Matematika

11-sinf · Algebra · 25 daqiqa

Urinma tenglamasi

Hosila: mavzular

Bu nima va nega kerak

Egri chiziqning biror nuqtasi atrofini kattalashtirib qarasak, u to'g'ri chiziqqa yaqinlashadi. Bu to'g'ri chiziq egri chiziqning urinmasi deyiladi.

Amaliyotda murakkab funksiyalarning yaqin nuqtalardagi qiymatlarini tez va taqribiy hisoblashda urinma to'g'ri chizig'idan foydalaniladi (chiziqlashtirish).

DTM testlarida urinma tenglamasini tuzish, parallel yoki perpendikulyar urinmalarni topish va koordinata o'qlari bilan hosil bo'lgan uchburchak yuzini aniqlash masalalari muntazam tushadi.

Ushbu darsda urinma tenglamasini tuzish algoritmi va barcha turdagi amaliy masalalarni o'rganamiz.

Asosiy g'oya

1. Urinma tenglamasining keltirib chiqarilishi

(x0;y0)(x_0; y_0) nuqtadan o'tuvchi va burchak koeffitsiyenti kk bo'lgan to'g'ri chiziq tenglamasi: y−y0=k(x−x0)y - y_0 = k(x - x_0)

Hosilaning geometrik ma'nosiga ko'ra y0=f(x0)y_0 = f(x_0) va k=f′(x0)k = f'(x_0). Shunday qilib, urinma tenglamasi:

y=f(x0)+f′(x0)(x−x0)y = f(x_0) + f'(x_0)(x - x_0)

2. Urinma tenglamasini tuzish algoritmi

  1. Urinish nuqtasidagi funksiya qiymatini topish: y0=f(x0)y_0 = f(x_0).
  2. Funksiyaning hosilasini topish: f′(x)f'(x).
  3. Urinish nuqtasidagi burchak koeffitsiyentini topish: k=f′(x0)k = f'(x_0).
  4. Qiymatlarni y=y0+k(x−x0)y = y_0 + k(x - x_0) formulasiga qo'yib, y=kx+by = kx + b ko'rinishiga keltirish.

3. Maxsus holatlar va geometrik xossalar

  • Parallel to'g'ri chiziqlar: Agar urinma y=k1x+b1y = k_1 x + b_1 to'g'ri chiziqqa parallel bo'lsa: f′(x0)=k1f'(x_0) = k_1
  • Perpendikulyar to'g'ri chiziqlar: Agar urinma y=k1x+b1y = k_1 x + b_1 to'g'ri chiziqqa perpendikulyar bo'lsa: f′(x0)⋅k1=−1  ⟹  f′(x0)=−1k1f'(x_0) \cdot k_1 = -1 \implies f'(x_0) = -\frac{1}{k_1}
  • Koordinata o'qlari bilan uchburchak yuzi: y=kx+by = kx + b to'g'ri chiziq o'qlarni (0;b)(0; b) va (−bk;0)\left(-\frac{b}{k}; 0\right) nuqtalarda kesadi. Uchburchak yuzi: S=b22∣k∣S = \frac{b^2}{2|k|}

Yechilgan misollar

1-misol. Urinma tenglamasini tuzish

f(x)=x2−4x+5f(x) = x^2 - 4x + 5 parabola grafigiga x0=3x_0 = 3 nuqtada o'tkazilgan urinma tenglamasini tuzing.

Yechish:

  1. f(3)=32−4(3)+5=2f(3) = 3^2 - 4(3) + 5 = 2.
  2. f′(x)=2x−4f'(x) = 2x - 4.
  3. k=f′(3)=2(3)−4=2k = f'(3) = 2(3) - 4 = 2.
  4. y=2+2(x−3)=2x−4y = 2 + 2(x - 3) = 2x - 4.

Javob: y=2x−4y = 2x - 4.

