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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(1e/5,1e/5).
Is this correct, and what are the bounds?
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(1e/5,1e/5). Is this correct, and what are the bounds?

This Question is Closed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6One way to test if it is the correct parameterization is : plug in \(t = 0\) and you should get the starting point

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, I get point 1,1, so I assume this is correct then

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Assuming 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)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0`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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it does not state the t value. So If time is a one way road, is my equation supposed to be the other way round?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Yes, at the minimum, I think your parameterization must agree on starting and ending points. start = (e/5, e/5) end = (1, 1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, so how can I find the bounds ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6If you fix the bounds to be \(0\le t\le 1\), then you can have an unique linear parameteriation : \(r(t) = (e/5,e/5) +t(1e/5,1e/5)\)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0The form is... I believe \(\sf \mathbf {\vec r} = P_0 +t\mathbf {\vec v}\) where \(\sf P_0\) is the initial starting point.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so do you just choose 0 and 1 ad fix, or there should be some way to go about it?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Thats a more natural and easiest way. You could also mess with your original parametric form and get suitable bounds for \(t\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Below parameterization works equally well too : \(r(t) = (1,1) +t(1e/5,1e/5)\) \(1\le t\le 0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Jhannybean ,@Empty ,@Astrophysics ,@imqwerty and @jim_thompson5910 thank you so much for taking some and looking at this. And all the help. Thank you!
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