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Parametric equation and vector equation

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Find Parametric equation and vector equation for the portion of parabola y = 1+(x^2) from (1,2) to (2,5)

    2. Relevant equations


    3. The attempt at a solution
    m = (2-1) i +(5-2)j = i +3j
    x = t , y = 1+t^2
    r(t) = ti + (2=3t)j, where 0<t<1
    , but the ans given is r(t) = ti + (1+t^2) j , 1<t<2

    Is my ans acceptable ?
     
  2. jcsd
  3. Oct 30, 2016 #2

    PeroK

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    No. It's not clear to me what you are doing in your solution.
     
  4. Oct 30, 2016 #3
    Which part you don't understand ?
    My ans is
    r(t) = ti + (2=3t)j, where 0<t<1
    , but the ans given is r(t) = ti + (1+t^2) j , 1<t<2
     
  5. Oct 30, 2016 #4

    PeroK

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    I don't understand any of it. Not least ##(2=3t)j##.
     
  6. Oct 30, 2016 #5
    My ans is
    r(t) = ti + (2+3t)j, where 0<t<1
    , but the ans given is r(t) = ti + (1+t^2) j , 1<t<2
     
  7. Oct 30, 2016 #6

    PeroK

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    For ##t = 0##, you have ##r(0) = 2j##, which is not correct.

    Moreover, what you have is a straight line ##y = 3x + 2##. Not a parabola.
     
  8. Nov 1, 2016 #7
    Find a vector equation for the line segment that joins A(1, -1 , 2) and B (4,1,7)

    My working is :
    m = (4-1)i + (1+1 ) j + (7-2)k , so m = 3i + 2j +5k

    so , x= 1+2t , y = -1+ 2t z = (2+5t)k
    so t = (x-2)/2 = (3-y)/6 = (z+1)/3

    So , i use the same concept to do the question in post #1 , is it wrong ? why cant i do so ?
     
  9. Nov 1, 2016 #8

    PeroK

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    Is a parabola a straight line?
     
  10. Nov 1, 2016 #9
    No
     
  11. Nov 1, 2016 #10

    PeroK

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    All you are doing is finding the straight line between the end points.
     
  12. Nov 1, 2016 #11
    Ok
     
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