A water trough is 5 m long and has a cross-section in the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 70 cm wide at the top, and has height 40 cm. If the trough is being filled with water at the rate of 0.1 m3/min how fast is the water level rising when the water is 20 cm deep?
Q = incoming flow rate = .1 m^3/min
dh/ dt = Q A
where
A = surface area
= length * width at 20 cm depth which depth is (1/2) height
width = 30 + 1/2(70-30) = 30+20 = 50 cm = .5 m wide water surface
so
A = 5 * .50 =2.5 m^3
so finally
dh/dt = .1 * 2.5 = .25 m/min = 25 cm/min
when the water has depth x, the cross-section is a trapezoid with bases
30 and 30+x. So the volume of water at depth x is
v = (60+x)/2 * x * 500 cm^3
= 250x^2 + 15000x
so, knowing that
dv/dt = (500x + 15000) dx/dt
just solve that for dx/dt when x=20
LOL - Guess which of us is the mathematician and which is the Engineer :)
To find the rate at which the water level is rising, we need to determine the derivative of the water level with respect to time.
Let's first define the variables:
- Length of the water trough (L) = 5 m
- Width of the bottom of the trapezoid (b1) = 30 cm = 0.3 m
- Width of the top of the trapezoid (b2) = 70 cm = 0.7 m
- Height of the trapezoid (h) = 40 cm = 0.4 m
- Rate of water flow (dV/dt) = 0.1 m^3/min (given)
- Depth of the water (y) = 20 cm = 0.2 m (this is what we want to find)
We can create a variable "A" to represent the cross-sectional area of the trapezoid. The area can be calculated using the formula for the area of a trapezoid:
A = (b1 + b2) * h / 2
Now let's calculate the variable A using the given dimensions:
A = (0.3 + 0.7) * 0.4 / 2
= 1 * 0.4 / 2
= 0.2 m^2
Next, we need to determine the relationship between the depth of water and the volume of water in the trough. The volume of water (V) can be calculated by multiplying the cross-sectional area (A) by the length (L):
V = A * L
To find the rate at which the water level is rising, we need to take the derivative of V with respect to time (t):
dV/dt = d(A * L) / dt
Using the product rule of differentiation:
dV/dt = dA/dt * L + A * dL/dt
We are given the rate of water flow, which is equivalent to the rate of change of volume:
dV/dt = 0.1 m^3/min
We also know the length of the trough is constant:
dL/dt = 0 (since L is fixed)
Now we can calculate the derivative of the cross-sectional area with respect to time (dA/dt):
dV/dt = dA/dt * L + A * dL/dt
0.1 = dA/dt * 5 + 0
Since dL/dt = 0, the equation simplifies to:
dA/dt = 0.1 / 5
dA/dt = 0.02 m^2/min
Now we have the rate at which the cross-sectional area is changing with time. To find the rate at which the water level is rising (dy/dt), we divide dA/dt by the width of the water trough (b2 - b1).
dy/dt = dA/dt / (b2 - b1)
dy/dt = 0.02 / (0.7 - 0.3)
dy/dt = 0.02 / 0.4
dy/dt = 0.05 m/min
Therefore, the water level is rising at a rate of 0.05 meters per minute when the water is 20 cm deep.