Easy Math Question - SOLVED- thanks

[jordank32]

Just trying to develop a quadratic equation from the following:

when x=0, y=0
when x=N, y=0
when x=N/2, y = ymax

I know that I knew how to do this at some point in my life, any help would be much appreciated, thanks.

 

[aplefka]

I thought you never really used math in real life. That's what I was always told.

 

[jordank32]

yeah, I wish

 

[ArmenK]

x(x-n)=0

 

[notfred]

WTF is ymax?

Quadratic equations are continuous on the set of all real numbers.

 

[jordank32]

ymax is just any constant.
assuming that the constant ymax is given, i need an equation for y in terms of x and ymax that satisfies those requirements. It will be a continuous equation but I will only be looking at the range x=0 to x=N

 

[ArmenK]

Originally posted by: notfred
WTF is ymax?

Quadratic equations are continuous on the set of all real numbers.

I think he meant ymin

 

[jordank32]

I meant ymax, the parabola should reach a peak at N/2

 

[ArmenK]

Originally posted by: jordank32
I meant ymax, the parabola should reach a peak at N/2

I see.

So the equation you want is -x(x-N)=0

 

[jordank32]

But I need an equation for y in terms of x and ymax.

-x(x-N)=0 has neither a y or a ymax in it. thanks though

 

[jordank32]

maybe i was unclear. the parabola should reach its peak of ymax, when x=N/2.

 

[ArmenK]

Originally posted by: jordank32
But I need an equation for y in terms of x and ymax.

-x(x-N)=0 has neither a y or a ymax in it. thanks though

Oops. Here you go:

y=-x(x-N)=Nx-x^2

ymax is determined by N: -(N/2)(N/2-N)=(N^2)/4

 

[jordank32]

Thanks, but isn't there a way where I can specify what ymax is? I do not want ymax to be (N^2)/4. I want to somehow include ymax in the equation so I can pick what the ymax value will be. Is this not possible?

 

[ArmenK]

Originally posted by: jordank32
Thanks, but isn't there a way where I can specify what ymax is? I do not want ymax to be (N^2)/4. I want to somehow include ymax in the equation so I can pick what the ymax value will be. Is this not possible?

What you could do try is:

y-((N^2)/4)+ymax=-x(x-N)=Nx-x^2

that way you subtract off the max which puts the peak at 0, then add on whatver ymax you want

I think that should work

 

[jordank32]

that almost does the trick, however, doing that makes it so that y does not equal 0 when x=0 and x=N, any other ideas? thanks for all of your help

 

[RaynorWolfcastle]

for a quadratic eq
f(x) = Ax^2 + Bx + C
f'(x) = 2Ax + B
f''(x) = 2A

rewrite your constraints as
f(0) = 0
f(N) = 0
f'(N/2) = 0

solve for A, B, and C using these constraints

 

[jordank32]

I think I got it

y = -(4*ymax/N^2)*x*(x-N)

do you see any problem with this??

 

[notfred]

You people need to go verify my answer in this thread.

 

[jordank32]

yup, it works for what I needed. thanks for your help guys, i appreciate it

 

[ChuckHsiao]

1. y(x=0) = 0
2. y(x=N) = 0
3. y(x=N/2) = ymax (ymin, it doesn't really matter, it's just a variable name)

Quadratic equations take the form of y = ax^2 + bx + c, where a, b, and c are constants to be determined. There are three equations, and three unknowns, so this is (should be "technically", cuz not always) solvable. To do so, just plug them in one by one:

1. 0 = a*0 + b*0 + c => c = 0
2. 0 = a*N^2 + b*N => a*N^2 = -b*N => a*N = -b
3. ymax = (a*N^2)/4 + b*N/2 = -b*N/4 + b*N/2 = b*N/4, so ymax = b*N/4 => b = 4*ymax/N
Plugging in with the above (last part of #2), a = -4*ymax/N^2
Thus, the final solution is y = -4*ymax*(x/N)^2 + 4*ymax*x/N


Note that once you knew that c = 0 via condition 1, you could reduce this down to the more intuitive y = ax*(x+b) form. In this case, you get:
2. 0 = aN*(N+b) => a = 0 and/or b = -N
3. ymax = aN/2 * (N/2 + b) => a is NOT 0 (unless ymax = 0, in which case the whole equation is y = 0), so you know that b = -N
Plugging this in, ymax = aN/2 * (N/2-N) = aN/2*(-N/2) => ymax = -a*(N/2)^2 => a = -4*ymax/(N^2)
Thus it's y = -4*ymax*x/(N^2) * (x-N)

You can expand this out to see that both are equivalent.

Chuck Hsiao
Amptron

 

[ChuckHsiao]

Wow 8 posts while I was writing my reply, with multiple ones answering it. Boooooooo. I'm too slow.

Chuck Hsiao
Amptron

 

[jordank32]

thanks Chuck, one quick question, your second formulation gives y = -4*ymax*(x/N)^2 * (x-N) . If this is multiplied out, we get a x^3 term. Is this just a typo? Thanks for your help

 

[jordank32]

Nevermind, typo was corrected. Thanks, this reconfirms my answer. I appreciate it

 

[ChuckHsiao]

Yea I was writing it and double-checking that it's equivalent with the above, and stuck the (x/N)^2 there by mistake. Had too much fun sticking x and N in ( ).

Chuck Hsiao
Amptron