14.1-bo'lim : Tangens tekisliklar va chiziqli yaqinlashishlar
1. \(\left( {2, - 1} \right)\) da \(\displaystyle z = {x^2}\cos \left( {\pi y} \right) - \frac{6}{{x{y^2}}}\) ga teguvchi tekislik tenglamasini toping.
Yechim
Birinchidan, bizga ikkita 1 kerak bo'lishini bilamizst xususiy hosilalarni buyurtma qiling. Mana ular,
\[{f_x} = 2x\cos \left( {\pi y} \right) + \frac{6}{{{x^2}{y^2}}}\hspace{0.5in}{f_y} = - \pi {x^2}\sin \left( {\pi y} \right) + \frac{{12}}{{x{y^3}}}\]2-qadam
Endi bizga birinchi bosqichdagi ikkita lotin va \(\left( {2, - 1} \right)\) da baholangan funksiya kerak. Mana bu baholar,
\[f\left( {2, - 1} \right) = - 7\hspace{0.5in}{f_x}\left( {2, - 1} \right) = - \frac{5}{2}\hspace{0.5in}{f_y}\left( {2, - 1} \right) = - 6\]3-qadam
Tangens tekislik u holda,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{z = - 7 - \frac{5}{2}\left( {x - 2} \right) - 6\left( {y + 1} \right) = - \frac{5}{2}x - 6y - 8}}\]