14.1-bo'lim : Tangens tekisliklar va chiziqli yaqinlashishlar
2. \(\left( { - 4,3} \right)\) da \(z = x\sqrt {{x^2} + {y^2}} + {y^3}\) ga teguvchi tekislikning tenglamasini toping.
Yechim
Birinchidan, bizga ikkita 1 kerak bo'lishini bilamizst xususiy hosilalarni buyurtma qiling. Mana ular,
\[{f_x} = \sqrt {{x^2} + {y^2}} + \frac{{{x^2}}}{{\sqrt {{x^2} + {y^2}} }}\hspace{0.5in}{f_y} = \frac{{xy}}{{\sqrt {{x^2} + {y^2}} }} + 3{y^2}\]2-qadam
Endi bizga birinchi bosqichdagi ikkita lotin va \(\left( { - 4,3} \right)\) da baholangan funksiya kerak. Mana bu baholar,
\[f\left( { - 4,3} \right) = 7\hspace{0.5in}{f_x}\left( { - 4,3} \right) = \frac{{41}}{5}\hspace{0.5in}{f_y}\left( { - 4,3} \right) = \frac{{123}}{5}\]3-qadam
Tangens tekislik u holda,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{z = 7 + \frac{{41}}{5}\left( {x + 4} \right) + \frac{{123}}{5}\left( {y - 3} \right) = \frac{{41}}{5}x + \frac{{123}}{5}y - 34}}\]