Generalisations of area between two curves

[Gary Smart]

Ahhh. I see. So there isn't a way of finding an area of 1 with these restrictions using algebra?

 

[Gary Smart]

I've carried out some investigations involving altering values of the quadratic then change the gradient of the line to get an area of 1.

I think all the combinations of the quadratic can fit into the investigations carried out in that table. I haven't really seen any patterns though...

 

[momoko]

Simpson's rule can be used to calculate integrals of polynomial up to degree 3 exactly. It may make your life easier.

 

[LCKurtz]

I see lots of replies since I was here last. But, @Gary Smart, I see where you want to do it algebraically but you haven't followed my original suggestion to get started. I suggested you consider a special, easier, case to get started. Consider the upward opening parabola $y = (x-a)^2$ and the line $y=mx$ with $a\ge 0,~m\ge 0$. This parabola's vertex is at $(a,0)$ and the line passes through the origin into the first quadrant. If the intersection points $x$ values are $u$ and $v$, the area between the line and the parabola is$$
A = \int_u^v mx - (x-a)^2~dx$$You want some algebra? Here's some algebra for you. Solve for the $x$ values of $u$ and $v$ where the line intersects the parabola. Then put those in for the limits in the integral above and set the integral equal to $1$. Work it through and it will give you an equation for $m$ in terms of $a$. So each such parabola has a corresponding line and slope. That could be a start for your analysis.

 

[Gary Smart]

I see lots of replies since I was here last. But, @Gary Smart, I see where you want to do it algebraically but you haven't followed my original suggestion to get started. I suggested you consider a special, easier, case to get started. Consider the upward opening parabola $y = (x-a)^2$ and the line $y=mx$ with $a\ge 0,~m\ge 0$. This parabola's vertex is at $(a,0)$ and the line passes through the origin into the first quadrant. If the intersection points $x$ values are $u$ and $v$, the area between the line and the parabola is$$
A = \int_u^v mx - (x-a)^2~dx$$You want some algebra? Here's some algebra for you. Solve for the $x$ values of $u$ and $v$ where the line intersects the parabola. Then put those in for the limits in the integral above and set the integral equal to $1$. Work it through and it will give you an equation for $m$ in terms of $a$. So each such parabola has a corresponding line and slope. That could be a start for your analysis.

Just to make sure I'm clear. That equation involves integrating the general form of a line and parabola simultaneously. The u and v values are the points of interception as we are trying to find the area between these bounds.

How do I solve for x though with no values?

 

[LCKurtz]

Just to make sure I'm clear. That equation involves integrating the general form of a line and parabola simultaneously. The u and v values are the points of interception as we are trying to find the area between these bounds.

How do I solve for x though with no values?

It might be helpful to me to know what level of math you are studying. High school or college? What courses have you had or are taking?

Take the particular parabola where $a=1$ for example. Can you figure out the $x$ values where the line $y=mx$ intersects $y = (x-1)^2$? The answers will depend on $m$ of course. That will give you $u$ and $v$ for this particular parabola. Then use those values of $u$ and $v$ in the integral$$
\int_u^v mx-(x-1)^2~dx = 1$$When you are done integrating the $x$ variable you will have nothing but $m$ left and you can solve for $m$. That will give you the particular line $y=mx$ that cuts off an area of $1$ with $y = (x-1)^2$. Do that much and check back.

 

[Gary Smart]

The level of maths I have studied is GCSE (high school), that was a few years ago now though. This challenge I came across individually and I'm really interesting in investigating it and solving it.

So, I took the parabola: y = (x - 1)^2. I chose a line that intercepts it. The line was 2x. The points of intersection were: 0.27 and 3.73. Where: U = 0.27 and V = 3.73. I inputted these values in the lower and upper bounds of:

∫vumx−(x−1)2 dx=1

However, It did not leave me with the gradient of the line. I must have gone wrong in the process?

 

[LCKurtz]

The level of maths I have studied is GCSE (high school), that was a few years ago now though. This challenge I came across individually and I'm really interesting in investigating it and solving it.

