weloo

Consider the two parabolas :y1=2x^2-3x-1 and y2=x^2+7x+20.
(a) Find the points of intersection of the parabolas and decide which one is greater than the other between the intersection points.

(b) Compute the area enclosed by the two parabolas.

(c) Use Mathcad to draw the graphs of the parabolas and find their intersection points graphically .

(d) Solve the differential equation dy/dx = -exp(5-x)+3/5x^2 ; y(5)=3.

for their points of intersection...

2x^2 - 3x - 1 = x^2 + 7x + 20
x^2 - 10x - 21 = 0
x^2 - 10x + 25 = 21 + 25 , I completed the square
(x-5)^2 = 46
x-5 = ± √46
x = 5 ± √46
so x = 5+√46 and x = 5-√46
or appr. 11.8 and -1.8

so I will take any value of x between these two values, x = 0 seems like a good choice, and find their y values
if x=0, y1 = -1, and if x=0 , y2 = 20
so the parabola y2 = .... is greater for all values of x between the two values we found
So the height of the region is
x^2+7x+20-(2x^2-3x-1) = -x^2 + 10x + 21

Area = ∫(-x^2 + 10x + 21) dx from x = 5-√46 to 5+√46
= [(-1/3)x^3 + 5x^2 + 21x] from x = 5-√46 to 5+√46
= ...

I will let you do the button-pushing arithmetic

Obviously we cannot demonstrate Mathcad on this webpage.

To find the points of intersection between the two parabolas y1 = 2x^2 - 3x - 1 and y2 = x^2 + 7x + 20, we need to solve the quadratic equation formed by equating the two equations:

2x^2 - 3x - 1 = x^2 + 7x + 20

To solve this equation, we can rearrange terms to get:

2x^2 - 4x^2 - 3x - 7x - 1 - 20 = 0

-2x^2 - 10x - 21 = 0

Next, we can either factorize this quadratic equation or use the quadratic formula to find its roots. Let's use the quadratic formula to calculate the roots:

x = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = -2, b = -10, and c = -21. Plugging these values into the quadratic formula, we get:

x = (-(-10) ± √((-10)^2 - 4(-2)(-21))) / (2(-2))
x = (10 ± √(100 - 168)) / (-4)
x = (10 ± √(-68)) / (-4)

Since the discriminant (b^2 - 4ac) is negative, the equation has no real roots. Therefore, the two parabolas do not intersect, and there are no points of intersection between them.

Moving on to the area enclosed by the two parabolas, we need to find the points where one parabola is greater than the other. Since they do not intersect, we cannot determine a definite area enclosed by the two curves.

For part (c), you can use graphing software like Mathcad or any other graphing calculator to plot the two parabolas and find their intersection points visually. Simply enter the equations into the graphing software and look for the points where the two curves meet.

Lastly, for part (d), the given equation is a first-order ordinary differential equation (ODE). To solve it, we need to separate variables and integrate both sides:

dy/dx = -e^(5-x) + (3/5)x^2

dy = (-e^(5-x) + (3/5)x^2) dx

Integrating both sides:

∫ dy = ∫ (-e^(5-x) + (3/5)x^2) dx

y = ∫ (-e^(5-x)) dx + ∫ ((3/5)x^2) dx

To evaluate the integrals, we can use the properties of exponential functions and the power rule of integration. After evaluating the integrals, we will have an equation involving the constant of integration.

However, to determine the value of the constant of integration, y(5) = 3 is given. Substituting this value into the equation, we can solve for the constant and obtain the particular solution for the differential equation.