Generalisations of area between two curves

[Gary Smart]

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).

2. The attempt at a solution
My investigation up to now has found the following:
Function 1: 6x^2 - 2x + 8
Function 2: 4x + 8

These give points of intersection of: (0,8) and (1,12). These are the lower and upper bounds when integrating.

Integration produces:

Curve: 3x^2 - x^2 + 8x [Boundaries of 0 and 1]: Area under curve = 9 (between 0 and 1).

Line: 2x^2 + 8x [Boundaries of 0 and 1]; Area under line = 10 (Between 0 and 1).

=> Area between line and curve is: 10 - 9 = 1

3. Further investigations
I have already found that the coefficient of A (Parabola) has to be a multiple of 3, otherwise there will be a recurring decimal when integrated.

My next steps involve:
: Producing any generalisations that affect finding this specific area with these two functions.
: Proving/generalising using Algebra.

 

[LCKurtz]

There are a lot of things you haven't specified about this problem. May the parabola open either upwards or downwards? Must the intersections of the parabola and line be in the first quadrant? Is the area you want to calculate the area bounded by the two functions between the intersection points? Must the complete region "between" the functions lie in the first quadrant?

Also please show us how you got A must be a multiple of 3.

[Edit, added]: I would suggest starting with a simple subcase. For example, consider the two functions $(x-a)^2$ and $mx$. See if you can show that for any $a\ge 0$ there is a corresponding $m$ that works. Get the relationship between $m$ and $a$. Then start on more complicated cases.

 

[Gary Smart]

The Parabola can open either upwards or downwards. The intersections of the parabola and the line must be in the first quarter. The area is bounded by the functions between the points and the complete region between the functions must lie in the first quadrant.

I got that A must be a multiple of 3 because when the quadratic is integrated, the A value is divided by 3. If the value of A isn't a multiple of 3 then the result will always give a recurring decimal and will never produce an area of one.

(x-a)^2 = x^2 − 2xa + a^2 is not a parabola though is it?

 

[LCKurtz]

The Parabola can open either upwards or downwards. The intersections of the parabola and the line must be in the first quarter. The area is bounded by the functions between the points and the complete region between the functions must lie in the first quadrant.

I got that A must be a multiple of 3 because when the quadratic is integrated, the A value is divided by 3. If the value of A isn't a multiple of 3 then the result will always give a recurring decimal and will never produce an area of one.

(x-a)^2 = x^2 − 2xa + a^2 is not a parabola though is it?

That is a parabolic function. The graph of $y= (x-a)^2$ is a parabola with vertex $(a,0)$ opening upwards. $A$ does not have to equal $3$. Like I suggested, start with a real simple case. Take $y=(x-1)^2$ and $y = mx$. It is obvious geometrically that there is some positive $m$ so that the area between the line and parabola is $1$.

 

[Gary Smart]

Hello LCKurtz,

I not hundred percent sure I understand the process here. I have tried different m values when y = (x-1)^2 and it produces an m of approximately 0.704 (3 dp). What is the aim of the investigation?

 

[theodoros.mihos]

May you can use the absolute value of $$ \int_0^1f_1(x)dx-\int_0^1f_2(x) $$

 

[Gary Smart]

The first function being the line and the second function being the parabola?

 

[theodoros.mihos]

If you need just area, take absolute value. If you need sign then check for this on initial problem.

 

[Gary Smart]

Just to check I understand what you are saying:
I only need the area: specifically an area of 1 and if there are any generalisations that can be made.

What is the absolute value?

 

[theodoros.mihos]

Integral $\int f\,dx$ comes from area calculation between function graph and x axis.

 

[Gary Smart]

Integral $\int f\,dx$ comes from area calculation between function graph and x axis.

So, using that information.
The area of the function that is furthest away from the x-axis can't be less than 1?

 

[theodoros.mihos]

The integral have sign from f(x). The geometrical area is positive defined, like as geometrical distance. For area take the positive answer.

 

[Gary Smart]

The integral have sign from f(x). The geometrical area is positive defined, like as geometrical distance. For area take the positive answer.

This is because the area between the curves is above the x-axis and the fact there is no such thing as a negative geometrical distance?

 

[theodoros.mihos]

Area is positive, integral may not. Distance is positive, position coordinate value may not.

 

[Gary Smart]

Area is positive, integral may not. Distance is positive, position coordinate value may not.

I can't have negative integrals though? because the area is bounded in the first quadrant?

 

[theodoros.mihos]

Your parabola is positive defined so every $\int_a^bf_1(x)\,dx$ must be positive. The other function may give negative integral on some limits.

 

[Gary Smart]

Why must the parabola be positive? is that because we are subtracting the other function away from it, thus a negative subtracting a negative will always be a negative. Therefore will never give an area of 1?

Does the parabola have to be the first function? or can the line be function 1?

 

[theodoros.mihos]

Make a simple google graph to see the shapes.

 

[Gary Smart]

Both of these sets of functions are allowed in the constraints of this question.

 

[theodoros.mihos]

Function 1: 6x^2 - 2x + 8
Function 2: 4x + 8
This problem represented by the second graph.

 

[Gary Smart]

Function 1: 6x^2 - 2x + 8
Function 2: 4x + 8
This problem represented by the second graph.

These functions give an area of 1 but the only reason I know why is when function 2's area is subtracted by function's 1 it gives an area of 1 (Parabolic segment). I don't know why else it works.

 

[theodoros.mihos]

This is correct. See the axis units.

 

[Gary Smart]

This is correct. See the axis units.

I know the x-axis units of intersection are 0 and 1. But using these units doesn't always work.

 

[theodoros.mihos]

You make a coefficient error on your post, but I think this is typo, because you make the correct calculation

Curve: 3x^2 - x^2 + 8x [Boundaries of 0 and 1]: Area under curve = 9 (between 0 and 1). ---> 2x^3 ...

Line: 2x^2 + 8x [Boundaries of 0 and 1]; Area under line = 10 (Between 0 and 1).

 

[Gary Smart]

Okay, I get that. Is there a way of investigating this algebraically? Maybe proofing it e.g. if I wanted to find more areas of 1 in the first quadrant using a parabola and a line?

 

[theodoros.mihos]

See this

 

[Gary Smart]

The trapezoidal rule will only give me the area the area under the line though?

 

[theodoros.mihos]

By the same way for all functions.

 

[Gary Smart]

So, I need to apply the trapezoidal rule in general to general forms of both types of functions?

 

[theodoros.mihos]

No, trapezoidal rule is the discrete case (used on computational calculations) for the same think we make with calculus by integrals. Integrals are limits with $\Delta{x}\to0$ for the trapezoidal rule.

 

[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 ?