A water trough is 9 m long and has a cross-section in the shape of an isosceles trapezoid that is 40 cm wide at the bottom, 80 cm wide at the top, and has height 40 cm. If the trough is being filled with water at the rate of 0.3 m3/min how fast is the water level rising when the water is 30 cm deep?
A water trough is 6 m long and has a cross-section in the shape of an isosceles trapezoid that is 20 cm wide at the bottom, 80 cm wide at the top, and has height 60 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 20 cm deep?
____m/min
Well, well, well! It seems like we have a "trapezoidal" situation here, quite literally. Let's dive into the problem and sprinkle some humor along the way, shall we?
So, the water trough is 9 m long, has a width of 40 cm at the bottom, 80 cm at the top, and a height of 40 cm. It's like a fashion show for water troughs!
Now, you want to know how fast the water level is rising when it's 30 cm deep. Ah, the suspense is killing me! *Drumroll please* Let's go!
First things first, we need to convert all our measurements to the same units. Since we were kind enough to share that the trough was long-long, let's convert everything to meters. So, the bottom width is 0.4 m, the top width is 0.8 m, and the height remains 0.4 m. Nice and metric!
Now, we can calculate the area of the water in the trough. It's like measuring the size of a pool for our elegant swimmers. The formula for the area of an isosceles trapezoid is (A = 1/2 * (b1 + b2) * h), where b1 and b2 are the bases (bottom and top widths) and h is the height.
So, when the water is 30 cm deep (0.3 m), our area (A) is:
A = 1/2 * (0.4 + 0.8) * 0.3
Now, we have the area of the water, but we need to know how fast it's rising. Thankfully, we know the rate at which the water is being filled: 0.3 m3/min. But we want to find how fast the water level is rising in cm/min.
To do that, we'll use our trusty proportionality equation: Volume / Time = Area * Speed.
Now, we know the volume of water being filled per minute is 0.3 m3/min, but we want to find the speed (rate of rise) in units of cm/min. So, we need to convert liters to cubic meters because 1 liter = 0.001 m3. It's like helping our intelligent water scholars learn to swim!
Now, let's plug in the values and solve for the speed:
0.3 m3/min / 1 min = A * Speed
We know A (the area when the water is 30 cm deep), and we want to find the Speed (rate of rise). Just divide both sides of the equation by A, and voila!
Speed = 0.3 m3/min / A
Now, just substitute the value of A that you calculated earlier, and you'll have your answer. Ta-da!
Remember, as with any good joke, it's always important to see if the answer makes sense. Check the units, dimensions, and make sure they jive. And there you have it, the rate at which the water level is rising. Now, go forth and spread joy, my friend!
To find out how fast the water level is rising, we need to determine the rate at which the volume of water is increasing with respect to time.
Let's start by calculating the volume of the water in the trough. The volume of a trapezoidal cross-section can be found by applying the formula for the area of a trapezoid and then multiplying it by the length of the trough.
The formula for the area of a trapezoid is:
A = (a + b) * h / 2,
where a and b are the lengths of the parallel sides (the bottom and top of the trapezoid), and h is the height.
In this case, the bottom length (a) is 40 cm, the top length (b) is 80 cm, and the height (h) is 40 cm. However, we will convert these lengths to meters to maintain consistent units throughout the problem.
a = 40 cm = 0.4 m,
b = 80 cm = 0.8 m, and
h = 40 cm = 0.4 m.
Now we can calculate the area of the cross-section:
A = (0.4 + 0.8) * 0.4 / 2 = 0.6 * 0.4 / 2 = 0.12 m^2.
Since the trough has a constant cross-sectional shape and the water level is rising uniformly, the volume of water (V) in the trough is directly proportional to the height (h) of the water.
V = A * h,
where A is the area of the cross-section and h is the height of the water.
In this case, the height is given as 30 cm, which is 0.3 m. So,
V = 0.12 m^2 * 0.3 m = 0.036 m^3.
Given that the trough is being filled at a rate of 0.3 m^3/min, we can now find the rate at which the water level is rising.
Let's consider dh/dt as the rate at which the height (h) of the water is changing with respect to time (t). The derivative of the volume (V) of water with respect to time (t) can be represented as dV/dt. Since V = A * h, we can express dV/dt as dV/dt = A * (dh/dt).
We know that dV/dt = 0.3 m^3/min, A = 0.12 m^2, and we need to find dh/dt when h = 0.3 m.
Substituting the known values into the equation, we have:
0.3 m^3/min = 0.12 m^2 * (dh/dt).
To find dh/dt, divide both sides of the equation by 0.12 m^2:
0.3 m^3/min / 0.12 m^2 = dh/dt.
Simplifying the equation, we get:
2.5 m/min = dh/dt.
Therefore, the rate at which the water level is rising when the water is 30 cm deep is 2.5 m/min.
dv = surface area * dh
so
dv/dt = surface area * dh/dt
width at surface = 40 + (80-40)(30/40)
= 40 + 30 = 70 cm = 0.70 m
so
surface area = 9 * .7 = 6.3 m^2
so
.3 m^3/min = 6.3 m^2 * dh/dt
and
dh/dt = .047 meters/min or 4.7 cm/min