The ends of a water trough have the shape of the region bounded by the graphs of y = x^2 and y = 4 with x and y both measured in feet. To what depth must the trough be filled with water so that the force exerted by the water on either end of the trough is 779.423 lb? (Density of water is 62.5 lb per cubic foot).
3.528 feet
please help if possible !
What'd you get?
Let h be the water level
Each small slice of area from one end of parabola to the other end = 2*sqrt(y) dy
Pressure exerting on small slice = density * (h-y) ; h-y is the depth of water from the surface.
So, F = 62.5 integral of (h-y) 2*sqrt(y) dy from 0 to h
F = 125 (2/3*h*y^3/2 - 2/5*y^5/2) from 0 to h
Substitue F with 779.423
h = 3.528 ft.
Well, to find the depth of the water needed in the trough, let's just dive right into it!
First, let's find the area of the region bounded by the two graphs. The intersection points occur when y = x^2 and y = 4. By setting these two equations equal to each other, we can solve for x:
x^2 = 4
Taking the square root of both sides, we get:
x = ±2
Since we are only interested in the positive values, we have x = 2 as the right endpoint.
Now, we can find the area of the region bounded by the graphs by calculating the integral of (4 - x^2) from x = -2 to x = 2:
A = ∫[-2,2] (4 - x^2) dx
A = ∫[-2,2] 4 dx - ∫[-2,2] x^2 dx
A = [4x] from -2 to 2 - [(x^3)/3] from -2 to 2
A = 8 - 8/3
A = 16/3 square feet
Now, to find the volume of water needed to exert a force of 779.423 lb on each end, we can use the formula:
V = F / ρ
where V is the volume, F is the force, and ρ is the density of water.
V = 779.423 lb / (62.5 lb/ft^3)
V = 779.423 lb * (1/62.5) ft^3
V = 12.47 ft^3
And finally, to find the depth of the water, we can divide the volume by the area of the region:
Depth = V / A
Depth = (12.47 ft^3) / (16/3 ft^2)
Depth = 0.78 ft
So, the trough must be filled with water to a depth of approximately 0.78 feet to exert a force of 779.423 lb on each end. Water way to get to the bottom of it, huh?
To determine the depth to which the trough must be filled with water, we need to first find the volume of water in the trough.
Given that the trough has the shape of the region bounded by the graphs of y = x^2 and y = 4, we can find the depth by integrating the area between these two curves with respect to x.
The depth, denoted as h, can be expressed as:
h = ∫[a, b] (4 - x^2) dx
where [a, b] represents the interval over which the curves intersect.
To find the interval, we need to determine where the curves intersect.
Setting the equations of the curves equal to each other, we have:
x^2 = 4
Taking the square root of both sides, we get:
x = ±√4 = ±2
So, the intersection points are at x = -2 and x = 2.
To avoid negative values, we consider the interval [a, b] as [0, 2].
Now, we can integrate the difference of the curves over this interval:
h = ∫[0, 2] (4 - x^2) dx
Integrating will give us:
h = [4x - (x^3)/3] from 0 to 2
Plugging in the upper and lower limits, we have:
h = [4(2) - (2^3)/3] - [4(0) - (0^3)/3]
Simplifying, we get:
h = [8 - 8/3] - [0 - 0]
h = [(24 - 8)/3]
h = 16/3
So, the depth to which the trough must be filled with water is 16/3 feet.
Now, to determine the force exerted by the water, we need to calculate the weight of the water in the trough.
The weight of water can be calculated using the equation:
weight = density * volume
Given the density of water as 62.5 lb per cubic foot, and the volume of water in the trough as the depth h times the width (which we assume to be 1 foot), we have:
weight = (62.5 lb/ft^3) * (16/3 ft) * 1 ft
Calculating this, we get:
weight ≈ 347.222 lb
Therefore, to exert a force of approximately 779.423 lb on either end of the trough, it should be filled with water to a depth of 16/3 feet.