Solución

Primero, calcula fx(x,y)f_x(x, y) y fy(x,y)f_y(x, y), luego usa la ecuación 4.24 con x0=π/3x_0 = \pi/3 y y0=π/4y_0 = \pi/4:

fx(x,y)=2cos(2x)cos(3y)fy(x,y)=−3sen(2x)sen(3y)f(π3,π4)=sen(2(π3))cos(3(π4))=(32)(−22)=−64fx(π3,π4)=2cos(2(π3))cos(3(π4))=2(−12)(−22)=22fy(π3,π4)=−3sen(2(π3))sen(3(π4))=−3(32)(22)=−364\begin{aligned} f_x(x, y) &= 2cos(2x)cos(3y)\\ f_y(x, y) &= −3sen(2x)sen(3y)\\ f\bigg(\frac{\pi}{3}, \frac{\pi}{4}\bigg) &= sen\bigg(2\bigg(\frac{\pi}{3}\bigg)\bigg)cos\bigg(3\bigg(\frac{\pi}{4}\bigg)\bigg) = \bigg(\frac{\sqrt{3}}{2}\bigg)\bigg(-\frac{\sqrt{2}}{2}\bigg) = -\frac{\sqrt{6}}{4}\\ f_x\bigg(\frac{\pi}{3}, \frac{\pi}{4}\bigg) &= 2cos\bigg(2\bigg(\frac{\pi}{3}\bigg)\bigg)cos\bigg(3\bigg(\frac{\pi}{4}\bigg)\bigg) = 2\bigg(-\frac12\bigg)\bigg(-\frac{\sqrt{2}}{2}\bigg) = \frac{\sqrt{2}}{2}\\ f_y\bigg(\frac{\pi}{3}, \frac{\pi}{4}\bigg) &= -3sen\bigg(2\bigg(\frac{\pi}{3}\bigg)\bigg)sen\bigg(3\bigg(\frac{\pi}{4}\bigg)\bigg) = -3\bigg(\frac{\sqrt{3}}{2}\bigg)\bigg(\frac{\sqrt{2}}{2}\bigg) = -\frac{3\sqrt{6}}{4} \end{aligned}

Entonces la ecuación 4.24 se convierte en

z=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)z=−65+22(x−π3)−364(y−π4)x=22x−364y−64−π26+3π616\begin{aligned} z &= f(x_0, y_0) + f_x(x_0, y_0)(x − x_0) + f_y(x_0, y_0)(y − y_0)\\ z &= -\frac{\sqrt{6}}{5}+\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{3}\bigg) - \frac{3\sqrt{6}}{4}\bigg(y-\frac{\pi}{4}\bigg)\\ x &= \frac{\sqrt{2}}{2}x - \frac{3\sqrt{6}}{4}y - \frac{\sqrt{6}}{4} - \frac{\pi\sqrt{2}}{6} + \frac{3\pi\sqrt{6}}{16} \end{aligned}