Burchakning radian o'lchovi

Bu nima va nega kerak

Burchaklar qadimdan graduslarda o'lchab kelingan: aylana $360$ teng bo'lakka bo'linadi. Bu qadimgi Bobil astronomiyasidan qolgan shartli o'lchovdir.

Biroq matematik analiz va funksiyalar nazariyasida gradus noqulaylik tug'diradi:

1. $360$ soni sun'iy tanlangan bo'lib, matematik tabiiy asosga ega emas.
2. $y = x + \sin x$ kabi ifodalarda $x$ sonini gradusdagi burchak bilan qo'shib bo'lmaydi (o'lchov birliklari mos emas).
3. Hosila va integrallarda gradus doimo ortiqcha $\dfrac{\pi}{180^\circ}$ koeffitsiyentini keltirib chiqaradi.

Shu sababli fanda burchakning tabiiy o'lchovi — radian qabul qilingan. Radian burchakni radius va yoy uzunligi orqali to'g'ridan-to'g'ri haqiqiy sonlar bilan bog'laydi.

Asosiy g'oya

1. Radian ta'rifi

Ta'rif: Radiusi $R$ bo'lgan aylanada uzunligi $R$ ga teng yoyga tiralgan markaziy burchak $1$ radian deb ataladi ($1\text{ rad}$).

Aylanada yoy uzunligining radiusga nisbati o'zgarmasdir: $\alpha = \frac{l}{R}$

Bu yerda $\alpha$ — markaziy burchak (radianda), $l$ — yoy uzunligi, $R$ — aylana radiusi. Radian o'lchovsiz birlik bo'lgani uchun ko'pincha "rad" yozilmaydi: $\alpha = \pi$ yoki $\alpha = 2$.

2. Radian va gradus bog'lanishi

To'liq aylananing uzunligi $C = 2\pi R$. Unga mos to'liq burchak gradusda $360^\circ$, radianda esa $\dfrac{2\pi R}{R} = 2\pi\text{ rad}$.

Asosiy tenglik: $180^\circ = \pi\text{ rad}$

Bundan o'tish qoidalari kelib chiqadi:

• Gradusdan radianga: $a^\circ = a \cdot \dfrac{\pi}{180^\circ}\text{ rad}$ ($1^\circ \approx 0{,}0175\text{ rad}$)
• Radiandan gradusga: $\alpha\text{ rad} = \alpha \cdot \dfrac{180^\circ}{\pi}$ ($1\text{ rad} \approx 57{,}3^\circ$)

3. Asosiy burchaklar jadvali

Gradus ($^\circ$)	Radian (rad)	Gradus ($^\circ$)	Radian (rad)
$0^\circ$	$0$	$120^\circ$	$\dfrac{2\pi}{3}$
$30^\circ$	$\dfrac{\pi}{6}$	$135^\circ$	$\dfrac{3\pi}{4}$
$45^\circ$	$\dfrac{\pi}{4}$	$150^\circ$	$\dfrac{5\pi}{6}$
$60^\circ$	$\dfrac{\pi}{3}$	$180^\circ$	$\pi$
$90^\circ$	$\dfrac{\pi}{2}$	$270^\circ$	$\dfrac{3\pi}{2}$
$360^\circ$	$2\pi$	—	—

4. Yoy uzunligi va sektor yuzi formulalari

Burchak radianda ifodalanganda formulalar juda ixcham bo'ladi:

1. Aylana yoyining uzunligi: $l = \alpha \cdot R$
2. Doiraviy sektor yuzi: $S = \dfrac{1}{2}\alpha R^2 = \dfrac{l \cdot R}{2}$

Yechilgan misollar

1-misol. Gradus o'lchovini radianga aylantirish

Quyidagi burchaklarni radianda ifodalang: a) $75^\circ$; b) $240^\circ$; c) $-150^\circ$.

Yechilishi: $\alpha = a^\circ \cdot \dfrac{\pi}{180^\circ}$ formulasidan foydalanamiz.

a) $75^\circ = 75 \cdot \dfrac{\pi}{180} = \dfrac{75}{180}\pi$: Surat va maxrajni $15$ ga qisqartiramiz: $\frac{75}{180} = \frac{5}{12} \implies 75^\circ = \frac{5\pi}{12}\text{ rad}.$

b) $240^\circ = 240 \cdot \dfrac{\pi}{180} = \dfrac{240}{180}\pi$: $60$ ga qisqartirsak: $\frac{240}{180} = \frac{4}{3} \implies 240^\circ = \frac{4\pi}{3}\text{ rad}.$

c) Manfiy burchaklar ham shu tarzda o'tkaziladi: $-150^\circ = -150 \cdot \frac{\pi}{180} = -\frac{5\pi}{6}\text{ rad}.$

Javob: a) $\dfrac{5\pi}{12}$; b) $\dfrac{4\pi}{3}$; c) $-\dfrac{5\pi}{6}$.

