Solución

Primero, debemos calcular f(x0,y0),fx(x0,y0)f(x_0, y_0), f_x(x_0, y_0) y fy(x0,y0)f_y(x_0, y_0) usando x0=2x_0 = 2 e y0=−3y_0 = −3:

f(x0,y0)=f(2,−3)=3(2)2−2(2)(−3)+(−3)2=12+12+9=33fx(x,y)=6x−2yfy(x,y)=−2x+2yfx(x0,y0)=fx(2,−3)=6(2)−2(−3)=12+6=18fy(x0,y0)=fy(2,−3)=−2(2)+2(−3)=−4−6=−10.\begin{aligned} f(x_0, y_0) &= f(2, −3) = 3(2)^2 − 2(2)(−3) + (−3)^2 = 12 + 12 + 9 = 33\\ f_x(x, y) &= 6x − 2y\\ f_y(x, y) &= −2x + 2y\\ f_x(x_0, y_0) &= f_x(2, −3) = 6(2) − 2(−3) = 12 + 6 = 18\\ f_y(x_0, y_0) &= f_y(2, −3) = −2(2) + 2(−3) = −4 − 6 = −10. \end{aligned}

Luego, sustituimos estas cantidades en la Ecuación 4.27:

dz=fx(x0,y0)dx+fy(x0,y0)dydz=18(0.1)−10(−0.05)=1.8+0.5=2.3\begin{aligned} dz &= f_x(x_0, y_0)dx + f_y(x_0, y_0)dy\\ dz &= 18(0.1) − 10(−0.05) = 1.8 + 0.5 = 2.3 \end{aligned}

Esta es la aproximación a Δz=f(x0+Δx,y0+Δy)−f(x0,y0)\Delta z = f(x_0 + \Delta x, y_0 + \Delta y) - f(x_0, y_0)

El valor exacto de Δz\Delta z viene dado por

Δz=f(x0+Δx,y0+Δy)−f(x0,y0)=f(2+0.1,−3−0.05)−f(2,−3)=f(2.1,−3.05)−f(2,−3)=2.3425\begin{aligned} \Delta z &= f(x_0 + \Delta x, y_0 + \Delta y) − f(x_0, y_0)\\ &= f(2 + 0.1, −3 − 0.05) − f(2, −3)\\ &= f(2.1, −3.05) − f(2, −3)\\ &= 2.3425 \end{aligned}