# Tangent Plane in 4-dimensional Euclidean space.

• March 10th 2011, 04:13 PM
matt.qmar
Tangent Plane in 4-dimensional Euclidean space.
Hello,

My question is, given a mapping $F: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ defined by $F(x,y) = (sin(x-y),cos(x-y))$,

the problem is to find the linear equations of the tangent plane in $\mathbb{R}^4$ to the graph of F at the point $(\pi/2, \pi/2,0,0)$

I am confused because the image of F has two components, so how could it have a four-dimensional tangent plane? Wouldn't there be infinetly tangent planes in $\mathbb{R}^4$? We can't even evaluate F at the given point...

Sorry I am so hopelessly confused - any clarity or guidance much appriciated.
• March 10th 2011, 04:36 PM
HallsofIvy
The graph of the function y= f(x) is a curve in $R^2$ with points denoted by (x, y). The graph of the function z= f(x,y) is a surface in $R^3$ with points denoted by (x, y, z). The graph of the vector function (u,v)= f(x,y) is in $R^4$ with points denoted by (x, y, u, v).

But you are right about evaluating F at the "given point": at $x= \pi/2$, $y= \pi/2$, $F(x,y)= F(\pi/2, \pi/2)= (sin(\pi/2-\pi/2, cos(\pi/2- \pi/2))= (sin(0), cos(0))= (0, 1)$, NOT (0, 0) as implied by $(\pi/2, \pi/2, 0, 0)$. Are you sure you have copied that correctly? If it were f(x,y)= (sin(x-y), cos(x+y)) then $f(\pi/2,\pi/2)= (0, 0)$ would be correct.

If we think of the curve y= f(x) as a "level curve" of F(x,y)= y- f(x)= 0, then the two dimensional vector $\nabla F= -f'(x)\vec{i}+ \vec{j}$ is normal to the curve. If we think of the surface z= f(x,y) as a "level surface" of F(x,y,z)= z- f(x,y), then the three dimensional vector $\nabla F= -f_x\vec{i}- f_y\vec{j}+ \vec{k}$ is normal to the level surface. Finally, if we think of the graph (u,v)= f(x,y) as a level curve of $F(x,y,u,v)= (u,v)- f(x,y)$, then $\nabla F= -f_x\vec{i}- f_y\vec{j}+ \vec{k}+ \vec{l}$ will be normal to it.
• March 10th 2011, 04:56 PM
matt.qmar
Aha! I understand... the graphof the function includes the arguments as well as the image... (x, y, F(x,y)) where F(x,y) has two components!

And yes, you are correct, cos(x+y) is correct, I mis-copied.

Thank you!