2-misol. Berilgan to'g'ri chiziqqa parallel urinma

y=x3−3x+2y = x^3 - 3x + 2 egri chiziqqa o'tkazilgan qaysi urinmalar y=9x−5y = 9x - 5 to'g'ri chiziqqa parallel bo'ladi?

Yechish:

  1. Burchak koeffitsiyentlarini tenglaymiz: f′(x0)=9f'(x_0) = 9.
  2. 3x02−3=9  ⟹  3x02=12  ⟹  x0=±23x_0^2 - 3 = 9 \implies 3x_0^2 = 12 \implies x_0 = \pm 2.
  3. Har bir nuqta uchun urinma tenglamasini tuzamiz:
    • x1=2  ⟹  y1=23−3(2)+2=4x_1 = 2 \implies y_1 = 2^3 - 3(2) + 2 = 4. Urinma: y=4+9(x−2)=9x−14y = 4 + 9(x - 2) = 9x - 14.
    • x2=−2  ⟹  y2=(−2)3−3(−2)+2=0x_2 = -2 \implies y_2 = (-2)^3 - 3(-2) + 2 = 0. Urinma: y=0+9(x+2)=9x+18y = 0 + 9(x + 2) = 9x + 18.

Javob: y=9x−14y = 9x - 14 va y=9x+18y = 9x + 18.

3-misol. Urinma hosil qilgan uchburchak yuzi

f(x)=4xf(x) = \dfrac{4}{x} giperbola grafigiga x0=2x_0 = 2 nuqtada o'tkazilgan urinmaning koordinata o'qlari bilan chegaralangan uchburchagi yuzini toping.

Yechish:

  1. f(2)=2f(2) = 2.
  2. f′(x)=−4x2  ⟹  f′(2)=−1f'(x) = -\dfrac{4}{x^2} \implies f'(2) = -1.
  3. Urinma: y=2−1(x−2)=−x+4y = 2 - 1(x - 2) = -x + 4.
  4. O'qlar bilan kesishish nuqtalari: (0;4)(0; 4) va (4;0)(4; 0). Katetlari a=4,b=4a = 4, b = 4.
  5. Yuza: S=12⋅4⋅4=8S = \dfrac{1}{2} \cdot 4 \cdot 4 = 8.

Javob: 88.

4-misol. Tashqi nuqtadan o'tuvchi urinmalar

A(0;−4)A(0; -4) nuqtadan y=x2y = x^2 parabolaga o'tkazilgan urinmalar tenglamalarini tuzing.

Yechish:

  1. Paraboladagi noma'lum urinish nuqtasi x0x_0 bo'lsin. Unda y0=x02y_0 = x_0^2 va f′(x0)=2x0f'(x_0) = 2x_0.
  2. Urinma tenglamasi: y=x02+2x0(x−x0)=2x0x−x02y = x_0^2 + 2x_0(x - x_0) = 2x_0 x - x_0^2.
  3. Urinma A(0;−4)A(0; -4) nuqtadan o'tadi: −4=2x0(0)−x02  ⟹  x02=4  ⟹  x0=±2.-4 = 2x_0(0) - x_0^2 \implies x_0^2 = 4 \implies x_0 = \pm 2.
  4. Urinmalar:
    • x0=2  ⟹  y=4x−4x_0 = 2 \implies y = 4x - 4;
    • x0=−2  ⟹  y=−4x−4x_0 = -2 \implies y = -4x - 4.

Javob: y=4x−4y = 4x - 4 va y=−4x−4y = -4x - 4.

