the cross section of a 5-ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3-foot upper base, and an altitude of 2 feet. water is running into the trough at a rate of 1 cubic foot per minute. how fast is the water level rising when the water is 1 foot deep?
Well, if the water is already 1 foot deep, I guess we can say it's in a "deep" situation! Now, let's solve this math problem, shall we?
To find the rate at which the water level is rising, we can use the formula for the volume of a trapezoidal trough: V = (1/2)(h)(b1 + b2)(L), where h is the height, b1 is the lower base, b2 is the upper base, and L is the length.
Given that the lower base (b1) is 2 feet, the upper base (b2) is 3 feet, and the length (L) is 5 feet, we can substitute these values into the formula.
V = (1/2)(2 + 3)(5) = 12.5 cubic feet
Now, since the water is flowing at a rate of 1 cubic foot per minute, we can say that dV/dt = 1.
To find dh/dt, the rate at which the water level is rising, we can differentiate both sides of the equation with respect to time:
dV/dt = (1/2)(b1 + b2)(dL/dt)
Since dV/dt = 1 and the rate at which the length is changing (dL/dt) is 0 (because the trough's length is not changing), we can substitute these values into the equation.
1 = (1/2)(2 + 3)(0)
1 = 0
Oh, dear! It seems like we've encountered a mathematical conundrum here. The rate at which the water level is rising when the water is 1 foot deep is actually zero.
Looks like our trough is giving us the silent treatment. Better find a new way to entertain ourselves!
To find the rate at which the water level in the trough is rising, we can use related rates.
Let's denote the depth of water in the trough as "h" (in feet) and the volume of water in the trough as "V" (in cubic feet). We are given that the water is running into the trough at a constant rate of 1 cubic foot per minute.
The volume of a trapezoidal prism can be calculated using the formula:
V = (1/2) * (b1 + b2) * h * L
where b1 and b2 are the lengths of the bases, h is the height (altitude) of the trapezoid, and L is the length of the trough.
In this case, we have b1 = 2 ft, b2 = 3 ft, h = 2 ft, and L = 5 ft. So the volume of water in the trough is given by:
V = (1/2) * (2 + 3) * 2 * 5
V = 35 cubic feet
Now, let's differentiate both sides of the equation with respect to time (t):
dV/dt = (1/2) * (db1/dt + db2/dt) * h * L + (1/2) * (b1 + b2) * dh/dt * L
We are interested in finding dh/dt, the rate at which the water level is rising, when the water is 1 foot deep. At this point, h = 1 ft.
Since the lengths of the bases (b1 and b2) are not changing with time, their rates of change, db1/dt and db2/dt, are both zero.
Substituting the known values (h = 1 ft, V = 35 cubic feet, L = 5 ft) into the equation, we have:
1 = (1/2) * (0 + 0) * 1 * 5 + (1/2) * (2 + 3) * dh/dt * 5
1 = (5/2) * dh/dt * 5
dh/dt = 2/25 ft/min
Therefore, the water level is rising at a rate of 2/25 feet per minute when the water is 1 foot deep.
To find how fast the water level is rising when the water is 1 foot deep, we need to find the rate at which the volume of water in the trough is changing with respect to time.
Let's first determine the volume of the water in the trough. Since the cross-section of the trough is an isosceles trapezoid, we can use the formula for the area of a trapezoid to find the area of the cross-section.
The formula for the area of a trapezoid is:
A = (1/2) * (b1 + b2) * h
where A is the area, b1 and b2 are the lengths of the two bases, and h is the altitude.
Given the lower base = 2 ft, upper base = 3 ft, and altitude = 2 ft, we can substitute these values into the formula:
A = (1/2) * (2 + 3) * 2 = 5 ft²
Since the length of the trough is 5 ft, the volume of water in the trough can be given by:
V = A * l = 5 ft² * 5 ft = 25 ft³
Now, let's differentiate the volume equation with respect to time to find the rate of change of the volume:
dV/dt = d/dt (25 ft³)
Since the rate at which water is flowing into the trough is 1 cubic foot per minute, we can substitute this into the equation:
dV/dt = 1 ft³/min
This tells us that the volume is changing at a rate of 1 ft³ per minute.
Next, we need to relate the rate of change of volume to the rate at which the water level is rising. The water level is the height or depth of the water in the trough.
We can use the relationship between volume and the water level to find the rate at which the water level is rising. Since the trough is rectangular in shape, the volume of water can be given by:
V = l * w * h
where V is the volume, l is the length of the trough, w is the width of the trough, and h is the height or depth of the water.
Given the length of the trough is 5 ft, the width can be calculated by using the formula for the area of the cross-section:
A = b2 * h
5 ft² = 3 ft * h
h = 5/3 ft
Now, we have enough information to find the width w:
V = l * w * h
25 ft³ = 5 ft * w * 5/3 ft
w = (25 ft³) / (5 ft * 5/3 ft)
w = 3 ft
So, the width of the trough is 3 ft.
Now, we need to differentiate the volume equation with respect to time and solve for dh/dt, the rate at which the water level is rising:
dV/dt = (dw/dt) * l * h + w * dh/dt
Since the width of the trough w is constant, the derivative of w with respect to time is 0 (dw/dt = 0). Thus, the equation becomes:
1 ft³/min = 0 * 5 ft * (5/3) ft + 3 ft * dh/dt
1 ft³/min = 3 ft * dh/dt
Now, we can solve for dh/dt:
dh/dt = (1 ft³/min) / (3 ft) = 1/3 ft/min
Therefore, when the water is 1 foot deep, the water level is rising at a rate of 1/3 ft/min.
Another trapezoid problem?? Oh well, here goes:
Volume is area of base * length
When the water is h feet deep The trapezoid has lower base = 2, upper base = 2 + h/2 (draw a diagram to see why this is so)
So, the volume is 5 * (2 + 2+h/2)/2 * h
= 5/4 * (h^2 + 8h)
dV = 5/4 * (2h + 8) dh
1 = 5/4 * (2+8) * dh
dh = 1/12.5 ft/min
This makes sense. The base of the trough has area 5x2 = 10 ft^2. If it had straight sides, the height would rise 1/10 ft/min.
The top of the trough has area 15 ft^2. If it had straight sides from the top, the height would rise 1/15 ft/min