14.2-bo'lim: Gradient vektor, tangens tekisliklar va normal chiziqlar
Ushbu bo'limda biz tangens tekisliklarni qayta ko'rib chiqmoqchimiz, bu safar biz ularni gradient vektori nuqtai nazaridan ko'rib chiqamiz. Jarayonda biz sirtga oddiy chiziqni ham ko'rib chiqamiz.
Avval \(\vec n = \left\langle {a,b,c} \right\rangle \) normal vektorli \(\left( {{x_0},{y_0},{z_0}} \right)\) nuqtasini o'z ichiga olgan tekislikning tenglamasini eslaylik:
\[a\left( {x - {x_0}} \right) + b\left( {y - {y_0}} \right) + c\left( {z - {z_0}} \right) = 0\]Bo'limda gradient vektorini kiritganimizda yo'nalishli hosilalar quyidagi faktni keltirdik.
Gradient darajali egri chiziqlarga ortogonaldir
\(\nabla f\left( {{x_0},{y_0}} \right)\) gradient vektori ortogonal (yoki perpendikulyar) daraja egri chizig'i \(f\left( {x,y} \right) = k\) \(\left( {{x_0},{y_0}} \right)\) nuqtasida. Xuddi shunday, \(\nabla f\left( {{x_0},{y_0},{z_0}} \right)\) gradient vektori \(\left( {{x_0},{y_0},{z_0}} \right)\) nuqtasida \(f\left( {x,y,z} \right) = k\) sath yuzasiga ortogonaldir.
Aslida, bu erda bizga kerak bo'lgan narsa bu haqiqatning oxirgi qismidir. Bu shuni anglatadiki, gradient vektori har doim ortogonal yoki normal, bir nuqtada sirtga.
Shuni ham yodda tutingki, gradient vektori,
\[\nabla f = \left\langle {{f_x},{f_y},{f_z}} \right\rangle \]Shunday qilib, \(f\left( {x,y,z} \right) = k\) tomonidan \(\left( {{x_0},{y_0},{z_0}} \right)\) da berilgan sirtga teginish tekisligi tenglamaga ega:
\[{f_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) + {f_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0\]Bu oldingi bo'limda olinganidan ko'ra, tangens tekislik tenglamasining ancha umumiy shaklidir.
Ammo shuni yodda tutingki, biz ushbu umumiy formuladan foydalanib, avvalgi bo'limdagi tenglamani ham olishimiz mumkin. Buni ko'rish uchun \(z = f\left( {x,y} \right)\) tenglamasidan boshlaylik va biz \({z_0} = f\left( {{x_0},{y_0}} \right)\) bo'lgan \(\left( {{x_0},{y_0},{z_0}} \right)\) nuqtasida \(z = f\left( {x,y} \right)\) tomonidan berilgan sirtga teginish tekisligini topmoqchimiz. Yuqoridagi formuladan foydalanish uchun barcha o'zgaruvchilar bir tomonda bo'lishi kerak. Buni qilish yetarlicha oson. Buning uchun har ikki tomondan \(z\) ni ayirsak kifoya,
\[f\left( {x,y} \right) - z = 0\]Endi, agar biz yangi funktsiyani aniqlasak
\[F\left( {x,y,z} \right) = f\left( {x,y} \right) - z\]biz \(z = f\left( {x,y} \right)\) tomonidan berilgan sirt \(F\left( {x,y,z} \right) = 0\) tomonidan berilgan sirt bilan bir xil ekanligini va bu yangi ekvivalent tenglama biz ushbu bo'limda olingan tangens tekislik tenglamasi uchun to'g'ri shaklda ekanligini ko'rishimiz mumkin.
Shunday qilib, biz qilishimiz kerak bo'lgan birinchi narsa \(F\) uchun gradient vektorini topishdir.
\[\nabla F = \left\langle {{F_x},{F_y},{F_z}} \right\rangle = \left\langle {{f_x},{f_y}, - 1} \right\rangle \]E'tibor bering
\[\begin{array}{c}{F_x} = \frac{\partial }{{\partial x}}\left( {f\left( {x,y} \right) - z} \right) = {f_x}\hspace{0.75in}{F_y} = \frac{\partial }{{\partial y}}\left( {f\left( {x,y} \right) - z} \right) = {f_y}\\ {F_z} = \frac{\partial }{{\partial z}}\left( {f\left( {x,y} \right) - z} \right) = - 1\end{array}\]Tangens tekislikning tenglamasi u holda,
\[{f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right) - \left( {z - {z_0}} \right) = 0\]Yoki \(z\) ni hal qilib, biz quyidagilarni olamiz:
\[z = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right)\]Bu biz oldingi bo'limda olingan tenglama bilan bir xil.
Biz gradient vektoridan yana bir yaxshi ma'lumot olishimiz mumkin. Biz ba'zan bir nuqtada sirtga ortogonal bo'lgan chiziqni xohlashimiz mumkin, ba'zan esa deb ataladi oddiy chiziq. Agar biz buni eslasak, buni olish juda oson chiziq tenglamasi faqat nuqta va parallel vektorga ega bo'lishni talab qiladi. Biz \(\left( {{x_0},{y_0},{z_0}} \right)\) nuqtasida bo'lgan chiziqni xohlayotganimiz sababli, bu nuqta ham chiziqda bo'lishi kerakligini bilamiz va biz \(\nabla f\left( {{x_0},{y_0},{z_0}} \right)\) sirt uchun normal vektor ekanligini bilamiz va shuning uchun shunday bo'ladi. chiziqqa parallel. Demak, normal chiziq tenglamasi:
\[\vec r\left( t \right) = \left\langle {{x_0},{y_0},{z_0}} \right\rangle + t\,\nabla f\left( {{x_0},{y_0},{z_0}} \right)\]Yechim
Bu holda biz ishlamoqchi bo'lgan funksiya,
\[F\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}\]va teng belgisining bir tomonida nolga ega bo'lishimiz shart emasligiga e'tibor bering. Bizga kerak bo'lgan yagona narsa doimiydir. Ushbu muammoni hal qilish uchun bizga nuqtada baholangan gradient kerak bo'ladi.
\[\begin{align*}\nabla F & = \left\langle {2x,2y,2z} \right\rangle \\ \nabla F\left( {1, - 2,5} \right) & = \left\langle {2, - 4,10} \right\rangle \end{align*}\]Tangens tekislik u holda,
\[2\left( {x - 1} \right) - 4\left( {y + 2} \right) + 10\left( {z - 5} \right) = 0\]Oddiy chiziq,
\[\vec r\left( t \right) = \left\langle {1, - 2,5} \right\rangle + t\left\langle {2, - 4,10} \right\rangle = \left\langle {1 + 2t, - 2 - 4t,5 + 10t} \right\rangle \]