Mashq

  1. f(x)=x2+2x−3f(x) = x^2 + 2x - 3 parabolaga x0=1x_0 = 1 nuqtada o'tkazilgan urinma tenglamasini tuzing.
  2. f(x)=x3−2xf(x) = x^3 - 2x egri chiziqqa x0=−1x_0 = -1 nuqtada o'tkazilgan urinmaning burchak koeffitsiyentini toping.
  3. y=x2−6x+5y = x^2 - 6x + 5 parabolaga x0=2x_0 = 2 nuqtada o'tkazilgan urinmaning OyOy o'qi bilan kesishish nuqtasi ordinatasini toping.
  4. y=2x2−3x+1y = 2x^2 - 3x + 1 parabolaga qaysi nuqtada o'tkazilgan urinma y=5x+3y = 5x + 3 to'g'ri chiziqqa parallel bo'ladi?
  5. y=6xy = \dfrac{6}{x} egri chiziqqa x0=3x_0 = 3 nuqtada o'tkazilgan urinmaning koordinata o'qlari bilan hosil qilgan uchburchak yuzini hisoblang.
  6. f(x)=xf(x) = \sqrt{x} funksiya grafigiga x0=4x_0 = 4 nuqtada o'tkazilgan urinma tenglamasini yozing.
  7. y=x3−3x2+4y = x^3 - 3x^2 + 4 egri chiziqqa o'tkazilgan gorizontal urinmalar tenglamalarini toping.
  8. y=x2−3x+2y = x^2 - 3x + 2 parabolaga qaysi nuqtada o'tkazilgan urinma OxOx o'qi bilan 45∘45^\circ burchak hosil qiladi?
  9. Koordinatalar boshidan y=x2+4y = x^2 + 4 parabolaga o'tkazilgan urinmalarning burchak koeffitsiyentlarini toping.
  10. y=13x3−xy = \dfrac{1}{3}x^3 - x egri chiziqning x0=3x_0 = 3 nuqtasidagi urinmasi va normali burchak koeffitsiyentlari ko'paytmasini toping.
Javoblar va yechim yo'llari
  1. f(1)=0f(1) = 0, f′(1)=4  ⟹  y=0+4(x−1)=4x−4f'(1) = 4 \implies y = 0 + 4(x - 1) = 4x - 4. Javob: y=4x−4y = 4x - 4.

  2. k=f′(−1)=3(−1)2−2=1k = f'(-1) = 3(-1)^2 - 2 = 1. Javob: 11.

  3. f(2)=−3f(2) = -3, f′(2)=−2  ⟹  y=−2x+1f'(2) = -2 \implies y = -2x + 1. x=0x = 0 da y=1y = 1. Javob: 11.

  4. f′(x0)=5  ⟹  4x0−3=5  ⟹  x0=2f'(x_0) = 5 \implies 4x_0 - 3 = 5 \implies x_0 = 2. y(2)=3y(2) = 3. Nuqta: (2;3)(2; 3). Javob: (2;3)(2; 3).

  5. f(3)=2f(3) = 2, f′(3)=−23  ⟹  y=−23x+4f'(3) = -\dfrac{2}{3} \implies y = -\dfrac{2}{3}x + 4. Kesishish nuqtalari x=6,y=4  ⟹  S=12⋅6⋅4=12x = 6, y = 4 \implies S = \dfrac{1}{2} \cdot 6 \cdot 4 = 12. Javob: 1212.

  6. f(4)=2f(4) = 2, f′(4)=14  ⟹  y=2+14(x−4)=14x+1f'(4) = \dfrac{1}{4} \implies y = 2 + \dfrac{1}{4}(x - 4) = \dfrac{1}{4}x + 1. Javob: y=14x+1y = \dfrac{1}{4}x + 1.

  7. k=0  ⟹  3x2−6x=0  ⟹  x1=0,x2=2k = 0 \implies 3x^2 - 6x = 0 \implies x_1 = 0, x_2 = 2. y(0)=4  ⟹  y=4y(0) = 4 \implies y = 4; y(2)=0  ⟹  y=0y(2) = 0 \implies y = 0. Javob: y=4y = 4 va y=0y = 0.

  8. f′(x0)=tg 45∘=1  ⟹  2x0−3=1  ⟹  x0=2f'(x_0) = \text{tg}\,45^\circ = 1 \implies 2x_0 - 3 = 1 \implies x_0 = 2. y(2)=0y(2) = 0. Nuqta: (2;0)(2; 0). Javob: (2;0)(2; 0).

