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You are currently reading a thread in /sci/ - Science & Math

How to find the two graphs for which the distance between two of their points is 2 when the line joining those points is angled so that the angle is pi/2 at 0 and linearly approaches 0 when approaching c, and the same applies symmetrically when approaching -c?
The graphs only need to have this property in (-c-1, c+1)
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You rotate one axe, and then the other one.
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>>7841767
although my bottom curve is wrong, should be the other way (an arc of circle from the center of rotation)...
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>>7841767
>>7841774
What?
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>>7841792
well instead of starting from the graphs you start from a solid line of length 2 that you move through the plane. The numbers represent time :
at time 1 (ok should be 0) it is vertical,
then it rotates on its bottom end until the top point reaches the right place (time 1 2 3)
then it rotates on its top end until the bottom point reaches the right place (time 3 4 5)
then it translates horizontally (time 5 6)
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>>7841809
Or even simpler, you rotate at the center until the line is horizontal, and then you translate it (so the graph is just a circle with an horizontal line.
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>>7841671
>>7841767
>>7841774
>>7841792
>>7841809
>>7841817
What?

I don't really understand what you're trying to do but it looks like it might have something to do with bézier curves.
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>>7841809
>>7841817
I need it specifically like it's in the picture. Originally I wanted to find out the volume traced by a rotating, translating circle but got fixated on this side view - those line segments of length 2 are the profile of a unit circle that rotates pi radians while translating a distance d. (d=2c in pic)
The curves are VERY close to simple manipulations of sin (or cos) which is interesting.
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>>7841861
You could get the parametric equations for a point rotating around a point with radius one, and a point moving from -c to c, add them, and then solve for t to get it in terms of y = f(x)
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>>7841861
oh ok I didn't see the angle was changing linearly with translation.

>>7841869
yes that's it, so the the angle is $a(t) = \frac{\pi}{2}(1 - \frac{t}{c}$, so upper curve would be $y = t + sin a(t)$ and bottom $y = t + sin -a(t)$
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>>7841869
I don't know those parametrics.
>>7841879
Doesn't work.
For the linear angle function a(x)=(pi/2)-((pi*abs(x))/2c), the upper curve takes the value of sin(a(x)) at x+cos(a(x)) and the lower curve takes the value of sin(a(x)) at x-cos(a(x)), when x>0... Reverse the + and - for x<0.
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>>7841671

Someone already solved this back in June; check the archives
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>>7842712
Really nice. Can't find it though.
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I cant understand what youre trying to do
You asked a question with six lines of text in just one sentence.
What the fuck dudr
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>>7842726
The picture should help.
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>>7842712

And can this result in itself be applied to find the volume traced by a rotating, translating circle?
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>>7842712
This is not the same as OP's question. In his drawing, the middle of the line doesn't change.
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>>7841991
You shouldn't take the absolute value of x for the angle.
But you're right there are (bad) mistakes.
So you can describe a curve by an equation $y = f(x)$, (that is, the curve will be all the points $(x, f(x))$), but you can also describe it using two fonctions $x(t), y(t)$ of a parameter $t$ (the points of the curve will be $(x(t), y(t))$.

So you can use this parametrisation to describe the linear movement of the center of the line, and another parametrisation to describe the movement of an extremity of the line (which will give a circle), and then you add the two.
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>>7843276
Are these very simple parametrisations that I should be able to figure out?
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>>7843699
Yes they are rather simple. Tell me if you're stuck, or have any question.
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>>7843736
Apparently the upper curve isn't [x=cos(t)-((d*t)/pi)+(d/2), y=sin(t), t=0..pi], for a translation of d and rotation of pi?
I don't know how to verify because eliminating t from that is way over my head.
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>>7844471
What do you want to do with that curve ? To eliminate t, you just need to arccos the x (with checking you are in the right domains for arccos).
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>>7845954
I can't do it, can someone please do it?
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>>7845954
Can it really be as simple as arccosing the x?
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>>7848000
Ok sorry it's not that simple. Assuming the circle is of radius 1, and t goes from 0 to pi/2, and x's endpoint is c*pi*2 + 1 (so as to have simpler equations).

I've got $x(t) = \cos(\pi/2 - t) + ct = \sin t + ct$ and $y(t) = cos (t)$ for the upper curve. You could, using the identity $\cos^2 + \sin^2 = 1$, write a quadratic equation in t, solve it, which would give $t = 1/c(x +- \sqrt{1-y^2})$.
To know wether it's plus or minus, you check what happens for t = pi/2, and find it's minus.

You can then inject this into y, giving you a much nastier equation, altough t has disappeared... I don't know much more, but I don't think this could be simplified.