# Find the tangent plane to the graph of $f:\mathbb R^2\to \mathbb R^2$, $f(x,y)=(\sin(x-y),\cos(x+y))$

Let $f:\mathbb R^2\to \mathbb R^2$, $f(x,y)=(\sin(x-y),\cos(x+y))$, find the tangent plane to the graph of $f$ in $\mathbb R^4$ at $({\pi\over 4},{\pi\over 4},0,0)$.

What I did:

The equation of the tangent plane is given by $P(x,y)=f(x_0,y_0)+Df(x_0,y_0)\cdot (x-x_0,y-y_0)$ where $Df(x_0,y_0)$ is the jacobian matrix of $f$ at $(x_0,y_0)$.

Computing the partial derivatives:

$\displaystyle{\partial f_1(x,y)\over \partial x}=\cos(x-y)$

$\displaystyle{\partial f_2(x,y)\over \partial x}=-\sin(x+y)$

$\displaystyle{\partial f_1(x,y)\over \partial y}=-\cos(x-y)$

$\displaystyle{\partial f_2(x,y)\over \partial y}=-\sin(x+y)$

then evaluating at $(\dfrac\pi4,\dfrac\pi4)$ :

$\displaystyle{\partial f_1({\pi\over 4},{\pi\over 4})\over \partial x}=1$; $\displaystyle{\partial f_2({\pi\over 4},{\pi\over 4})\over \partial x}=-1$; $\displaystyle{\partial f_1({\pi\over 4},{\pi\over 4})\over \partial y}=-1$; $\displaystyle{\partial f_2({\pi\over 4},{\pi\over 4})\over \partial y}=-1$

Then we have that:

$$Df({\pi\over 4},{\pi\over 4})=\begin{bmatrix} 1 & -1 \\ -1 & -1\\ \end{bmatrix} \text{ and }f({\pi\over 4},{\pi\over 4})=(0,0)$$

Then we have that the tangent plane to the graph of $f$ at $\displaystyle({\pi\over 4},{\pi\over 4},0,0)$ is

$$P(x,y)=(0,0)+ \begin{pmatrix} 1 & -1 \\ -1 & -1\\ \end{pmatrix}\begin{pmatrix} x-{\pi\over 4} \\ y-{\pi\over 4}\\ \end{pmatrix}= (x-y,-x-y+{\pi\over 2})$$

I would really appreciate if you can tell me if this is the correct approach

• For $Df(\frac{\pi}4,\frac{\pi}4)$, it should be the determinant of the matrix you printed, not the matrix itself. I have tried to edit this accordingly. – Cookie Feb 17 '15 at 2:51
• why is the determinant instead of the matrix? – user128422 Feb 17 '15 at 2:55
• I thought $Df$ is typically the determinant. However, I will undo my edit since I don't remember too well about this myself. – Cookie Feb 17 '15 at 2:56
• Why do you have the point $(\pi/4,0)$? – science Feb 17 '15 at 3:07
• I´m sorry is the point $({\pi\over 4},{\pi\over 4},0,0)$ – user128422 Feb 17 '15 at 3:13

Here is an alternative method that verifies your result.

First of all, your Jacobian matrix is evaluated correctly— $$\frac{\partial \vec F}{\partial(x,y)} = \left[\begin{matrix} \partial F_x/\partial x & \partial F_x/\partial y \\ \partial F_y/\partial x & \partial F_y/\partial y \end{matrix}\right] = \left[\begin{matrix} \cos(x-y) & -\cos(x-y) \\ -\sin(x+y) & -\sin(x+y) \end{matrix}\right]$$ produces $$\left.\frac{\partial \vec F}{\partial(x,y)}\right|_{(\frac\pi4,\frac\pi4)} = \left[\begin{matrix} 1 & -1 \\ -1 & -1 \\ \end{matrix}\right].$$

In $\mathbb R^4$, the tangent plane is the intersection of 2 hyperplanes. Let $$ax+by+cu+dv+e=0,$$ be one such hyperplane, where we $u$ and $v$ represent the coordinate along the third and fourth dimension. A normal vector for the hyperplane would be $\vec n=(a,b,c,d)$, and is perpendicular to all lines that lie in the tangent plane.

From the entries of the Jacobian matrix, we know that the following two vectors are parallel to the tangent plane: $\vec s = (1,0,1,-1)$ and $\vec t = (0,1,-1,-1)$. Therefore, $$\begin{cases} \vec n \cdot \vec s = a + c - d = 0, \\ \vec n \cdot \vec t = b - c - d = 0. \end{cases}.$$

Furthermore, the hyperplanes must pass through $(\dfrac\pi4,\dfrac\pi4,0,0)$: $$\frac\pi4 a + \frac\pi4 b + e = 0 \implies e = -\frac\pi4 (a+b).$$

Since there are infinitely many pairs of hyperplanes that intersect at the desired plane, we arbitrary let $(a,b)=(1,1)$ for the first hyperplane and let $(a,b)=(1,-1)$ for the second hyperplane. This produces the equation of two hyperplanes: $$\begin{cases} x + y + v - \frac\pi2 &= 0, \\ x - y - u &= 0. \end{cases}$$

This is identical to your results of $$P(x,y) = (u,v) = (x-y, -x-y+\frac\pi2).$$