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2-misol. Radian o'lchovini gradusga aylantirish

Quyidagi burchaklarni gradusda ifodalang: a) $\dfrac{5\pi}{4}$; b) $-\dfrac{7\pi}{12}$; c) $2\text{ rad}$.

Yechilishi:

a) $\pi$ o'rniga $180^\circ$ qo'yamiz: $\frac{5\pi}{4} = \frac{5 \cdot 180^\circ}{4} = 225^\circ.$

b) $-\dfrac{7\pi}{12} = -\dfrac{7 \cdot 180^\circ}{12} = -105^\circ$.

c) $1\text{ rad} \approx 57{,}3^\circ$ dan foydalansak: $2\text{ rad} = 2 \cdot \frac{180^\circ}{\pi} \approx 114{,}6^\circ.$

Javob: a) $225^\circ$; b) $-105^\circ$; c) $\approx 114{,}6^\circ$.

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3-misol. Aylana yoyi uzunligini topish

Radiusi $R = 15\text{ sm}$ bo'lgan aylanada $144^\circ$ li markaziy burchakka tiralgan yoy uzunligini toping.

Yechilishi:

1. Burchakni radianga o'tkazamiz: $\alpha = 144^\circ \cdot \frac{\pi}{180^\circ} = \frac{4\pi}{5}\text{ rad}.$

2. Yoy uzunligini hisoblaymiz: $l = \alpha \cdot R = \frac{4\pi}{5} \cdot 15 = 12\pi\text{ sm}.$

Javob: $12\pi\text{ sm}$ (yoki $\approx 37{,}68\text{ sm}$).

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4-misol. Doiraviy sektor parametrlarini hisoblash

Doiraviy sektorning perimetri $28\text{ sm}$, yuzi esa $48\text{ sm}^2$ ga teng. Sektorning radiusi va burchagini toping.

Yechilishi: Perimetr $P = 2R + l = 28 \implies l = 28 - 2R$.

Sektor yuzi: $S = \frac{l \cdot R}{2} = \frac{(28 - 2R)R}{2} = (14 - R)R = 48.$

Kvadrat tenglamani yechamiz: $R^2 - 14R + 48 = 0 \implies R_1 = 6, \; R_2 = 8.$

1. $R = 6\text{ sm}$ bo'lsa: $l = 28 - 12 = 16\text{ sm}$. Burchak: $\alpha = \dfrac{16}{6} = \dfrac{8}{3}\text{ rad}$.
2. $R = 8\text{ sm}$ bo'lsa: $l = 28 - 16 = 12\text{ sm}$. Burchak: $\alpha = \dfrac{12}{8} = 1{,}5\text{ rad}$.

Ikkala holat ham shartga mos keladi.

Javob: $R = 6\text{ sm}, \alpha = \dfrac{8}{3}\text{ rad}$ yoki $R = 8\text{ sm}, \alpha = 1{,}5\text{ rad}$.

Mashq

Quyidagi masalalarni mustaqil yeching va javobingizni tekshiring:

1. Quyidagi burchaklarni radianda ifodalang: a) $20^\circ$; b) $150^\circ$; c) $315^\circ$; d) $-210^\circ$.

2. Quyidagi burchaklarni gradusda ifodalang: a) $\dfrac{5\pi}{6}$; b) $-\dfrac{4\pi}{3}$; c) $\dfrac{7\pi}{2}$; d) $\dfrac{3\pi}{10}$.

3. Radiusi $R = 10\text{ sm}$ bo'lgan aylanada $\alpha = \dfrac{3\pi}{5}\text{ rad}$ bo'lgan yoy uzunligini toping.

4. Aylananing uzunligi $21\text{ sm}$ bo'lgan yoyiga $3\text{ radian}$ li markaziy burchak tiralgan. Aylana radiusini toping.

5. Radiusi $R = 4\text{ sm}$, markaziy burchagi $\alpha = \dfrac{\pi}{4}\text{ rad}$ bo'lgan doiraviy sektor yuzini hisoblang.

6. Sektor yuzi $12\pi\text{ sm}^2$, radiusi esa $6\text{ sm}$. Shu sektorning markaziy burchagini radian va gradusda toping.

7. Soatning minut mili uzunligi $6\text{ sm}$. Minut mili 20 minut davomida burilganda, uning uchi qanday masofani bosib o'tadi?

8. Sektorning perimetri $20\text{ sm}$, yuzi $16\text{ sm}^2$. Sektorning radiusi va markaziy burchagini toping.

Javoblar va yechimlar

1. Yechilishi: a) $20 \cdot \dfrac{\pi}{180} = \dfrac{\pi}{9}\text{ rad}$; b) $150 \cdot \dfrac{\pi}{180} = \dfrac{5\pi}{6}\text{ rad}$; c) $315 \cdot \dfrac{\pi}{180} = \dfrac{7\pi}{4}\text{ rad}$; d) $-210 \cdot \dfrac{\pi}{180} = -\dfrac{7\pi}{6}\text{ rad}$.

