## anonymous one year ago If a particle moves from point (e/5,e/5) to point (1,1), what is the parametric equation and its bounds? So I subtracted the two and and said: r(t) = (1,1) +t(1-e/5,1-e/5). Is this correct, and what are the bounds?

1. anonymous

@Empty

2. ganeshie8

One way to test if it is the correct parameterization is : plug in $$t = 0$$ and you should get the starting point

3. anonymous

oh, I get point 1,1, so I assume this is correct then

4. ganeshie8

Assuming time is a one way road, going from (e/5, e/5) to (1, 1) is not same as going from (1, 1) to (e/5, e/5)

5. jim_thompson5910

If a particle moves from point (e/5,e/5) to point (1,1), what is the parametric equation and its bounds? How long does it take to go from (1,1) to (e/5, e/5) ? It doesn't state the time t value.

6. anonymous

it does not state the t value. So If time is a one way road, is my equation supposed to be the other way round?

7. ganeshie8

Yes, at the minimum, I think your parameterization must agree on starting and ending points. start = (e/5, e/5) end = (1, 1)

8. anonymous

ok, so how can I find the bounds ?

9. ganeshie8

If you fix the bounds to be $$0\le t\le 1$$, then you can have an unique linear parameteriation : $$r(t) = (e/5,e/5) +t(1-e/5,1-e/5)$$

10. anonymous

The form is... I believe $$\sf \mathbf {\vec r} = P_0 +t\mathbf {\vec v}$$ where $$\sf P_0$$ is the initial starting point.

11. anonymous

so do you just choose 0 and 1 ad fix, or there should be some way to go about it?

12. ganeshie8

Thats a more natural and easiest way. You could also mess with your original parametric form and get suitable bounds for $$t$$

13. ganeshie8

Below parameterization works equally well too : $$r(t) = (1,1) +t(1-e/5,1-e/5)$$ $$-1\le t\le 0$$

14. anonymous

@Jhannybean ,@Empty ,@Astrophysics ,@imqwerty and @jim_thompson5910 thank you so much for taking some and looking at this. And all the help. Thank you!

15. anonymous

Np :)