the water level h as a function of the volume V:

V = (1/2) * π * r^2 * h
h = 2V / (π * r^2)

Now, let's differentiate this equation with respect to time to find the rate at which the water level is rising:

dh/dt = d(2V / (π * r^2)) / dt

To find dh/dt, we need to find dV/dt. The volume V is given to be increasing at a constant rate of 300 cm^3/s. Since the volume is related to the water level h, we can find dV/dt in terms of dh/dt:

dV/dt = (dV/dh) * (dh/dt)

We know that dV/dt = 300 cm^3/s and we need to find dh/dt.

To find dV/dh, let's differentiate the volume equation with respect to h:

V = (1/2) * π * r^2 * h
dV/dh = (1/2) * π * r^2

Now we have all the necessary information to find dh/dt:

dV/dt = (dV/dh) * (dh/dt)
300 cm^3/s = (1/2) * π * (0.2 m)^2 * (dh/dt)

Solving for dh/dt:
(dh/dt) = 300 cm^3/s / [(1/2) * π * (0.2 m)^2]

Converting the units to cm:
(dh/dt) = 300 cm^3/s / [(1/2) * π * (20 cm)^2]

Simplifying:
(dh/dt) = 300 cm^3/s / [2 * π * 400 cm^2]
(dh/dt) = 0.094 cm/s

Therefore, the level of the water in the trough is rising at a rate of 0.094 cm/s at the moment when the level of the water is 10 cm below the upper surface of the trough.