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 = โ(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 ยฑ โ(b^2 - 4ac))/(2a)
Plugging in a = 9, b = -15, and c = 9/4:
x = (15 ยฑ โ(15^2 - 4 * 9 * 9/4))/(2 * 9)
x = (15 ยฑ โ(225 - 81))/18
x = (15 ยฑ โ144)/18
x = (15 ยฑ 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 = โซ(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 = 2ฯ * 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 = โซ(2ฯ * y * x) dx, from x = 0 to x = 2/3
Substituting y = 3x:
V = โซ(2ฯ * 3x * x) dx, from x = 0 to x = 2/3
V = 6ฯ * โซ(x^2) dx, from x = 0 to x = 2/3
Integrating:
V = 6ฯ * (x^3/3) from x = 0 to x = 2/3
V = 6ฯ * [(2/3)^3/3 - 0^3/3]
V = 6ฯ * (8/27) * 1/3
V = 16ฯ/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 16ฯ/9.