User Name Remember Me? Password

 Calculus Calculus Math Forum

 February 25th, 2019, 01:01 PM #1 Member   Joined: Feb 2019 From: United Kingdom Posts: 44 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: somewhere Posts: 634 Thanks: 91 With vectorial calculus . February 25th, 2019, 02:55 PM   #3
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,266
Thanks: 934

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by NineDivines 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?
Do you mean that you have a particle at x such that $\displaystyle x = \sin( \theta )$ and you need a $\displaystyle \theta (t)$ such that $\displaystyle \theta$ goes from 0 to $\displaystyle \pi$ to 0? Any function $\displaystyle \theta (t)$ that goes from 0 to $\displaystyle \pi$ to 0 (such as a parabola, or even a sinusoidal curve if you have that kind of sense of humor) will do.

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 Member   Joined: Feb 2019 From: United Kingdom Posts: 44 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 $\displaystyle \sin(2\pi t/T)$. Thanks from topsquark and idontknow Last edited by skipjack; February 26th, 2019 at 09:02 AM. February 26th, 2019, 11:43 AM   #6
Member

Joined: Feb 2019
From: United Kingdom

Posts: 44
Thanks: 3

Quote:
 Originally Posted by Country Boy 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 $\displaystyle \sin(2\pi t/T)$.
Yeah I see, that makes sense.

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 Member   Joined: Feb 2019 From: United Kingdom Posts: 44 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. Tags equations, parametric Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jiasyuen Calculus 8 January 13th, 2015 05:07 AM milanstar Algebra 8 May 17th, 2012 04:23 PM mathslog Algebra 4 May 4th, 2012 08:00 AM Yukina Calculus 2 July 14th, 2011 01:45 PM Yukina Calculus 1 July 13th, 2011 02:09 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      