13.3-bo'lim : Xususiy hosilalarning talqini
3. \(\left( {2,0} \right)\) da \(f\left( {x,y} \right) = x\,{{\bf{e}}^{2x - {y^{\,2}}}}\) izlariga tangens chiziqlarning vektor tenglamalarini yozing.
Yechim
Biz bilamizki, ikkita iz bor. \(x = 2\) uchun bitta (i.e. \(x\) o'rnatildi va \(y\) o'zgarishi mumkin) va \(y = 0\) uchun bitta (i.e. \(y\) sobit va \(x\) o'zgarishiga ruxsat beriladi). Bundan tashqari, biz \({f_y}\left( {2,0} \right)\) birinchi iz uchun qiyalik bo'lishini bilamiz (\(y\) o'zgaradi va \(x\) aniq!) va \({f_x}\left( {2,0} \right)\) ikkinchi iz uchun qiyalik bo'ladi (\(x\) o'zgaradi va \(y\) aniq!).
Shunday qilib, bizga ushbu ikkala xususiy hosilalarning qiymati kerak bo'ladi. Mana bu ish,
\[\begin{align*}{f_x}\left( {x,y} \right) & = \,{{\bf{e}}^{2x - {y^{\,2}}}} + 2x\,{{\bf{e}}^{2x - {y^{\,2}}}} & \hspace{0.05in} & \to \hspace{0.25in}{f_x}\left( {2,0} \right) = 5{{\bf{e}}^4} = 272.9908\\ {f_y}\left( {x,y} \right) & = - 2yx\,{{\bf{e}}^{2x - {y^{\,2}}}} & \hspace{0.05in} & \to \hspace{0.25in}{f_y}\left( {2,0} \right) = 0\end{align*}\]2-qadam
Endi biz chiziqning vektor tenglamasini yozishimiz kerak va shuning uchun biz (ba'zi darajada) oldingi bosqichda sanab o'tilgan "qiyaliklar" ga muhtoj emasmiz. Bizga bu qiyaliklarni beradigan tangens vektorlar kerak.
Eslatmalardan birinchi iz uchun tangens vektori ekanligini eslang,
\[\left\langle {0,1,{f_y}\left( {2,0} \right)} \right\rangle = \left\langle {0,1,0} \right\rangle \]Xuddi shunday, ikkinchi iz uchun tangens vektori,
\[\left\langle {1,0,{f_x}\left( {2,0} \right)} \right\rangle = \left\langle {1,0,5{{\bf{e}}^4}} \right\rangle \]3-qadam
Keyinchalik, biz ko'rib chiqayotgan sirtdagi nuqta uchun pozitsiya vektori ham kerak bo'ladi. Bu,
\[\left\langle {2,0,f\left( {2,0} \right)} \right\rangle = \left\langle {2,0,2{{\bf{e}}^4}} \right\rangle \]4-qadam
Nihoyat, birinchi iz uchun tangens chiziqning vektor tenglamasi:
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left\langle {2,0,2{{\bf{e}}^4}} \right\rangle + t\left\langle {0,1,0} \right\rangle = \left\langle {2,t,2{{\bf{e}}^4}} \right\rangle }}\]ikkinchi izning izi esa,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left\langle {2,0,2{{\bf{e}}^4}} \right\rangle + t\left\langle {1,0,5{{\bf{e}}^4}} \right\rangle = \left\langle {2 + t,0,2{{\bf{e}}^4} + 5{{\bf{e}}^4}t} \right\rangle }}\]