ChainRule - 30-masala yechimi | MathModel.uz

3.9-bo'lim : Zanjir qoidasi

30. Ob'ektning o'rni $s\left( t \right) = \sin \left( {3t} \right) - 2t + 4$ tomonidan berilgan. $\left[ {0,3} \right]$ oralig'ida ob'ekt qayerga o'ngga va chapga harakatlanayotganini aniqlang.

Yechim

Bizga birinchi navbatda hosila kerak bo'ladi, chunki hosila bizga ob'ekt tezligini berishini bilamiz. Mana lotin.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{s'\left( t \right) = 3\cos \left( {3t} \right) - 2}}\]

2-qadam

Keyinchalik, ob'ekt qayerda harakatlanishni to'xtatayotganini bilishimiz kerak va shuning uchun biz qilishimiz kerak bo'lgan narsa lotinni nolga tenglashtirish va hal qilishdir.

\[3\cos \left( {3t} \right) - 2 = 0\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\cos \left( {3t} \right) = \frac{2}{3}\]

Tez kalkulyator hisobi bizga shuni aytadi:

\[3t = {\cos ^{ - 1}}\left( {\frac{2}{3}} \right) = 0.8411\]

"Ko'rib chiqish" bo'limidagi ishimizni eslab, birlik aylanasini tez tekshirish orqali ikkinchi burchak uchun $3t = 0.8411$ yoki $3t = 2\pi - 0.8411 = 5.4421$ dan foydalanish mumkinligini ko'rishimiz mumkin. E'tibor bering, ikkalasi ham ishlaydi, lekin biz ikkinchisidan foydalanamiz, chunki bu ijobiy burchak.

Bularning barchasini birlashtirib, 3 ga bo'linib, hosila nolga teng bo'lishini ko'ramiz,

\[\begin{align*}3t & = 0.8411 + 2\pi n\hspace{0.25in}{\mbox{and}}\hspace{0.25in}3t = 5.4421 + 2\pi n\hspace{0.25in}n = 0, \pm 1, \pm 2, \pm 3, \ldots \\ t & = 0.2804 + \frac{{2\pi n}}{3}\hspace{0.25in}{\mbox{and}}\hspace{0.25in}t = 1.8140 + \frac{{2\pi n}}{3}\hspace{0.25in}n = 0, \pm 1, \pm 2, \pm 3, \ldots \end{align*}\]

Va nihoyat, biz ishlayotgan $\left[ {0,3} \right]$ oralig'ida qaysi yechimlar to'g'ri kelishini aniqlash uchun ba'zi $n$-larni ulashimiz kerak.

Shunday qilib, $\left[ {0,3} \right]$ oralig'ida ob'ekt quyidagi uchta nuqtada harakatini to'xtatadi.

\[t = 0.2804,\,\,\,\,\,1.8140,\,\,\,\,2.3748\]

3-qadam

Bu muammoga javob olish uchun biz bilishimiz kerak bo'lgan narsa - hosila ijobiy (va shuning uchun ob'ekt o'ngga siljiydi) yoki salbiy (shuning uchun ob'ekt chapga siljiydi). Hosil uzluksiz bo'lgani uchun biz bilamizki, uning belgisini o'zgartirishi mumkin bo'lgan yagona joy hosila nolga teng bo'lgan joydir. Shunday qilib, biz ushbu bo'limda qilganimizdek, tezkor raqamlar qatori bizga turli intervallar uchun hosilaning belgisini beradi.

Mana bu muammo uchun raqam chizig'i.

$A number line for the derivative of $s\left( t \right) = \sin \left( {3t} \right) - 2t + 4$ with dashed vertical lines at $t = 0.2804$, $t = 1.8140$ and $t = 2.3748$ dividing the interval into four regions. In the first region $s'\left( 0 \right) = 1$ so $s'\left( t \right) > 0$, in the second $s'\left( 1 \right) = - 4.97$ so $s'\left( t \right) < 0$, in the third, whose computation is boxed and joined to its region by an arrow, $s'\left( 2 \right) = 0.88$ so $s'\left( t \right) > 0$, and in the fourth $s'\left( 3 \right) = - 4.73$ so $s'\left( t \right) < 0$.$

Bundan biz quyidagi harakatlanuvchi o'ngga/chapga harakatlanuvchi ma'lumotlarni olamiz.

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{array}{ll}{\mbox{Moving Right :}} & \,\,0 \le t < 0.2804,\,\,\,\,1.8140 < t < 2.3748\\ {\mbox{Moving Left :}} & \,\,0.2804 < t < 1.8140,\,\,\,\,2.3748 < t \le 3\end{array}}\]

E'tibor bering, biz faqat $\left[ {0,3} \right]$ oralig'ida nima sodir bo'layotganini ko'rib chiqdik, chunki biz ushbu intervaldan tashqarida ob'ektning o'ngga/chapga harakatlanishi haqida hech narsa deya olmaymiz.