I'm having a little problem with the equation for a graph. I'm looking for something where the line is completely horizontal at point 1,1 and completely vertical at 0,0. So far I have y=-x^2+2x which gives a horizontal line at 1,1, but not a vertical line at point 0,0 - I don't think it can be done with a parabola, but I have no idea where to go from here. Pardon my noobishness. Some help please?
Name:
Anonymous2007-12-03 23:32
So for a line (by which I assume you mean tangent) to be completely vertical means that the limit of derivative must be ∞, so the easiest graph I can think of which satisfies this is a circle of radius 1 centered at (1,0) I belive that is (x-1)^2+y^2=1.
Name:
Anonymous2007-12-03 23:35
It can't be done with a parabola (or any polynomial for that matter), since their derivatives are always defined everywhere, making a vertical tangent impossible.
f(x) = x^(1/3) + x/3 satisfies the requirement.
Name:
Anonymous2007-12-03 23:55
You win multiple internets.
Name:
Anonymous2007-12-04 9:23
>>3
That doesn't even go through (1,1). Circle works though.
Name:
Anonymous2007-12-04 12:36
>>5
Good call, I forgot to check that. 2/3 * x^(1/3) + x/3 works fine, however.
It's a circle, radius 1, shifted 1 unit to the right -_-
Name:
Anonymous2007-12-04 15:34
By "horizontal" and "vertical" do you mean dy/dx=0 at 1,1 and dx/dy=0 at 0,0? If so the easiest way would be to set up a some kind of differential equation. dy/dx could easily be some kind of rational expression f(x)/g(x), where f(x)=0 at 1, and g(x)=0 at 0, then just integrate it (the tricky part) to find your function, then ensure that it goes through the points 1,1 and 0,0 (adding c as applicable). Trial and error I think. Just think of different f(x) and g(x) that are 0 and 1 and 0 respectively...
Name:
Anonymous2007-12-04 16:36
>>8
This is correct.
Unless the function needs to be injective.
Name:
Anonymous2007-12-04 17:03
I think this would be a bijective solution:
f(x) =
-sqrt(-x), if x <= 0
sqrt(1-(x-1)^2), if 0 < x < 1
1+(x-1)^2, if x >= 1
Name:
Anonymous2007-12-04 17:42
>>11
Now make it infinitely differentiable on R \ {0}.
Name:
Anonymous2007-12-04 18:45
(x-1)^2 + y^2 = 1
Differentiate
2(x - 1) + 2y (dy/dx) = 0
dy/dx = 2(1 - x) / 2y
dy/dx @ 1,1 = 0/2 = Horizontal
dy/dx @ 0,0 = 2/0 = Vertical
All he said he needed was an equation, not a function. (x - 1)^2 + y^2 = 1 works.
Name:
Anonymous2007-12-04 18:48
(x-1)^2 + y^2 = 1
Differentiate
2(x - 1) + 2y (dy/dx) = 0
dy/dx = 2(1 - x) / 2y
dy/dx @ 1,1 = 0/2 = Horizontal
dy/dx @ 0,0 = 2/0 = Vertical
All he said he needed was an equation, not a function. (x - 1)^2 + y^2 = 1 works.