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Unread 16-03-2014, 21:08
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Re: Math Quiz: Parabola Path

I don't have an answer to your new question, Ether, I am about to explain the answer to your first question because none exists on this thread and I'd hate it if I couldn't figure it out and no one explained how they got to the answer. I don't have the paper I figured this on and do not have a decent calculator on me.

This requires an understanding of calculus 1, the normal line.

f(x) = upper curve, y = original curve, and g(x) = lower curve.

Note: f' != g' != y' on [-5,5] except at x = 0.

I realized that f - g != 2 at all x on the interval [-5, 5] except at x = 0, but rather the distance of the normal line from f to g (or vise versa) = 2. I also figured that the normal line +/- a distance of 1 on the normal line of f will get you to f and g.

Given that y = 0.0433x^2, then y' = .0866x. Then the normal line = -1/y'

The normal line of y(x) = -1/(.0866x) = ynorm.

The problem is still find f and g, but this equation gives a line at every point of x that is exactly a unit of 1 away from both f and g.

It will be easier to solve for the x and y components of f and g instead of a function f, which is exactly what ryan did.

With any circle centered at y, where it touches f.x > y.x > g.x and f.y > y.y > g.y, where .x and .y are the x and y components respectively.

so let's make a triangle.
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| /
|/

The top is change in x (dx) and the left is change in y, dy. The hypotenuse(h) is 1.

So what is the angle between dy and h?

The slope of h = ynorm.

the slope of dy = undefined (straight up).
The slope of dx = 0 (flat).

I do not feel like typing out the equation, the angle between two lines is described here: http://mathforum.org/library/drmath/view/68285.html

and from those equations you get the angle the triangle, which I do not remember the equation for f and g's triangles. I do apologize. From this, you can use dy^2 + dx^2 = 1 (c^2 where c = 1) and simple trig to get a and b for both triangles, and that is the answer.
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