Chief Delphi - Math Quiz: Parabola Path

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Re: Math Quiz: Parabola Path

Quote:

Originally Posted by maths222 (Post 1357708)

As a function of x, it will describe the upper boundary on one side and the lower on the other.

Yes, there will be 2 separate functions - one for the upper boundary and one for the lower (just like Ryan's parametric equations).

Quote:

It does exist, but it is very ugly. I will try to remember to post it later.

Excellent. Would you please post both a "typeset" version and a "cut-and-paste programming" version? (see example in attachment)

Thank you.

Re: Math Quiz: Parabola Path

The display ones will come later.

Eq 1:

Code:

−11.547344110855*sqrt(((x^(2)+133.34115601449)/(2.*x*sqrt(x^(2)+133.34115601449)*abs(x)+x^(4)+267.68231202898*x^(2)+17779.86388728)))*sign(abs(x)+x*sqrt(x^(2)+133.34115601449))-((0.0433*(abs(x)+x*sqrt(x^(2)+133.34115601449))^(2))/(x^(2)+133.34115601449))

Eq 2:

Code:

(((11.547344110855*x*(x^(2)+133.34115601449)-11.547344110855*sqrt(x^(2)+133.34115601449)*abs(x))*sqrt(((1)/(−2.*x*sqrt(x^(2)+133.34115601449)*abs(x)+x^(4)+267.68231202898*x^(2)+17779.86388728))))/(abs(abs(x)-x*sqrt(x^(2)+133.34115601449))))-((0.0433*(abs(x)-x*sqrt(x^(2)+133.34115601449))^(2))/(x^(2)+133.34115601449))

Re: Math Quiz: Parabola Path

Quote:

Originally Posted by maths222 (Post 1358001)

The display ones will come later.

Eq 1:

Code:

−11.547344110855*sqrt(((x^(2)+133.34115601449)/(2.*x*sqrt(x^(2)+133.34115601449)*abs(x)+x^(4)+267.68231202898*x^(2)+17779.86388728)))*sign(abs(x)+x*sqrt(x^(2)+133.34115601449))-((0.0433*(abs(x)+x*sqrt(x^(2)+133.34115601449))^(2))/(x^(2)+133.34115601449))

Eq 2:

Code:

(((11.547344110855*x*(x^(2)+133.34115601449)-11.547344110855*sqrt(x^(2)+133.34115601449)*abs(x))*sqrt(((1)/(−2.*x*sqrt(x^(2)+133.34115601449)*abs(x)+x^(4)+267.68231202898*x^(2)+17779.86388728))))/(abs(abs(x)-x*sqrt(x^(2)+133.34115601449))))-((0.0433*(abs(x)-x*sqrt(x^(2)+133.34115601449))^(2))/(x^(2)+133.34115601449))

Hmm.

Your equations differ from Ryan's by about 0.01 near x=5

Also, the x-intercept of your upper-boundary equation differs from Ryan's by about 0.03

Given the number of decimal places in your equations, I would have expected them to be closer.

Re: Math Quiz: Parabola Path

Jacob,

Can you re-run Mathematica using the following parametric equations instead?

Code:





x_lower = t + 2*a*t / sqrt(1 + (2*a*t)^2);



y_lower = a*t^2 - 1 / sqrt(1 + (2*a*t)^2);





x_upper = t - 2*a*t / sqrt(1 + (2*a*t)^2);



y_upper = a*t^2 + 1 / sqrt(1 + (2*a*t)^2);

Re: Math Quiz: Parabola Path

I used a TI-Nspire and logic to arrive at my equations, so they are probably wrong. I will take a look at them this weekend.

Re: Math Quiz: Parabola Path

Quote:

Originally Posted by maths222 (Post 1358702)

I used a TI-Nspire and logic to arrive at my equations, so they are probably wrong. I will take a look at them this weekend.

Just to be clear: I am looking for an exact analytical solution, not an approximating function.

Finding a polynomial to fit the numerical parametric functions data is trivial. For example, Here's y=f(x) for the upper boundary:

1-4x²/100

Re: Math Quiz: Parabola Path

I had thought it was exact. I need to think through my logic that I used to develop my equation and figure out what I missed. I had gotten the same parametrics as Ryan; I just have not yet figured out a good way to represent them as a function of x.

Re: Math Quiz: Parabola Path

Are there any Mathematica gurus out there?

Wondering if there is a solution for the following problem:

Find explicit functions y=y_lower(x) and y=y_upper(x) for the following parametric equations:

Code:





x_lower = t + 2*a*t / sqrt(1 + (2*a*t)^2);



y_lower = a*t^2 - 1 / sqrt(1 + (2*a*t)^2);





x_upper = t - 2*a*t / sqrt(1 + (2*a*t)^2);



y_upper = a*t^2 + 1 / sqrt(1 + (2*a*t)^2);

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.
___
| /
|/

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.

Re: Math Quiz: Parabola Path

Quote:

Originally Posted by faust1706 (Post 1359895)

...and that is the answer.

Not sure what you are saying is the answer.

I don't see an explicit function of x anywhere in your post, which is what is being sought.

I'm not convinced such a function even exists. Prove me wrong :)

Re: Math Quiz: Parabola Path

I post all the remaining work when I find it XD. I am not sure either. I'll work through it more when I have time. I'll ask my calc teacher as well to see if she can come up with an explicit function of x. Quite the interesting problem. Thank you for posting it! I look forward to another one if there will be one.

Re: Math Quiz: Parabola Path

Quote:

Originally Posted by faust1706 (Post 1359895)

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

This is definitely the simpler/faster way to derive the equations, but requires a separate proof that the shortest distance between the original curve and the boundary curve can be found along the normal line of the original curve. For bonus points:

Let y(x) be the original function [ y(x) = 0.0433x² ]
Suppose f(x) is the equation describing boundary curve that we're trying to find.
Let D(x, x₀) = distance between [x, y(x)] and [x₀, f(x₀)]
D(x, x₀) = √((x - x₀)² + (y(x) - f(x₀))²)

Do the derivation as the intersection between the hyperplane where the distance between the curves is 1 [ D(x, x₀) = 1 ] and the hyperplane where the distance between the curves at the solution points is minimal [ d D(x, x₀) / d x₀ = 0 ]

Re: Math Quiz: Parabola Path

Curve E is the locus of the center of the osculating circle of the involute of a Curve C in the XY plane. What is the relationship between Curve E and Curve C?

Re: Math Quiz: Parabola Path

I didn't even think about proving that statement. I am a sloppy mathematician. thank you, this was really interesting. Time to read up on "osculating circle of the involute."

Re: Math Quiz: Parabola Path

Quote:

Originally Posted by Ether (Post 1360019)

Curve E is the locus of the center of the osculating circle of the involute of a Curve C in the XY plane. What is the relationship between Curve E and Curve C?

Are the curves closed?