anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Jhannybean
  • Jhannybean
@Empty
ganeshie8
  • ganeshie8
One way to test if it is the correct parameterization is : plug in \(t = 0\) and you should get the starting point
anonymous
  • anonymous
oh, I get point 1,1, so I assume this is correct then

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ganeshie8
  • 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)
jim_thompson5910
  • 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.
anonymous
  • 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?
ganeshie8
  • ganeshie8
Yes, at the minimum, I think your parameterization must agree on starting and ending points. start = (e/5, e/5) end = (1, 1)
anonymous
  • anonymous
ok, so how can I find the bounds ?
ganeshie8
  • ganeshie8
If you fix the bounds to be \(0\le t\le 1\), then you can have an unique linear parameteriation : \(r(t) = (e/5,e/5) +t(1-e/5,1-e/5)\)
Jhannybean
  • Jhannybean
The form is... I believe \(\sf \mathbf {\vec r} = P_0 +t\mathbf {\vec v}\) where \(\sf P_0\) is the initial starting point.
anonymous
  • anonymous
so do you just choose 0 and 1 ad fix, or there should be some way to go about it?
ganeshie8
  • ganeshie8
Thats a more natural and easiest way. You could also mess with your original parametric form and get suitable bounds for \(t\)
ganeshie8
  • ganeshie8
Below parameterization works equally well too : \(r(t) = (1,1) +t(1-e/5,1-e/5)\) \(-1\le t\le 0\)
anonymous
  • 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!
Jhannybean
  • Jhannybean
Np :)

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