ChangeOfVariables - 5-masala yechimi | MathModel.uz

15.8-bo'lim : O'zgaruvchilarni almashtirish

5. Agar $R$ $\left( {1,0} \right)$, $\left( {4,3} \right)$, $\left( {1,6} \right)$ va $\left( { - 2,3} \right)$ cho'qqilari bilan parallelogramm bo'lsa, mintaqani aniqlaydi, biz $x = \frac{1}{2}\left( {v - u} \right)$, $y = \frac{1}{2}\left( {v + u} \right)$ni $R$ ga o'zgartirgan bo'lardik.

Yechim

Keling, $R$ eskizi bilan boshlaylik.

$A sketch of the region $R$. It is a diamond shaped parallelogram with vertices $\left( {1,0} \right)$, $\left( {4,3} \right)$, $\left( {1,6} \right)$ and $\left( { - 2,3} \right)$ and the inside is shaded green. The four sides are labeled $y = x + 5$ at the upper left, $y = - x + 7$ at the upper right, $y = x - 1$ at the lower right and $y = - x + 1$ at the lower left.$

Hududning har bir chegarasining tenglamalari eskizda berilgan. Biz har bir cho'qqining koordinatalarini bilganimizni hisobga olsak, uning oddiy algebrasi chiziqda ikkita nuqta berilgan chiziq tenglamasini aniqlash uchun tenglamalarni tekshirishni sizga qoldiramiz.

2-qadam

OK, endi biz qilishimiz kerak bo'lgan yagona narsa chegara tenglamalarining har biriga o'zgartirishni qo'llashdir. Buning uchun biz chegara tenglamalarining har biriga konvertatsiyani ulashimiz kifoya.

$y = x + 5$ dan boshlaylik. Ushbu tenglamaga o'zgartirishni qo'llash,

\[\begin{align*}\frac{1}{2}\left( {v + u} \right) & = \frac{1}{2}\left( {v - u} \right) + 5\\ v + u & = v - u + 10\\ 2u &= 10\hspace{0.5in} \Rightarrow \hspace{0.5in}\underline {u = 5} \end{align*}\]

Shunday qilib, bu chegara $u = 5$ tenglamasiga aylanadi.

3-qadam

Keling, konvertatsiyani $y = - x + 7$ ga qo'llaylik.

\[\begin{align*}\frac{1}{2}\left( {v + u} \right) & = - \frac{1}{2}\left( {v - u} \right) + 7\\ v + u & = - v + u + 14\\ 2v & = 14\hspace{0.5in} \Rightarrow \hspace{0.5in}\underline {v = 7} \end{align*}\]

Bu chegara $v = 7$ tenglamasiga aylanadi.

4-qadam

Keyin o'zgartirishni $y = x - 1$ ga qo'llaymiz.

\[\begin{align*}\frac{1}{2}\left( {v + u} \right) & = \frac{1}{2}\left( {v - u} \right) - 1\\ v + u & = v - u - 2\\ 2u & = - 2\hspace{0.5in} \Rightarrow \hspace{0.5in}\underline {u = - 1} \end{align*}\]

Shunday qilib, bu chegara $u = - 1$ tenglamasiga aylanadi.

5-qadam

Nihoyat, konvertatsiyani $y = - x + 1$ ga qo'llaylik.

\[\begin{align*}\frac{1}{2}\left( {v + u} \right) & = - \frac{1}{2}\left( {v - u} \right) + 1\\ v + u & = - v + u + 2\\ 2v & = 2\hspace{0.5in} \Rightarrow \hspace{0.5in}\underline {v = 1} \end{align*}\]

Bu chegara $v = 1$ tenglamasiga aylanadi.

6-qadam

O'zgartirilgan tenglamalarni chizish quyidagi mintaqani beradi.

$A sketch of the transformed region, drawn in the uv-plane. It is a rectangle whose sides are the lines $u = - 1$ on the left, $u = 5$ on the right, $v = 7$ across the top and $v = 1$ across the bottom, and the inside is shaded.$

Shunday qilib, biz olmos shaklidagi hududni transformatsiya ostida to'rtburchakga aylantiramiz.

E'tibor bering, biz $u$ ni gorizontal o'q sifatida tanladik va $v$ ni vertikal o'q sifatida o'zgartirilgan hududda ushbu muammo uchun tanladik. Buni qilish uchun hech qanday sabab yo'q, bundan tashqari, biz har doim qilgan narsamiz. Sizning tanlovingizdan qat'i nazar, bu erda aniq bo'lishi uchun o'qlarni belgilab qo'ying!