So, I took the parabola: y = (x - 1)^2. I chose a line that intercepts it. The line was 2x. The points of intersection were: 0.27 and 3.73. Where: U = 0.27 and V = 3.73. I inputted these values in the lower and upper bounds of:

All lines like $y=x,~y=2x,~y=3x,~ y = \frac 1 2 x$ and $y=mx$ for any positive $m$ will intersect $y=(x-1)^2$ in two points. They all cut off different areas. Draw some graphs and see. Only one value of $m$ is just right to cut off an area of $1$. There is no reason to expect that $y=2x$ is the one. You have to treat $m$ as an unknown and figure it out. Like I asked in post #36, can you figure out the $x$ values where $y=mx$ intersects $y=(x-1)^2$? That's a first step. The answers will depend on $m$. If you can't do that, you are probably looking at a problem that is beyond your current math skill levels.

 

[Gary Smart]

I understand now, just had a little misunderstanding in the previous text. I

I have found the line that intersects the parabola [y = (x-1)^2]. It is approximately:
y = 0.702209x

The points of intersection are 2.25967 and 0.442543.

These values seem very far fetched though, are they supposed to be like this?

Also, out of interest. I solved that using trial and error. Is there an algebraic way e.g. setting an equation equal to 1?

 

[Mentallic]

I understand now, just had a little misunderstanding in the previous text. I

I have found the line that intersects the parabola [y = (x-1)^2]. It is approximately:
y = 0.702209x

The points of intersection are 2.25967 and 0.442543.

These values seem very far fetched though, are they supposed to be like this?

Also, out of interest. I solved that using trial and error. Is there an algebraic way e.g. setting an equation equal to 1?

Those numbers look good.

What LCKurtz has been saying IS an algebraic method. You have some expression which turns out to be an integral, and you then let it equal to 1. Would you have been happier if you were instead given a equation in m that was equal to 1? Because that is what the integral is! It's just sort of hiding the answer, waiting for you to solve it so it can become the equation that you're hoping for.

Now, considering it would be an equation that is tough to compute, you can always use numerical techniques to do it. This is different from trial and error as you can usually get arbitrarily close to the answer very quickly.

 

[Gary Smart]

The mental block that I'm finding difficult to get past is the idea of letting an integral = 1. It makes no sense to me.

 

[LCKurtz]

The mental block that I'm finding difficult to get past is the idea of letting an integral = 1. It makes no sense to me.

Have you had any calculus? Do you understand that the integral calculates the area? If you want an area of $1$ don't you have to have the integral come out equal to $1$?

 

[Gary Smart]

I've done some basic calculus. Ah, yes I definitely understand that but I don't understand "Would you have been happier if you were instead given a equation in m that was equal to 1? Because that is what the integral is! It's just sort of hiding the answer, waiting for you to solve it so it can become the equation that you're hoping for"

I understand I need the integral between the line and curve to be 1. I understand that I can play with values to get an area of 1. I just don't understand this last part of setting up and understanding an equation that helps approximate an area of 1 between a line and a parabola.

 

[theodoros.mihos]

Make the same work with symbols: $ax^2+bx+c = mx+c$. You can find a relation with $a,m,b$ (c can be anything).

 

[momoko]

I will sketch the way to do the general case mentioned in #1.
Let P(x)=|ax^2+(b-m)x|
Let p=(m-b)/a
The required area is given by the integral of P(x) dx between the roots x=0 and x=p
As this is quadratic, we can use Simpson's rule to integrate exactly.
P(0)=P(p)=0, which greatly simplify our work.

Area=(2/3) |p| P(p/2)
I will leave it for you to simplify.

 

[Gary Smart]

Momoko. I'm a little lost with that. Where do we get ax^2+(b-m)x from?

Could you simplify it a little further for me please?

Make the same work with symbols: $ax^2+bx+c = mx+c$. You can find a relation with $a,m,b$ (c can be anything).

I know that the parabola and the line have to intersect, so we set them equal to each other. The x values are the coordinates of intersection which then become the lower and upper bound of the integral. I don't know where to go apart this.

