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 0.2 m3/min how fast is the water level rising when the water is 10 cm deep?
Round your answer to 3 decimal digits.
see related questions below.
After seeing other examples on here, I tried to do it myself. I got .05 m/min but the online system I do my homework on, said it was wrong.
I plugged in your numbers to Damon's solution and got 40 cm/min
What work do you have to show?
I tried that but that was wrong too.
Q= 0.2 m^3/min
A=surface area= length(width) at 10 cm deep depth is 1/2 the height. Width= 30 + 1/2(70-30)= 50 cm or .5 m
dh/dt= QA
A= 5(.5)= 2.5 m^3
dh/dt= (.2)(2.5)= .05 m/min
To find the rate at which the water level is rising, we need to use related rates. Let's break down the problem into parts and find the necessary information.
We know the trough has an isosceles trapezoidal cross-section. From the problem, we are given the following dimensions:
- Width of the bottom of the trough (base): 30 cm
- Width of the top of the trough: 70 cm
- Height of the trough: 40 cm
To solve the problem, we need to convert the dimensions to the same units. Let's use meters since the rate of water filling is given in cubic meters per minute.
1 meter = 100 cm
So, the dimensions in meters are:
- Width of the bottom of the trough: 30 cm ÷ 100 = 0.3 m
- Width of the top of the trough: 70 cm ÷ 100 = 0.7 m
- Height of the trough: 40 cm ÷ 100 = 0.4 m
Now, let's find the formula for the volume of a trapezoidal trough since we need the volume to solve the problem.
The formula for the volume of a trapezoidal trough is:
Volume = (1/2) × (b1 + b2) × h × L
where:
- b1 and b2 are the lengths of the two parallel bases (bottom and top)
- h is the height of the trough
- L is the length of the trough
Plugging in our numbers:
Volume = (1/2) × (0.3 + 0.7) × 0.4 × 5
= 0.5 × 1 × 0.4 × 5
= 1 × 0.4 × 5
= 2 m³
Now, we know that the trough is being filled at a rate of 0.2 m³/min. Let's call the rate of change of the water level "dh/dt" (the derivative of the height with respect to time).
Using the chain rule, we have:
dV/dt = (dh/dt) × (dV/dh)
We already know dV/dt (the rate at which the trough is being filled) is 0.2 m³/min, and we want to find dh/dt (the rate at which the water level is rising) when h = 0.1 m.
Now, let's find dV/dh (the derivative of the volume with respect to the height).
Using the formula for the volume of a trapezoidal trough, we can differentiate it with respect to the height:
dV/dh = (1/2) × (b1 + b2) × L
Plugging in our numbers:
dV/dh = (1/2) × (0.3 + 0.7) × 5
= (1/2) × 1 × 5
= 0.5 × 5
= 2.5 m²
Now we have all the necessary information to find dh/dt.
From the equation:
dV/dt = (dh/dt) × (dV/dh)
We can solve for dh/dt:
dh/dt = (dV/dt) ÷ (dV/dh)
dh/dt = 0.2 ÷ 2.5
dh/dt = 0.08 m/min
So, the water level is rising at a rate of 0.08 meters per minute when the water is 10 cm deep.
Therefore, the answer to the question is 0.08 m/min.