MATH 1920 Lecture Notes - Lecture 12: Cross Product

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*2 and !2* are continuous near (a, b) then ! *2 =!2* then f(x, y) exists, if not, then f(x, y) does not exist. Thus, f does not exist at point (a, b) in the domain d. We want a plane that contains both of these lines. The cross product of the direction vectors gives the normal to the tangent plane (1) mn(o)= <4,5,!(4,5)>+o<1,0,! Equation of the tangent plane: ( <4,5,! (4,5)>) < ! Find the equation of the tangent line at tu( ,0v. Linearization of f(x, y) at (a, b): a(#,%)=!(4,5)+! Find the equation of the tangent plane and realize it as the graph of a linear function ( = u < k t# u( v+0 ( u < k t# u( v. Definition: f is differentiable at (a, b) if the linear approximation l(x, y) is a good approximation for f(x, y) near (a, b) So the "tangent plane" of f at (0, 0) is z = 0.

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