 

[momoko]

It is just |(ax^2+bx+c)-(mx+c)|

 

[SammyS]

Let's go back to the OP

Homework Statement

The problem consists of investigating the area between two functions of the forms (Parabolic segment):
: y = mx + c and y = ax^2 + bx + c

The investigation involves finding a combination that has one of each of the above functions and finding an area of one. The area between the functions has to be in the first quarter (positive x and y).
...

Let's call the linear function, ƒ₁ and the quadratic function ƒ₂.

Notice that both functions have the the same y-intercept, y = c. ( I don't know if that was intentional on your part, but it does make life much simpler.)

Therefore one of the points of intersection is (x,y) = (0, c) .

It's easy enough to find the other intercept for the general case.

You (Gary Smart) seemed to have some trouble with the idea of requiring some integral to be equal to 1. Let's see if we can clear that up.

Let's call the x-coordinate of other intercept x_R, since it's the right-most of the two intercepts.

Then you want the difference of the areas, A₁, under ƒ₁ and A₂, under ƒ₂ to be equal to 1. That is to say, you want $\ \left| A_1 - A_2\right|=1\ .$

In terms of integrals, that's $\displaystyle \ A_1-A_2=\int_0^{x_R} f_1(x)\,dx\ -\int_0^{x_R} f_2(x)\,dx \,, \ 0$ and you want the absolute value of this difference to be 1.

But basic calculus gives us $\displaystyle \ \int_0^{x_R} f_1(x)\,dx\ -\int_0^{x_R} f_2(x)\,dx =\int_0^{x_R} \left(f_1(x)-f_2(x)\right)\,dx \ .\$ It's somewhat easier to work with that one integral than to work with the two individual integrals. That's what was behind those suggestions to set that integral equal to 1.

See if this makes anything clearer.

See what you can do with $\displaystyle \ f_1(x)=mx+c\$ and $\displaystyle \ f_2(x)=ax^2+bx+c\$ .

 

[Gary Smart]

Let's go back to the OPLet's call the linear function, ƒ₁ and the quadratic function ƒ₂.

Notice that both functions have the the same y-intercept, y = c. ( I don't know if that was intentional on your part, but it does make life much simpler.)

Therefore one of the points of intersection is (x,y) = (0, c) .

It's easy enough to find the other intercept for the general case.

You (Gary Smart) seemed to have some trouble with the idea of requiring some integral to be equal to 1. Let's see if we can clear that up.

Let's call the x-coordinate of other intercept x_R, since it's the right-most of the two intercepts.

Then you want the difference of the areas, A₁, under ƒ₁ and A₂, under ƒ₂ to be equal to 1. That is to say, you want $\ \left| A_1 - A_2\right|=1\ .$

In terms of integrals, that's $\displaystyle \ A_1-A_2=\int_0^{x_R} f_1(x)\,dx\ -\int_0^{x_R} f_2(x)\,dx \,, \ 0$ and you want the absolute value of this difference to be 1.

But basic calculus gives us $\displaystyle \ \int_0^{x_R} f_1(x)\,dx\ -\int_0^{x_R} f_2(x)\,dx =\int_0^{x_R} \left(f_1(x)-f_2(x)\right)\,dx \ .\$ It's somewhat easier to work with that one integral than to work with the two individual integrals. That's what was behind those suggestions to set that integral equal to 1.

See if this makes anything clearer.

See what you can do with $\displaystyle \ f_1(x)=mx+c\$ and $\displaystyle \ f_2(x)=ax^2+bx+c\$ .

Hello Sammy, firstly thank you for helping me. I have been working with f1(x)=mx+c and f2(x)=ax2+bx+c and trying different values.
The problem I'm finding is, if I find the other point of intersection Xr (which is dependant on the functions) then when I alter the values of the equations slightly this alters the Xr value in which case, I run around in circles...

 

[SammyS]

Hello Sammy, firstly thank you for helping me. I have been working with f1(x)=mx+c and f2(x)=ax2+bx+c and trying different values.
The problem I'm finding is, if I find the other point of intersection Xr (which is dependent on the functions) then when I alter the values of the equations slightly this alters the Xr value in which case, I run around in circles...

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 ?

 

[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.