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Math Help - Tangent Plane in 4-dimensional Euclidean space.

  1. #1
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    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.
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  2. #2
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    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.
    Last edited by HallsofIvy; March 11th 2011 at 01:54 AM.
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  3. #3
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    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!
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