  9. Urinma: y=2x0x−x02+4y = 2x_0 x - x_0^2 + 4. (0;0)(0; 0) da −x02+4=0  ⟹  x0=±2-x_0^2 + 4 = 0 \implies x_0 = \pm 2. k=2x0  ⟹  k1=4,k2=−4k = 2x_0 \implies k_1 = 4, k_2 = -4. Javob: 44 va −4-4.

  10. O'zaro perpendikulyar to'g'ri chiziqlarning burchak koeffitsiyentlari ko'paytmasi k1⋅k2=−1k_1 \cdot k_2 = -1. Javob: −1-1.

Keng tarqalgan xatolar

1. Formuladagi ishorani adashtirish. ✗ y=f(x0)−f′(x0)(x−x0)y = f(x_0) - f'(x_0)(x - x_0) deb yozish. ✓ Urinma formulasida doimo qo'shish belgisi bo'ladi: y=f(x0)+f′(x0)(x−x0)y = f(x_0) + f'(x_0)(x - x_0).

2. Hosila ifodasini nuqtaga qo'ymasdan qoldirish. ✗ y=x2y = x^2 ga x0=2x_0 = 2 da urinma tuzishda y=4+2x(x−2)y = 4 + 2x(x - 2) deb yozish. ✓ Burchak koeffitsiyenti o'zgarmas son bo'lishi kerak: k=f′(2)=4k = f'(2) = 4. Urinma tenglamasi doimo chiziqli: y=4x−4y = 4x - 4.

3. Berilgan tashqi nuqtani urinish nuqtasi deb olish. ✗ (0;−4)(0; -4) nuqtadan o'tuvchi urinmani topishda x0=0x_0 = 0 deb olish. ✓ Nuqta funksiya grafigida yotmaydi. Avval urinish nuqtasi (x0;y0)(x_0; y_0) ni tenglama yordamida topish zarur.

4. Parallellik shartida funksiyalarning o'zini tenglash. ✗ y=x2y = x^2 ga y=4x+1y = 4x + 1 ga parallel urinma topishda x2=4x+1x^2 = 4x + 1 tenglamani yechish. ✓ Parallellik burchak koeffitsiyentlari tengligini bildiradi: f′(x0)=4  ⟹  2x0=4  ⟹  x0=2f'(x_0) = 4 \implies 2x_0 = 4 \implies x_0 = 2.

DTM uslubidagi test

Ushbu mavzu bo'yicha tayyorlangan 5 ta test savoli platformada interaktiv tarzda ishlaydi: o'z javobingizni tanlang va to'liq yechimni ko'ring.

Keyingi qadam

Biz hosilaning geometrik tatbiqi bo'lgan urinma tenglamasini to'liq o'rgandik. Keyingi darsda hosila yordamida funksiyalarning o'sish va kamayish oraliqlarini topish hamda funksiyani to'liq tekshirish usullari bilan tanishamiz.

Test

0 / 5 javob berildi

  1. 1.y=x2−2x+3y = x^2 - 2x + 3 parabolaga x0=2x_0 = 2 nuqtada o'tkazilgan urinma tenglamasini toping.

  2. 2.y=x3−3x2+5y = x^3 - 3x^2 + 5 egri chiziqqa x0=1x_0 = 1 nuqtada o'tkazilgan urinmaning burchak koeffitsiyentini toping.

  3. 3.y=x2+4xy = x^2 + 4x parabolaga o'tkazilgan urinma y=2x−5y = 2x - 5 to'g'ri chiziqqa parallel bo'lsa, urinish nuqtasini toping.

  4. 4.y=2xy = \dfrac{2}{x} giperbolaga x0=1x_0 = 1 nuqtada o'tkazilgan urinmaning koordinata o'qlari bilan hosil qilgan uchburchak yuzini toping.

  5. 5.y=x2−5x+6y = x^2 - 5x + 6 parabolaga x0=3x_0 = 3 nuqtada o'tkazilgan urinmaning OxOx o'qi musbat yo'nalishi bilan hosil qilgan burchagini toping.