A water trough is 6 m long, and its cross-section has the shape of an isosceles trapezoid that is 20 cm wide at the bottom, 50 cm wide at the top and 40 cm high. If the trough is being filled with water at the rate of 0.2 cubic meters/min, how fast is the water level rising when the water is 25 cm deep?
Draw a figure, noting that the trapezoid has 30-cm wide triangles at each end.
When the water is y cm high, its surface is 20+2*30*(y/40) cm across the cross-section. That's 20 + 3y/2.
So, at depth y, the volume
v = (20 + 20+3y/2)/2 * y * 600
= 450y^2 + 12000y
dv/dt = 900y dy/dt + 12000 dy/dt
since .2 m^3 = 200000 cm^3
200000 = 900*25 dy/dt + 12000 dy/dt
dy/dt = 5.80 cm/min
To find the rate at which the water level is rising, we can use the formula for the volume of a trapezoidal prism, which is given by:
V = (1/2) * (b1 + b2) * h * L
Where:
V = volume
b1 = width of the bottom base
b2 = width of the top base
h = height
L = length
First, let's convert the dimensions to meters, as the rate of water flow given is in cubic meters per minute:
b1 = 20 cm = 0.2 m
b2 = 50 cm = 0.5 m
h = 40 cm = 0.4 m
L = 6 m
Substituting these values into the formula, we have:
V = (1/2) * (0.2 + 0.5) * 0.4 * 6
V = 0.35 * 0.4 * 6
V = 0.84 cubic meters
Now, let's differentiate both sides of the equation to find the rate at which the volume is changing with respect to time (dV/dt):
dV/dt = [(1/2) * (b1 + b2) * L] * (dh/dt)
Substituting the given values into the equation, along with the rate of water flow:
0.2 = [(1/2) * (0.2 + 0.5) * 6] * (dh/dt)
Simplifying:
0.2 = 1.7 * (dh/dt)
Now, rearrange the equation to solve for the rate at which the water level is rising (dh/dt):
dh/dt = 0.2 / 1.7
dh/dt ≈ 0.1176 meters/minute
Therefore, the water level is rising at a rate of approximately 0.1176 meters per minute when the water is 25 cm deep.
To find the rate at which the water level is rising, we need to use the concept of similar triangles and the properties of volumes.
First, let's convert all the measurements to the same unit. Since we are given that water is being filled at a rate of 0.2 cubic meters/min, it would be helpful to have our measurements in meters.
Given:
Length of the trough (L) = 6 m
Width at the bottom (b1) = 20 cm = 0.2 m
Width at the top (b2) = 50 cm = 0.5 m
Height (h) = 40 cm = 0.4 m
Let's denote the water depth as y (in meters). Since we are looking for the rate at which the water level is rising, we need to express y as a function of time. We know that the volume of the water in the trough (V) is increasing at a rate of 0.2 cubic meters/min. V is given by the formula:
V = (1/2) * (b1 + b2) * h * y
Now, we can differentiate both sides of the equation with respect to time (t) to find the rate of change of V with respect to t:
dV/dt = (1/2) * (b1 + b2) * h * dy/dt
We can substitute the given values:
0.2 = (1/2) * (0.2 + 0.5) * 0.4 * dy/dt
Now, let's solve the equation for dy/dt:
dy/dt = (0.2) / [(1/2) * (0.7) * 0.4]
dy/dt = (0.2) / (0.14)
dy/dt = 1.4286 m/min
So, the water level is rising at a rate of approximately 1.4286 meters per minute when the depth of the water is 25 cm.