Recipe

Build the second-order Taylor polynomial in two variables

Use when: you need a quadratic, not just linear, local approximation to f(x,y) near a point (a,b), one that also matches the surface's curvature

Goal: the polynomial P_2(x,y), matching f's value, both first partials, and all three second partials at (a,b)

Taught in Taylor's theorem in two variables

1. 1

   Find f(a,b), f_x(a,b), f_y(a,b), exactly as for the tangent plane.

   Example

   For f(x,y)=x^2y-y^2 at (1,2): f(1,2)=1^2(2)-2^2=2-4=-2; f_x=2xy\Rightarrow f_x(1,2)=4; f_y=x^2-2y\Rightarrow f_y(1,2)=1-4=-3.

   Why: these three numbers build the linear part of P_2, the same tangent plane the earlier section already found.

2. 2

   Find f_{xx}, f_{yy}, and the mixed partial f_{xy}. By Clairaut's Theorem f_{xy}=f_{yx}, so only one of the two mixed partials needs computing.

   Example

   f_{xx}=2y\Rightarrow f_{xx}(1,2)=4; f_{yy}=-2 (constant); f_{xy}=2x\Rightarrow f_{xy}(1,2)=2.

   Why: Clairaut's Theorem is what guarantees there are only three distinct second partials to find here, not four.

3. 3

   Substitute every value into P_2(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)+\tfrac12f_{xx}(a,b)(x-a)^2+f_{xy}(a,b)(x-a)(y-b)+\tfrac12f_{yy}(a,b)(y-b)^2.

   Example

   P_2(x,y)=-2+4(x-1)-3(y-2)+2(x-1)^2+2(x-1)(y-2)-(y-2)^2

   Why: the \tfrac12 coefficients on the pure second-order terms mirror single-variable Taylor's \tfrac{f''(a)}{2!}. The mixed term carries no such factor, since it already appears once from each differentiation order.

4. 4

   The first three terms of P_2, everything except the three quadratic corrections, should be identical to the tangent-plane formula from the earlier section.

   Why: P_2 adds curvature corrections ON TOP OF the tangent plane rather than replacing it, so this check catches a dropped or miscopied linear term before it compounds into the quadratic part.