Math Quiz: Parabola Path

[Ether]

The center of a circle of radius=1 traverses the parabolic trajectory.. y = -0.0433*x2.. from x=-5 to x=+5 in the XY plane, and as it does so it "sweeps out" a path with upper and lower boundaries.

Find the equations for the upper and lower boundary curves.

[Ether]

Hint: the answer is not

y = -0.0433*x2 + 1
and
y = -0.0433*x2 - 1

[Qbot2640]

I get:

Top curve: y = -0351x^2 + 1

Bottom curve: y = -.0433x^2 - 1

https://s3.amazonaws.com/grapher/exports/eibpsui62f.png

[Ether]

I get:

Bottom curve: y = -.0433x^2 - 1


Not. See my previous post (http://www.chiefdelphi.com/forums/showpost.php?p=1344675&postcount=2).

I get:

Top curve: y = -0351x^2 + 1


I assume there's supposed to be a decimal point before the "0".

Nice try, but no cigar.

[Qbot2640]

Not. See my previous post (http://www.chiefdelphi.com/forums/showpost.php?p=1344675&postcount=2).



I assume there's supposed to be a decimal point before the "0".

Nice try, but no cigar.




My mistake...I had all three equations together, and grabbed your original as the "bottom". The top I got (as you corrected my missing decimal) is 0.0351x^2 + 1

The bottom I got was 0.0474x^2 -1

I calculated two normal lines and found points that were 1 unit away from points on your given original. Then calculated the quadratics from the two sets of three points (although the y intercepts being known, one really only needs two points).

[Ether]

My mistake...I had all three equations together, and grabbed your original as the "bottom". The top I got (as you corrected my missing decimal) is 0.0351x^2 + 1

The bottom I got was 0.0474x^2 -1


The top equation y = -0.0351x^2 + 1 is still not right (even with the required "-" sign added back in :))

The bottom equation y = -0.0474x^2 - 1 (with the "-" sign added) is a very good approximation in the given range, but as you can see in the attached graph, it diverges outside that range.

I calculated two normal lines and found points that were 1 unit away from points on your given original.

You've got the right general idea.


Then calculated the quadratics...

Hint#2: the top and bottom boundary curves are not true quadratics.

[Ether]

Hint#2: the top and bottom boundary curves are not true quadratics.

Graph of upper and lower boundary curves

[Ether]

Any math/physics teachers want to give this a try?

[RyanCahoon]

Can I give the answers in parametric form?

xlower(t) = t - 0.0866 t / √(1 + (0.0866 t)2)
ylower(t) = -0.0433 t2 - 1 / √(1 + (0.0866 t)2)

xupper(t) = t + 0.0866 t / √(1 + (0.0866 t)2)
yupper(t) = -0.0433 t2 + 1 / √(1 + (0.0866 t)2)

[Ether]

Can I give the answers in parametric form?

You got it:)

Reps to you !

[Ether]

Is it possible to represent Ryan's parametric equations in explicit form y=f(x) ?

Or even implicit f(x,y)=0 ?

[Ether]

Is it possible to represent Ryan's parametric equations in explicit form y=f(x) ?

Or even implicit f(x,y)=0 ?

Does anyone have access to Mathematica?

[cgmv123]

Does anyone have access to Mathematica?

http://www.wolfram.com/mathematica/trial/

[Ether]

http://www.wolfram.com/mathematica/trial/

Thank you, I am aware of that. For a variety of reasons, I don't install or use trial software.

Are there any Mathematica gurus out there in CD land?

[maths222]

As a function of x, it will describe the upper boundary on one side and the lower on the other. It does exist, but it is very ugly. I will try to remember to post it later.

[Ether]

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).

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.

[maths222]

The display ones will come later.

Eq 1:

−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:

(((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)+26 7.68231202898*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))


http://beastiebots.org/images/Moving%20Ball.gif

[Ether]

The display ones will come later.

Eq 1:

−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:

(((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)+26 7.68231202898*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.

[Ether]

Jacob,

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



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);

[maths222]

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.

[Ether]

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

[maths222]

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.

[Ether]

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:



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);

[faust1706]

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.

[Ether]

...and that is the answer.

Not sure what you are saying is the answer.

I don't see an explicit function (http://www.mathwords.com/e/explicit_function.htm) of x anywhere in your post, which is what is being sought.

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

[faust1706]

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.

[RyanCahoon]

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 ]

[Ether]

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?

[faust1706]

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."

[faust1706]

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?

[Ether]

Are the curves closed?

That's not the answer I was looking for.

[faust1706]

I was asking whether curves E and C are closed, that wasn't may answer. I am still trying to wrap my head around what the question is asking

[Ether]

I was asking whether curves E and C are closed,

Not necessarily.

[faust1706]

Curve E is the evolute of the osculating circle, so curve E is the evolute of the involute of curve C.

I do not have a lot of faith in this answer. This really stretched by brain. If it's wrong please correct me and thank you for supplying me with these awesome questions!

[Ether]

curve E is the evolute of the involute of curve C.

So far so good. Now take that one step further...

[faust1706]

The involute of an evolute is the original curve itself, and I believe the evolute of an involute is the original curve too, so that means that Curve E and Curve C are the same.

[Ether]

I believe the evolute of an involute is the original curve too, so that means that Curve E and Curve C are the same.


The evolute of an involute is the original curve, less portions of zero or undefined curvature. (http://en.wikipedia.org/wiki/Involute)


the original curve is the evolute of its involute (http://mathworld.wolfram.com/Involute.html)

[Ether]

Quoth the Kamen

Poe enthusiasts might enjoy the raven's version of what happened: