Comparing Parabolas

cloud255

Hi,

Not sure if this qualifies as advanced math, but I'm not sure how to approach this problem.

I use the standard formula:

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 y = ax^2 + bx + c where x<= 0, c=0, y>=0, b=0
 
so it can be simplified to y = ax^2

So basically its a parabola only existing in the 2nd (i think) quadrant.

Now I create a second parabola in the same quadrant. The x values for both are always the same (x represents time), but the Y value for parabola 2 is based on the Y value of Parabola 1.

so I came up with the following:

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Y1 = ax^2

Y2 = Y1 + ((Y1)(constant1) + ((a(x-1)^2)*constant2) + Constant3

In the second part I do a(x-1)^2 because we need to look at the Y1 value at the previous X value

My question is:
1 is this mathematically correct,
2 is there a better way of doing this
3 I find the difference between Y1 and Y2 for a given X, at the moment I just go Y2(x-1) - Y1(x), is it possible to do this with just one graph (I doubt this as Y2 relies on Y1, but maybe one of you guys knows some awesome way of doing this)

Thanks in advance

JosAH

Quote:

Originally Posted by cloud255

Expand|Select|Wrap|Line Numbers

 
Y1 = ax^2

Y2 = Y1 + ((Y1)(constant1) + ((a(x-1)^2)*constant2) + Constant3

In the second part I do a(x-1)^2 because we need to look at the Y1 value at the previous X value

My question is:
1 is this mathematically correct,
2 is there a better way of doing this
3 I find the difference between Y1 and Y2 for a given X, at the moment I just go Y2(x-1) - Y1(x), is it possible to do this with just one graph (I doubt this as Y2 relies on Y1, but maybe one of you guys knows some awesome way of doing this)

What does the term ((Y1)(constant1) mean? (it isn't even well-formed). What is your second parabola supposed to represent in terms of the first parabola? In short: I don't understand your maths.

kind regards,

Jos

cloud255

Quote:

Originally Posted by JosAH

What does the term ((Y1)(constant1) mean? (it isn't even well-formed). What is your second parabola supposed to represent in terms of the first parabola? In short: I don't understand your maths.

kind regards,

Jos

Apologies for that kind of hard to write the equation with a standard key board but (Y1)(constant1) is suppose to be the Y value for parabola 1 at a given X value multiplied by a constant value.

The first parabola shows a modeling of a perfect scenario, where the second one shows the actual data based on the results of the perfect scenario when affected by given constants if that makes any sense at all.

So parabola 1 shows results in a closed system, parabola 2 shows what happens at each X value based on a deviation from parabola 1 at the previous X value.

JosAH

Quote:

Originally Posted by cloud255

Apologies for that kind of hard to write the equation with a standard key board but (Y1)(constant1) is suppose to be the Y value for parabola 1 at a given X value multiplied by a constant value.

The first parabola shows a modeling of a perfect scenario, where the second one shows the actual data based on the results of the perfect scenario when affected by given constants if that makes any sense at all.

So parabola 1 shows results in a closed system, parabola 2 shows what happens at each X value based on a deviation from parabola 1 at the previous X value.

Am I right saying that you basically want to do this:

Let C1, C2 and C3 be constants.
Let Y1(x) == a*x^2 and Y2(x) == C1*Y1(x)+C2*Y1(x-1)+C3

If so Y2(x) is just another parabola.

kind regards,

Jos

cloud255

Quote:

Originally Posted by JosAH

Let C1, C2 and C3 be constants.
Let Y1(x) == a*x^2 and Y2(x) == C1*Y1(x)+C2*Y1(x-1)+C3

If so Y2(x) is just another parabola.

Hi,

Yes, these are the graphs I have now, but is it possible to create one graph which represents the difference between the parabolas for a given x value?

Thanks

JosAH

Quote:

Originally Posted by cloud255

Hi,

Yes, these are the graphs I have now, but is it possible to create one graph which represents the difference between the parabolas for a given x value?

Thanks

Sure, you want Y2(x)-Y1(x); it's a parabola too.

kind regards,

Jos

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