Taylor expansion

What the problems look like

A few statements from this skill. Answers, hints and worked solutions live in the practice session.

1. Warm-up

   The second-order Taylor polynomial of $f(x,y)$ about $(-3, -4)$ is $T(x,y) = 6 - (x+3) + 3(y+4) - 3(x+3)^{2} - 2(x+3)(y+4) + 5(y+4)^{2}$. Since the coefficient of $(x+3)^{2}$ equals $\frac{1}{2}f_{x x}(-3,-4)$ and the coefficient of $(y+4)^{2}$ equals $\frac{1}{2}f_{y y}(-3,-4)$, find $f_{x x}(-3,-4)$ and $f_{y y}(-3,-4)$.

2. Warm-up

   The second-order Taylor polynomial of $f(x,y)$ about $(3, 4)$ is $T(x,y) = 8 + 2(x-3) + (y-4) + 4(x-3)^{2} - 3(x-3)(y-4) - 3(y-4)^{2}$. Since the coefficient of $(x-3)^{2}$ equals $\frac{1}{2}f_{x x}(3,4)$ and the coefficient of $(y-4)^{2}$ equals $\frac{1}{2}f_{y y}(3,4)$, find $f_{x x}(3,4)$ and $f_{y y}(3,4)$.

3. Warm-up

   The second-order Taylor polynomial of $f(x,y)$ about $(-2, 4)$ is $T(x,y) = 3 - 5(x+2) + 5(y-4) + 6(x+2)^{2} - 4(x+2)(y-4) - 4(y-4)^{2}$. Since the coefficient of $(x+2)^{2}$ equals $\frac{1}{2}f_{x x}(-2,4)$ and the coefficient of $(y-4)^{2}$ equals $\frac{1}{2}f_{y y}(-2,4)$, find $f_{x x}(-2,4)$ and $f_{y y}(-2,4)$.

Problems shown under their original terms: Generated (template, solver-computed).