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Equation of the tangent plane in R^4

  1. Feb 17, 2015 #1
    Let [itex] f: \mathbb R^2 \to \mathbb R^2 [/itex] given by [itex] f=(sin(x-y),cos(x+y)) [/itex] : find the equation of the tangent plane to the graph of the function in [itex] \mathbb R^4 [/itex] at [itex] (\frac{\pi}{4}, \frac{\pi}{4}, 0 ,0 ) [/itex] and then find a parametric representation of the equation of the tangent plane

    What I did: the equation of the tangent plane from [itex] \mathbb R^2 \to \mathbb R^2 [/itex] is given by

    [itex] P(x,y)=f(x_0,y_0)+Df(x_0,y_0)\cdot (x-x_0,y-y_0)^T [/itex] where [itex] Df(x_0,y_0) [/itex] is the jacobian matrix of f at [itex] (x_0,y_0) [/itex]computing this matrix and evaluating at the point [itex] (\frac{\pi}{4}, \frac{\pi}{4}) [/itex] yields

    [tex] \begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix} [/tex]

    we also have that [itex] f(x_0,y_0)= f(\frac{\pi}{4}, \frac{\pi}{4})=(0,0) [/itex] then we have that the equation of the tangent plane at [itex] (\frac{\pi}{4}, \frac{\pi}{4}, 0 ,0 ) [/itex] is:

    [itex] P(x,y)= (x-y,-x-y+\frac{\pi}{2}) [/itex]

    but I don´t know if this is the correct approach; I also don´t know how to get the parametric representation of the tangent plane. I would really appreciate if you can help me with this problem :)
    Last edited by a moderator: Feb 21, 2015
  2. jcsd
  3. Feb 19, 2015 #2


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    You're doing fine mostly but two things to note :
    -You dropped the first two coordinates ; that's OK as long as you mean that as a shorthand notation but here it doesn't help. Points on the tangent plane (seen as a subset of ## R^4 ##) are of the form (x,y,P(x,y)), keeping P as you define it.
    -This is already a parametric representation, x and y being the parameters. An equation would be something of the form f(x,y,P(x,y))=0. In this case it wiill be two equations, since that tangent plane is a 2d thing in 4d space.
    Or maybe I misunderstood your question.
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