Find the number b such that the line y = b divides the region bounded by the curves y = 16x2 and y = 9 into two regions with equal area. (Round your answer to two decimal places.)
We want
Int(x dy)[0,b] = Int(x dy)[b,9]
x = 1/4 y^(1/2)
Int(y^(1/2))[0,b] = Int(y^1/2)[b,9]
2/3 y^(3/2)[0,b] = 2/3 y^(3/2)[b,9]
2/3 b^(3/2) = 2/3 9^(3/2) - 2/3 b^(3/2)
4/3 b^(3/2) = 2/3 9^(3/2)
2b^(3/2) = 9^(3/2)
2^(2/3) b = 9
b = 9/2^(2/3) = 9/cbrt(4)
To find the value of b, we need to determine the area bounded by the curves y = 16x^2 and y = 9. Let's start by setting up an equation to find the area between the curves.
We need to find the x-values where the curves intersect. Setting 16x^2 = 9, we have:
16x^2 - 9 = 0
This equation can be factored as a difference of squares:
(4x - 3)(4x + 3) = 0
Setting each factor equal to zero, we get:
4x - 3 = 0 --> x = 3/4
4x + 3 = 0 --> x = -3/4
Since we are interested in the positive values of x for the area, we will use x = 3/4.
To find the y-values corresponding to x = 3/4 in each curve, we substitute x = 3/4 into the equations:
y = 16(3/4)^2 --> y = 9
y = 9
So, the two curves intersect at the points (3/4, 9).
The area bounded by the curves can be found by integrating the difference between the upper curve (y = 16x^2) and the lower curve (y = 9) with respect to x, over the interval (0, 3/4):
A = ∫[0, 3/4] (16x^2 - 9) dx
Let's calculate the area:
A = ∫[0, 3/4] (16x^2 - 9) dx
= ∫[0, 3/4] (16x^2) dx - ∫[0, 3/4] (9) dx
= [16/3 * x^3] [0, 3/4] - [9x] [0, 3/4]
= (16/3 * (3/4)^3) - (9 * (3/4))
= (16/3 * 27/64) - (27/4)
= 27/16 - 27/4
= (27 - 108)/16
= -81/16
Since we are looking for the line y = b to divide the region into two equal areas, we need to find a value of b such that the area below the line is half of the total area.
So, to find the value of b, we need to set the area bounded by the line y = b and the curve y = 16x^2 equal to half of the total area (which is -81/16):
Area under the line y = b = (-81/16) / 2
= -81/32
The area under the line y = b can be found by integrating (16x^2 - b) with respect to x over the same interval (0, 3/4):
A = ∫[0, 3/4] (16x^2 - b) dx
Setting A equal to -81/32 and solving for b, we have:
-81/32 = ∫[0, 3/4] (16x^2 - b) dx
-81/32 = (16/3 * x^3 - bx) [0, 3/4]
-81/32 = 16/3 * (3/4)^3 - b * (3/4)
-81/32 = 16/3 * 27/64 - 3b/4
-81/32 = 27/16 - 3b/4
-81/32 = 27/16 - 12b/16
-81/32 - 27/16 = -12b/16
-162/32 = -12b/16
-162/32 * 16/12 = b
-3.375 = b
Therefore, the value of b such that the line y = b divides the region bounded by the curves y = 16x^2 and y = 9 into two regions with equal area is approximately -3.38 (rounded to two decimal places).
To find the number b such that the line y = b divides the region into two equal areas, we need to solve for the x-coordinate where the areas on both sides are equal.
We are given the two curves y = 16x^2 and y = 9. To determine the bounds for integration, we need to find the x-values where these curves intersect.
Setting the two equations equal to each other:
16x^2 = 9
Simplifying:
x^2 = 9/16
Taking the square root of both sides:
x = ±(3/4)
Since we are interested in the area between the two curves, we will take the positive value and integrate from x = 3/4 to the x-coordinate where the areas are equal.
To find the equation of the line y = b, we substitute y = b into the equation of each curve:
16x^2 = b
9 = b
Now we set up the integral to find the x-coordinate:
∫[3/4, x] (16x^2 - 9) dx = ∫[x, -3/4] (16x^2 - 9) dx
Since we want to find the x-coordinate where the areas are equal, we will set up the equation:
∫[3/4, x] (16x^2 - 9) dx = ∫[x, -3/4] (16x^2 - 9) dx
Now we can solve the integral on both sides.
∫[3/4, x] (16x^2 - 9) dx = ∫[x, -3/4] (16x^2 - 9) dx
[16(x^3/3) - 9x] [3/4, x] = [16(x^3/3) - 9x] [x, -3/4]
Now we evaluate the definite integrals:
[(16x^3/3 - 9x) - (16(3/4)^3/3 - 9(3/4))] = [(16x^3/3 - 9x) - (16(-3/4)^3/3 - 9(-3/4))]
Simplifying the expression:
(16x^3/3 - 9x) - (9/4 - 27/4) = (16x^3/3 - 9x) - (-63/4 + 81/4)
Further simplification gives us:
(16x^3/3 - 9x) + (54/4) = (16x^3/3 - 9x) + (18/4)
Combining like terms, we get:
16x^3/3 - 9x + 54/4 = 16x^3/3 - 9x + 18/4
Since we want to find the value of b, we set up an equation with this expression and solve for x:
16x^3/3 - 9x + 54/4 = 16x^3/3 - 9x + 18/4
Simplifying the equation:
54/4 = 18/4
This equation implies that both sides are equal, meaning that the value of x at which the areas are equal is 0.
To find the value of b, substitute x = 0 into either of the equation forms for y = b:
b = 16x^2 = 16(0)^2 = 0
Therefore, the number b such that the line y = b divides the region bounded by the curves y = 16x^2 and y = 9 into two regions with equal area is 0.