# Equation of the tangent plane in R^4

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1. Feb 17, 2015

### 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 :)

Last edited by a moderator: Feb 21, 2015
2. Feb 19, 2015

### 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.