☰ Reja · Matematika

Hosila: ta’rif, qoidalar, urinma

22-hafta · blok 1.13 · 3 dars
3

Urinma va normal tenglamasi

maktab darsligi: matematika 11-sinf, funksiya grafigiga o‘tkazilgan urinma
Maqsad: Berilgan nuqtada urinma va normal tenglamasini tuzish; berilgan og‘ishli urinma nuqtasini topish.
Qoida va g‘oya

y = f(x) grafigiga abssissasi x0 bo‘lgan nuqtadagi urinma tenglamasi: y = f(x0) + f'(x0) · (x - x0). Normal — urinmaga perpendikulyar bo‘lgan, o‘sha nuqtadan o‘tuvchi to‘g‘ri chiziq: uning og‘ish koeffitsiyenti -1/f'(x0): y = f(x0) - (x - x0)/f'(x0). Algoritm: 1) f(x0) ni toping; 2) f'(x) ni toping va f'(x0) ni hisoblang; 3) formulaga qo‘ying. Agar urinma berilgan y = kx + b ga parallel bo‘lsa, f'(x0) = k tenglamadan x0 topiladi. Agar urinma Ox o‘qi bilan 45° burchak tashkil qilsa, f'(x0) = tg 45° = 1. Urinmaning x o‘qiga parallel bo‘lishi f'(x0) = 0 ga teng.

Namunaviy masalalar
y = x^2 grafigiga x0 = 3 nuqtada urinma tenglamasini tuzing.
Yechim: f(3) = 9, f'(x) = 2x, f'(3) = 6. y = 9 + 6(x - 3) = 6x - 9.
Javob: y = 6x - 9
y = x^3 - 2x grafigiga x0 = 1 nuqtada urinma tenglamasini tuzing.
Yechim: f(1) = -1, f' = 3x^2 - 2, f'(1) = 1. y = -1 + (x - 1) = x - 2.
Javob: y = x - 2
y = x^2 - 4x grafigiga o‘tkazilgan urinma y = 2x ga parallel bo‘ladigan nuqtaning abssissasini toping.
Yechim: f' = 2x - 4 = 2, x = 3.
Javob: 3
Tipik xatolar
  • Urinma tenglamasida f'(x0) o‘rniga f(x0) ni qo‘llash.
  • Normalning og‘ishini urinma bilan bir xil olish: u -1/f'(x0).
  • Parallel urinmada f'(x0) = k ni emas, f(x0) = k ni yechish.
Mashq
1
y = x^2 + 1 ga x0 = 1 nuqtada urinma tenglamasini tuzing.
2
y = 1/x ga x0 = 2 nuqtada urinma tenglamasini tuzing.
3
y = e^x ga x0 = 0 nuqtada o‘tkazilgan urinmaning Ox bilan burchagini toping.
4
y = x^2 ga x0 = 1 nuqtada normal tenglamasini tuzing.