/sci/ - Science & Math

>>7589751
>>7588955
>>7588970

One way to get a single vector path/branch that describes the whole level curve in an infinite loop would be to try polar-like coordinates:

<span class="math"> x = r\,\cos(t+\delta),\ \ \ y = \sqrt{r}\,\sin(t+\delta)[/spoiler]

(WARNING: these are not really polar coords!)

Substituting these into your equation (*) above, and solving for r(t) gives
<div class="math">r(t) = \sqrt{\frac{2}{\cos^2 (t+\delta) + \sin^4 (t+\delta)}}</div>
So we obtain the path components
<div class="math"> x(t) = \sqrt{\frac{2}{\cos^2 (t+\delta) + \sin^4 (t+\delta)}}\,\cos(t+\delta)</div>
<div class="math"> y(t) = \Big(\frac{2}{\cos^2 (t+\delta) + \sin^4 (t+\delta)}\Big)^{\frac{1}{4}} \, \sin(t+\delta)</div>
<div class="math"> z(t) = \frac{1}{2}</div>

(see pic related)

I chose to include the <span class="math">\delta[/spoiler] so that the path passes through <span class="math">\vec{r_0} = (1,1,1/2)[/spoiler] when t=0. (You could drop it for simplicity if you want, thereby shifting the time values.)

Personally, I prefer this vector path, but the cost is in the complication of
<div class="math">\delta = \arccos \Big( \frac{\sqrt{5} - 1}{2} \Big) = \arcsin \Big( \sqrt{\frac{\sqrt{5} - 1}{2}} \Big)</div>
I'll leave it for you to show that when

<span class="math"> t=0:\ \ \ \ x(0)=1,\ \ y(0)=1 [/spoiler]

I double-checked that it does, but the calculation is a bit ugly.

>>7589751
>>7588955
>>7588970

One way to get a single vector path/branch that describes the whole level curve in an infinite loop would be to try polar-like coordinates:

<span class="math"> x = r\,\cos(t+\delta),\ \ \ y = \sqrt{r}\,\sin(t+\delta)[/spoiler]

(WARNING: these are not really polar coords!)

Substituting these into your equation (*) above, and solving for r(t) gives
<div class="math">r(t) = \sqrt{\frac{2}{\cos^2 (t+\delta) + \sin^4 (t+\delta)}}</div>
So we obtain the path components
<div class="math"> x(t) = \sqrt{\frac{2}{\cos^2 (t+\delta) + \sin^4 (t+\delta)}}\,\cos(t+\delta)</div>
<div class="math"> y(t) = \Big(\frac{2}{\cos^2 (t+\delta) + \sin^4 (t+\delta)}\Big)^{\frac{1}{4}}\,\sin (t+\delta)</div>
<div class="math"> z(t) = \frac{1}{2}</div>

(see pic related)

I chose to include the <span class="math">\delta[/spoiler] so that the path passes through <span class="math">\vec{r_0} = (1,1,1/2)[/spoiler] when t=0. (You could drop it for simplicity if you want, thereby shifting the time values.)

Personally, I prefer this vector path, but the cost is in the complication of
<div class="math">\delta = \arccos \Big( \frac{\sqrt{5} - 1}{2} \Big) = \arcsin \Big( \sqrt{\frac{\sqrt{5} - 1}{2}} \Big)</div>
I'll leave it for you to show that when

<span class="math"> t=0:\ \ \ \ x(0)=1,\ \ y(0)=1 [/spoiler]

I double-checked that it does, but the calculation is a bit ugly.