Equation of the tangent plane in R^4

[david34]

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

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

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

\begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix}

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

P(x,y)= (x-y,-x-y+\frac{\pi}{2})

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 :)

 

[wabbit]

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.