Generalisations of area between two curves

[Gary Smart]

It seems to me that the main goal here is to determine what are the required values for the parameters, a, b, c, and m so that the resulting functions have the needed characteristics.

The first thing to do (for the general case) is to find the intercepts - - in particular the x-coordinates of those intercepts. That's pretty straight forward.

What are those two x values ?

The general case for finding the x coordinates of the intercepts is calculated by the following:
Setting: ax^2 + bx + c equal to mx + c => ax^2 + bx + c = mx + c
Solving for x gives: x = 0 and x = m − b / a

This means that the points of interception between a parabola and a line will always be 0 and m - b / a?

 

[SammyS]

The general case for finding the x coordinates of the intercepts is calculated by the following:
Setting: ax^2 + bx + c equal to mx + c => ax^2 + bx + c = mx + c
Solving for x gives: x = 0 and x = m − b / a

This means that the points of interception between a parabola and a line will always be 0 and m - b / a?

Use parentheses!

x = (m − b) / a, . . . which is actually x_R.)

Notice that now you can say that for x_R to be positive, the sign of a must be the same as the sign of m - b . Right?

 

[Gary Smart]

Use parentheses!

x = (m − b) / a, . . . which is actually x_R.)

Notice that now you can say that for x_R to be positive, the sign of a must be the same as the sign of m - b . Right?

Yes, I understand that. This is because if it wasn't the x value would be negative and would not be in the first quadrant. Therefore, if the gradient of the line and the gradient of the circle subtract to give a negative number then a must be negative and vice versa?

 

[SammyS]

Yes, I understand that. This is because if it wasn't, the x value would be negative and would not be in the first quadrant. Therefore, if the gradient of the line and the gradient of the circle subtract to give a negative number then a must be negative and vice versa?

Sorry for not responding sooner. I've been out most of the day.

The slope (gradient) of the line tangent to the parabola at x=0, is b . (You said gradient of the circle.)

So, if a > 0, then you must have m > b . Also the area between the curves is A₁ - A₂, which is the area under the linear function minus the area under the parabola. This can be computed as follows.

$\displaystyle \ \int_0^{x_R} \left(f_1(x)-f_2(x)\right)\,dx \ \$

$\displaystyle \ =\int_0^{\frac{m-b}{a}} \left(-ax^2+(m-b)x\right)\,dx \ \$​

On the other hand, if a < 0, then you must have m < b . In this case, the parabolic function is above the linear function on a graph. The area between the curves is A₂ - A₁, which is the area under the parabola minus the area under the linear function. This can be computed as follows.

$\displaystyle \ \int_0^{x_R} \left(f_2(x)-f_1(x)\right)\,dx \ \$

$\displaystyle \ =\int_0^{\frac{m-b}{a}} \left(ax^2+(b-m)x\right)\,dx \ \$​

Evaluate each integral and set it equal to 1 . That should give you some relationship involving a, b, and m.

You also know that you must have c > 0.

Any other restrictions on the parameters comes from the requirement that the functions be restricted to the First Quadrant for 0 ≤ x ≤ x_R.

 

[Gary Smart]

Thank you very much for your help.

I'm looking at different y intercepts now, to test for a lower bound that is not 0.

I know the Y intercepts will be different => ax^2 + bx + c = mx + d

That equation can not be simplified.. does this mean there is not a general case when the y intercept aren't the same?

I do know that, the lower bound (left point of intersection) can not be less than 0.

 

[SammyS]

Thank you very much for your help.

I'm looking at different y intercepts now, to test for a lower bound that is not 0.

I know the Y intercepts will be different => ax^2 + bx + c = mx + d

That equation can not be simplified.. does this mean there is not a general case when the y intercept aren't the same?

I do know that, the lower bound (left point of intersection) can not be less than 0.

First of all, for the case of both functions having the same y-intercept, there can be complications due to the requirement that the functions be in the first quadrant between the two points at which they intersect. Having the same y-intercept does make one point of intersection be at x = 0. This is of considerable help. Of course, you need c ≥ 0. The biggest challenge comes if parameter, a, is positive and the parabolic function has a vertex between the two points of intersection.If you allow different y-intercepts, then there is more to consider. First of all, the y-intersepts, c, and d, no longer need to be positive. The x-values for the points of intersection become more complicated. The integration is similar, except that the result is more complicated, due to the more complicated limits of integration and the appearance of constant term in the integrand.

 

[Gary Smart]

First of all, for the case of both functions having the same y-intercept, there can be complications due to the requirement that the functions be in the first quadrant between the two points at which they intersect. Having the same y-intercept does make one point of intersection be at x = 0. This is of considerable help. Of course, you need c ≥ 0. The biggest challenge comes if parameter, a, is positive and the parabolic function has a vertex between the two points of intersection.If you allow different y-intercepts, then there is more to consider. First of all, the y-intersepts, c, and d, no longer need to be positive. The x-values for the points of intersection become more complicated. The integration is similar, except that the result is more complicated, due to the more complicated limits of integration and the appearance of constant term in the integrand.

If parameter a is positive and the parabolic function has a vertex the two points of intersection, the only way to find where it lies is by completing the square (using vertex form)?

 

[SammyS]

If parameter a is positive and the parabolic function has a vertex between the two points of intersection, the only way to find where it lies is by completing the square (using vertex form)?

You don't have to actually put it in Vertex form, but you would need to do most of the work necessary to put it in Vertex form.

You can also check the discriminant. If $\displaystyle\ b^2-4ac\le0\,,\$ then the y-coordinate of the vertex is positive. (Still assuming $\ a\ge0\$).

 

[Gary Smart]

You don't have to actually put it in Vertex form, but you would need to do most of the work necessary to put it in Vertex form.

You can also check the discriminant. If $\displaystyle\ b^2-4ac\le0\,,\$ then the y-coordinate of the vertex is positive. (Still assuming $\ a\ge0\$).

Does the discriminant check work because we are trying to find where the combined function of ax^2 + bx + c = mx + c crosses the x axis?

 

[SammyS]

Does the discriminant check work because we are trying to find where the combined function of ax^2 + bx + c = mx + c crosses the x axis?

Yes.

However, that's only an issue if the vertex is between the two points of intersection.