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Vector Functions

  1. Sep 18, 2014 #1
    1. The problem statement, all variables and given/known data
    (a) Sketch the plane with the given vector expression
    (b) find r'(t)
    (c) sketch the position vector r(t) and the tangent vector r'(t) for the given value of t

    r(t) = <t^2,t^3> t=1
    2. Relevant equations



    3. The attempt at a solution
    1) i derived the equation : r'(t) = <2t,3t^2>
    2) sub in t=1: r(t)=<1,1>
    r'(t)=<2,3>

    I've always been a terrible graphing person so if anyone can help me improve on this, it would be really helpful
     
  2. jcsd
  3. Sep 18, 2014 #2

    Simon Bridge

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    Please show your working and reasoning.
     
  4. Sep 19, 2014 #3

    HallsofIvy

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    Do you understand that, in the xy-plane, r= <t^2, t^3> covers the same curve as x= t^2, y= t^3? Have you just calculated a number of (x, y) points for different t to help in drawing the curve? For example when t= 0, x and y are both 0 so (0, 0) is a point on the curve. When t= 1 the point is (1, 1). When t= -1, (1, -1), when t= 2, (4, 8), when t= 1/2, (1/4, 1/8), etc.
     
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