Answer
${f_x}{|_A} \approx \frac{{\Delta f}}{{\Delta x}} = 10$
${f_y}{|_A} \approx \frac{{\Delta f}}{{\Delta y}} = - 20$
Work Step by Step
We see in Figure 8, the coordinate of $A = \left( {x,y} \right) = \left( {2, - 2} \right)$ is at the level curve $30$.
1. Estimate ${f_x}$ at point $A$.
We move to the right horizontally to the next level curve $40$ at the estimated coordinate $\left( {3, - 2} \right)$. Notice that by doing so, we have kept the $y$-coordinate constant.
So,
${f_x}{|_A} \approx \frac{{\Delta f}}{{\Delta x}} = \frac{{40 - 30}}{{3 - 2}} = \frac{{10}}{1} = 10$
2. Estimate ${f_y}$ at point $A$.
Similarly, we move up vertically from point $A$ to the next level curve $20$ at the estimated coordinate $\left( {2, - 1.5} \right)$. Notice that by doing so, we have kept the $x$-coordinate constant.
So,
${f_y}{|_A} \approx \frac{{\Delta f}}{{\Delta y}} = \frac{{20 - 30}}{{ - 1.5 - \left( { - 2} \right)}} = \frac{{ - 10}}{{0.5}} = - 20$
