A water trough is 10 m long and a cross section has the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top, and has height 50 cm. If trough is being filled with water at the rate of 0.2 m^3/min how fast is the water level rising when the water is 30 cm deep?
To find how fast the water level is rising when it is 30 cm deep, we first need to find the volume of water in the trough when the depth is 30 cm.
The formula for the volume of a trapezoidal prism is given by: Volume = (1/2) * (b1 + b2) * h * l, where b1 and b2 are the lengths of the parallel bases, h is the height, and l is the length.
Given that the bottom width of the trough (b1) is 30 cm, the top width of the trough (b2) is 80 cm, and the height of the trough (h) is 50 cm, and the length of the trough (l) is 10 m (convert to cm: 10 m * 100 cm/m = 1000 cm), we can substitute these values into the formula:
Volume = (1/2) * (30 cm + 80 cm) * 50 cm * 1000 cm
Now, we can calculate the volume of the water in the trough when the depth is 30 cm.
Volume = (1/2) * (110 cm) * 50 cm * 1000 cm
Volume = 2,750,000 cm^3
Volume = 2.75 m^3 (converting cm^3 to m^3)
Given that the water is being filled at a rate of 0.2 m^3/min, we can now determine how fast the water level is rising when the water is 30 cm deep.
Let's call the rate of change of water level as dh/dt, where dh represents the change in height and dt represents the change in time.
dh/dt = (dV/dt) / (Area), where dV/dt represents the rate of change of volume with respect to time, and Area represents the cross-sectional area of the trough.
The cross-sectional area of the trough can be calculated using the formula for the area of an isosceles trapezoid: Area = (1/2) * (b1 + b2) * h.
Given that the bottom width of the trough (b1) is 30 cm, the top width of the trough (b2) is 80 cm, and the height of the trough (h) is 50 cm, we can substitute these values into the formula:
Area = (1/2) * (30 cm + 80 cm) * 50 cm
Area = 3,250 cm^2
Now, we can calculate dh/dt:
dh/dt = (dV/dt) / (Area)
Substituting the values, dh/dt = 0.2 m^3/min / (3,250 cm^2)
Now, we need to convert the units to match - either convert 0.2 m^3/min to cm^3/min or convert 3,250 cm^2 to m^2, let's convert 0.2 m^3/min to cm^3/min:
1 m^3 = 1,000,000 cm^3
0.2 m^3/min = 0.2 * 1,000,000 cm^3/min = 200,000 cm^3/min
dh/dt = 200,000 cm^3/min / (3,250 cm^2)
Now, we can calculate the value of dh/dt:
dh/dt = 61.54 cm/min
Therefore, the water level is rising at a rate of 61.54 cm/min when the water is 30 cm deep.
To find the rate at which the water level is rising, we need to first determine the volume of water in the trough as a function of the water depth.
The cross-section of the trough can be divided into two rectangles and a triangle. Let's label the bottom width of the trapezoid as b1 (30 cm) and the top width as b2 (80 cm). The height of the trapezoid is h (50 cm).
The volume of water in the trough is given by the formula:
V = A * d
where A is the cross-sectional area of the water in the trough and d is the depth of the water.
In this case, the cross-sectional area A can be calculated as follows:
A = (b1 + (b2 - b1) * d / h) * d
To express the volume V in terms of meters, we need to convert the dimensions from centimeters to meters:
b1 = 30 cm = 0.3 m
b2 = 80 cm = 0.8 m
h = 50 cm = 0.5 m
Now, let's substitute the values and simplify the equation for volume:
V = (0.3 + (0.8 - 0.3) * d / 0.5) * d
V = (0.3 + 0.5 * d) * d
V = 0.3d + 0.5d^2
Since the water is being filled at a constant rate of 0.2 m^3/min, we find the rate of change of the volume with respect to time (dv/dt):
dv/dt = 0.2 m^3/min
Now, to find the rate at which the water level is rising, we need to determine dy/dt, where y is the depth of the water in the trough. We can use the chain rule of differentiation:
dv/dt = d(V)/dy * dy/dt
Differentiating the equation for V with respect to y gives us:
dv/dt = 0.3 + d(0.5y^2)/dy
dv/dt = 0.3 + 1y
dv/dt = 0.3 + y
Substituting the given condition that the water is 30 cm (0.3 m) deep, we can solve for dy/dt:
dv/dt = 0.3 + y
0.2 = 0.3 + 0.3
0.2 - 0.3 = y
-0.1 = y
Therefore, the water level is rising at a rate of -0.1 m/min when the water depth is 30 cm. The negative sign indicates that the water level is decreasing, which makes sense since the trough is being filled.