February 25th, 2019, 01:01 PM  #1 
Newbie Joined: Feb 2019 From: United Kingdom Posts: 15 Thanks: 3  Parametric equations!
Let's be honest, equations of curves don't become interesting until you start to throw parameters in there. Nothing satisfies me more than a point travelling along a curve in some defined manner. But... I run into problems of my own with what the equations of motion should be. For example, let's say I have a sine curve defined from zero to pi. I have a particle (object), that travels from zero to pi, but I want it to come back. How do you accomplish this? 
February 25th, 2019, 02:21 PM  #2 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
With vectorial calculus .

February 25th, 2019, 02:55 PM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,079 Thanks: 845 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
If you need it to do more than that, you'll need to give us more information. Dan Last edited by skipjack; February 26th, 2019 at 09:02 AM.  
February 26th, 2019, 03:27 AM  #4 
Newbie Joined: Feb 2019 From: United Kingdom Posts: 15 Thanks: 3 
I'm just using a basic example to say at time t where does the particle lie on the curve? Here's another example. Let's say I have a helix curve in the form of a spring. The particle travels forward 3 coils and travels backwards one coil at a constant velocity then repeats so at time t it's making its way along the spring but not in the usual way. Presumably it will have a form f(t) = h(t)  g(t) Last edited by NineDivines; February 26th, 2019 at 03:32 AM. 
February 26th, 2019, 07:13 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
A helix, with vertical axis and radius r, can be written as x= r cos(t), y= r sin(t), z= at. That will up three "coils" with t from 0 to $\displaystyle 6\pi$. If you want the particle to stop and go back to the initial point in time T, multiply each of those by $\displaystyle \sin(2\pi t/T)$.
Last edited by skipjack; February 26th, 2019 at 09:02 AM. 
February 26th, 2019, 11:43 AM  #6  
Newbie Joined: Feb 2019 From: United Kingdom Posts: 15 Thanks: 3  Quote:
But how do you account for the points on the curve where the velocity is zero? I could list them for obvious reasons by what you've shown say $\displaystyle 6 \pi n $ and $\displaystyle \sin(2n\pi t/T)$ but it would be much more concise if the derivative gave a magnitude of zero for the velocity vector. Last edited by skipjack; February 26th, 2019 at 01:45 PM.  
February 28th, 2019, 12:04 PM  #7 
Newbie Joined: Feb 2019 From: United Kingdom Posts: 15 Thanks: 3 
I’ve given this some thought and it requires a transformation to make the derivative for velocity equate to zero at these points. Any point I choose on whatever curve I choose, if the particle stops wherever it chooses only a transformation can handle it. I will figure this out.


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