2. Yechilishi: a) $\dfrac{5 \cdot 180^\circ}{6} = 150^\circ$; b) $-\dfrac{4 \cdot 180^\circ}{3} = -240^\circ$; c) $\dfrac{7 \cdot 180^\circ}{2} = 630^\circ$; d) $\dfrac{3 \cdot 180^\circ}{10} = 54^\circ$.

3. Yechilishi: $l = \alpha R = \dfrac{3\pi}{5} \cdot 10 = 6\pi\text{ sm} \approx 18{,}84\text{ sm}$.

4. Yechilishi: $l = \alpha R \implies R = \dfrac{l}{\alpha} = \dfrac{21}{3} = 7\text{ sm}$.

5. Yechilishi: $S = \dfrac{1}{2}\alpha R^2 = \dfrac{1}{2} \cdot \dfrac{\pi}{4} \cdot 4^2 = \dfrac{16\pi}{8} = 2\pi\text{ sm}^2$.

6. Yechilishi: $S = \dfrac{1}{2}\alpha R^2 \implies 12\pi = \dfrac{1}{2}\alpha \cdot 36 = 18\alpha \implies \alpha = \dfrac{12\pi}{18} = \dfrac{2\pi}{3}\text{ rad}$. Gradusda: $\dfrac{2 \cdot 180^\circ}{3} = 120^\circ$.

7. Yechilishi: Minut mili 60 minutda to'liq $2\pi$ aylanadi. 20 minutda esa aylananing $\dfrac{20}{60} = \dfrac{1}{3}$ qismini bosib o'tadi, ya'ni $\alpha = \dfrac{2\pi}{3}\text{ rad}$. Bosib o'tilgan yoy uzunligi: $l = \alpha R = \dfrac{2\pi}{3} \cdot 6 = 4\pi\text{ sm} \approx 12{,}56\text{ sm}$.

8. Yechilishi: $P = 2R + l = 20 \implies l = 20 - 2R$. $S = \dfrac{l R}{2} = \dfrac{(20 - 2R)R}{2} = (10 - R)R = 16 \implies R^2 - 10R + 16 = 0$. Ildizlar: $R_1 = 2$, $R_2 = 8$.

   • Agar $R = 2\text{ sm}$ bo'lsa: $l = 16\text{ sm}$, $\alpha = \dfrac{16}{2} = 8\text{ rad}$.
   • Agar $R = 8\text{ sm}$ bo'lsa: $l = 4\text{ sm}$, $\alpha = \dfrac{4}{8} = 0{,}5\text{ rad}$.

Xatolar ro'yxati

O'quvchilar burchakning radian o'lchovida eng ko'p yo'l qo'yadigan xatolar:

1. Yoy uzunligi formulasida burchakni gradusda qo'yish:

   • ✗ Noto'g'ri: $R = 5\text{ sm}$, burchak $60^\circ$ bo'lsa, $l = 60 \cdot 5 = 300\text{ sm}$.
   • ✓ To'g'ri: $l = \alpha R$ formulasida $\alpha$ faqat radianda bo'lishi shart! $60^\circ = \dfrac{\pi}{3}\text{ rad}$, demak $l = \dfrac{\pi}{3} \cdot 5 = \dfrac{5\pi}{3}\text{ sm}$.

2. $\pi$ ni $180^\circ$ ga teng deb tushunib, son o'rniga adashtirish:

   • ✗ Noto'g'ri: $\pi$ soni doim $180^\circ$ ga teng, shuning uchun doira yuzi $S = 180 \cdot R^2$.
   • ✓ To'g'ri: $\pi \approx 3{,}14159...$ — bu son. $\pi\text{ radian}$ burchak esa $180^\circ$ ga teng. Doira yuzi yoki kesma uzunligini hisoblashda $\pi \approx 3{,}14$ deb olinadi.

3. O'tish koeffitsiyentini teskari yozish:

   • ✗ Noto'g'ri: Gradusdan radianga o'tishda $180/\pi$ ga ko'paytirish: $45^\circ \cdot \dfrac{180}{\pi}$.
   • ✓ To'g'ri: Gradus qisqarib ketishi uchun $\pi$ suratda, $180^\circ$ maxrajda bo'lishi kerak: $45^\circ \cdot \dfrac{\pi}{180^\circ} = \dfrac{\pi}{4}$.

4. Sektor perimetrida radiuslarni unutib qoldirish:

   • ✗ Noto'g'ri: Sektor perimetri faqat yoy uzunligiga teng: $P = l$.
   • ✓ To'g'ri: Sektor chegarasi ikkita radius va bitta yoydan iborat: $P = 2R + l$.

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