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Hello,
My question is, given a mappingdefined by
,
the problem is to find the linear equations of the tangent plane into the graph of F at the point
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
Sorry I am so hopelessly confused - any clarity or guidance much appriciated.
The graph of the function y= f(x) is a curve inwith points denoted by (x, y). The graph of the function z= f(x,y) is a surface in
with points denoted by (x, y, z). The graph of the vector function (u,v)= f(x,y) is in
with points denoted by (x, y, u, v).
But you are right about evaluating F at the "given point": at,
,
, NOT (0, 0) as implied by
. Are you sure you have copied that correctly? If it were f(x,y)= (sin(x-y), cos(x+y)) then
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 vectoris 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
is normal to the level surface. Finally, if we think of the graph (u,v)= f(x,y) as a level curve of
, then
will be normal to it.
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!