Consider the graphs of y = 3x + c and
y^2 = 6x, where c is a real constant.
a. Determine all values of c for which the graphs intersect in two distinct points.
b. suppose c = -3/2. Find the area of the region enclosed by the two curves.
c. suppose c = 0. Find the volume of the solid formed when the region bounded by
y = 3x and y^2 = 6x is revolved around the x-axis.
a. Well, to find the values of c for which the graphs intersect in two distinct points, we need to set the equations equal to each other and solve for x. Let's do that:
3x + c = √(6x)
To make it a bit simpler, let's square both sides:
(3x + c)^2 = 6x
Expanding the left side:
9x^2 + 6cx + c^2 = 6x
Rearranging and simplifying:
9x^2 + (6c - 6)x + c^2 = 0
For the graphs to intersect in two distinct points, this quadratic equation should have two different solutions for x. In other words, its discriminant should be greater than zero:
(6c - 6)^2 - 4 * 9 * c^2 > 0
Simplifying:
36c^2 - 72c + 36 - 36c^2 > 0
-72c + 36 > 0
Now we just need to solve this inequality to find the range of values for c:
-72c > -36
c < 1/2
So all values of c less than 1/2 will give us two distinct points of intersection.
b. Let's consider c = -3/2. Plugging this value into the equations:
y = 3x - 3/2
y^2 = 6x
Now to find the points of intersection, let's set the equations equal to each other:
(3x - 3/2)^2 = 6x
Expanding:
9x^2 - 9x + 9/4 = 6x
Rearranging and simplifying:
9x^2 - 15x + 9/4 = 0
To find the roots of this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
Plugging in a = 9, b = -15, and c = 9/4:
x = (15 ± √(15^2 - 4 * 9 * 9/4))/(2 * 9)
x = (15 ± √(225 - 81))/18
x = (15 ± √144)/18
x = (15 ± 12)/18
So the values of x are (15 + 12)/18 = 27/18 = 3/2 and (15 - 12)/18 = 3/6 = 1/2.
Now substitute these values back into one of the equations to find the corresponding y-values:
For x = 3/2:
y = 3 * (3/2) - 3/2 = 9/2 - 3/2 = 6/2 = 3
For x = 1/2:
y = 3 * (1/2) - 3/2 = 3/2 - 3/2 = 0
So the two points of intersection are (3/2, 3) and (1/2, 0).
To find the area of the region enclosed by the two curves, we need to find the definite integral of the difference between the curves, between the x-values where they intersect:
Area = ∫(y^2 - (3x + c)) dx, from x = 3/2 to x = 1/2
c. Suppose c = 0. To find the volume of the solid formed when the region bounded by y = 3x and y^2 = 6x is revolved around the x-axis, we can use the method of cylindrical shells.
The volume of each cylindrical shell is given by the formula:
V = 2π * radius * height * thickness
First, let's find the limits of integration. We need to determine the x-values where the two curves intersect:
(3x)^2 = 6x
9x^2 = 6x
9x^2 - 6x = 0
3x(3x - 2) = 0
x = 0 or x = 2/3
Now, let's set up the integral:
V = ∫(2π * y * x) dx, from x = 0 to x = 2/3
Substituting y = 3x:
V = ∫(2π * 3x * x) dx, from x = 0 to x = 2/3
V = 6π * ∫(x^2) dx, from x = 0 to x = 2/3
Integrating:
V = 6π * (x^3/3) from x = 0 to x = 2/3
V = 6π * [(2/3)^3/3 - 0^3/3]
V = 6π * (8/27) * 1/3
V = 16π/9
So the volume of the solid formed when the region bounded by y = 3x and y^2 = 6x is revolved around the x-axis is 16π/9.
a. To find the values of c for which the graphs intersect in two distinct points, we need to set the equations equal to each other and solve for x.
Setting y = 3x + c equal to y^2 = 6x, we get:
(3x + c)^2 = 6x
Expanding and simplifying:
9x^2 + 6cx + c^2 = 6x
Rearranging:
9x^2 + (6c - 6)x + c^2 = 0
For the graphs to intersect in two distinct points, this quadratic equation must have two distinct real solutions for x. This means the discriminant (b^2 - 4ac) must be greater than 0.
Using the discriminant formula, we have:
(6c - 6)^2 - 4(9)(c^2) > 0
Simplifying:
36c^2 - 72c + 36 - 36c^2 > 0
-72c + 36 > 0
Dividing by -36 (and flipping the inequality sign):
c < 1/2
Therefore, the graphs will intersect in two distinct points for all values of c less than 1/2.
b. Now, suppose c = -3/2. We need to find the area of the region enclosed by the two curves.
The two curves are y = 3x - 3/2 and y^2 = 6x.
To find the points of intersection, we set the equations equal to each other:
3x - 3/2 = √(6x)
Squaring both sides and simplifying:
9x^2 - 9x + 9/4 = 6x
9x^2 - 15x + 9/4 = 0
Multiplying through by 4 to remove the fraction:
36x^2 - 60x + 9 = 0
Now we can solve for x using either factoring, completing the square, or quadratic formula. In this case, the quadratic factors into a perfect square:
(6x - 3)^2 = 0
Solving for x:
6x - 3 = 0
x = 1/2
So, the two curves intersect at x = 1/2.
To find the corresponding y-values, we substitute x = 1/2 into either equation:
y = 3(1/2) - 3/2 = 0
Therefore, the two curves intersect at the point (1/2, 0).
Now we can find the area enclosed by the curves by finding the definite integral of their difference over the interval where they intersect:
Area = ∫(3x - 3/2 - √(6x)) dx, from x = 0 to x = 1/2
Let's calculate that:
Area = ∫(3x - 3/2 - √(6x)) dx, from x = 0 to x = 1/2
= [3/2x^2 - 3/2x - (4/3)(6x^(3/2))] from 0 to 1/2
= [3/8 - 3/4 - (4/3)(6(1/2)^(3/2))] - [0]
= [3/8 - 3/4 - (4/3)(6(1/2)^(3/2))]
Using a calculator, the area is approximately equal to 0.3486.
c. Now, suppose c = 0. We need to find the volume of the solid formed when the region bounded by y = 3x and y^2 = 6x is revolved around the x-axis.
To find the volume, we will use the method of cylindrical shells.
Each differential element of the shell will have a height of y (which ranges from 0 to 3x) and a radius of x.
Using the formula for the volume of a cylindrical shell:
dV = 2πxy dx
The limits of integration for x are determined by the intersection points of the two curves.
We found earlier that the intersection point is (1/2, 0).
So, the volume is given by:
V = ∫(0 to 1/2) [2πxy dx]
V = 2π ∫(0 to 1/2) [x(3x) dx]
V = 6π ∫(0 to 1/2) [x^2 dx]
V = 6π [(1/3)x^3] from 0 to 1/2
V = 6π [(1/3)(1/8) - (1/3)(0)]
V = π/8 cubic units
Therefore, the volume of the solid formed when the region bounded by y = 3x and y^2 = 6x is revolved around the x-axis is π/8 cubic units.