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Vec Calc - tangent to path

  1. Jan 19, 2006 #1
    The Problem: No clue how to begin the following. The textbook isn't too clear.

    Let [itex]f: \mathbb{R}^2 \rightarrow \mathbb{R}^2 ; (x,y) \rightarrow (e^{x+y}, e^{x-y})[/itex] Let c(t) be a path with c(0)=(0,0) and c'(0)=(1,1). What is the tangent vector to the image of c(t) under f at t=0?

    Any help at all would be much appreciated.
  2. jcsd
  3. Jan 19, 2006 #2


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    Let [tex] \vec{c}(t) = \left< x(t), y(t)\right> [/tex]

    then the image of c(t) under f is

    [tex] f( \vec{c}(t) ) = f(x(t), y(t)) = \left( e^{x(t)+y(t)},e^{x(t)-y(t)}\right)[/tex]

    The tangent vector is

    [tex] \frac{d}{dt} f( \vec{c}(t) ) =\frac{d}{dt} f(x(t), y(t)) = \frac{d}{dt}\left( e^{x(t)+y(t)},e^{x(t)-y(t)}\right) = \left( \frac{d}{dt} e^{x(t)+y(t)}, \frac{d}{dt} e^{x(t)-y(t)}\right) [/tex]
    [tex]= \left( e^{x(t)+y(t)}\left( x^{\prime}(t)+y^{\prime}(t)\right) , e^{x(t)-y(t)}\left( x^{\prime}(t)-y^{\prime}(t)\right) \right)[/tex]

    evalute at t=0 to get

    [tex]\left( e^{x(0)+y(0)}\left( x^{\prime}(0)+y^{\prime}(0)\right) , e^{x(0)-y(0)}\left( x^{\prime}(0)-y^{\prime}(0)\right) \right)[/tex]

    recall that c(0)=(0,0) and c'(0)=(1,1) and hence x(0)=0, y(0)=0, x'(0)=1, and y'(0)=1 so that

    [tex]\left( e^{x(0)+y(0)}\left( x^{\prime}(0)+y^{\prime}(0)\right) , e^{x(0)-y(0)}\left( x^{\prime}(0)-y^{\prime}(0)\right) \right) = \left( e^{0+0}\left( 1+1\right) , e^{0-0}\left( 1-1\right) \right) = \left( 2 , 0\right)[/tex]

    is the tangent vector to the image of c(t) under f at t=0.

    Though I'm not sure... is f a vector valued function of vectors?
    Last edited: Jan 19, 2006
  4. Jan 19, 2006 #3


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    Yes, that's exactly what [itex]f: \mathbb{R}^2 \rightarrow \mathbb{R}^2 ; (x,y) \rightarrow (e^{x+y}, e^{x-y})[/itex] means.

    And your solution is completely correct. It's really an exercise in using the chain rule.
  5. Jan 19, 2006 #4


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    Thanx Ivy.
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