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12.7-bo'lim: Vektor funktsiyalari bilan hisoblash

Ushbu bo'limda vektor funktsiyalarning limitlari, hosilalari va integrallari haqida qisqacha gapirishimiz kerak. Ko'rib turganingizdek, ular oldindan aytib bo'ladigan tarzda harakat qilishadi. Biz barcha ishlarni \({\mathbb{R}^3}\) da bajaramiz, lekin tabiiy ravishda ushbu bo'limdagi formulalar/ishni \({\mathbb{R}^n}\) ga kengaytira olamiz (i.e. \(n\) o'lchovli fazo).

Keling, chegaralardan boshlaylik. Bu yerda vektor funksiyaning chegarasi.

Vektor funksiyasining chegarasi

\[\begin{align*}\mathop {\lim }\limits_{t \to a} \vec r\left( t \right) & = \mathop {\lim }\limits_{t \to a} \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle \\ & = \left\langle {\mathop {\lim }\limits_{t \to a} f\left( t \right),\mathop {\lim }\limits_{t \to a} g\left( t \right),\mathop {\lim }\limits_{t \to a} h\left( t \right)} \right\rangle \\ & = \mathop {\lim }\limits_{t \to a} f\left( t \right)\vec i + \mathop {\lim }\limits_{t \to a} g\left( t \right)\vec j + \mathop {\lim }\limits_{t \to a} h\left( t \right)\vec k\end{align*}\]

Shunday qilib, biz qiladigan narsa - komponentning har bir funktsiyasining chegarasini olish va uni vektor sifatida qoldirish.

1-misol \(\vec r\left( t \right) = \left\langle {{t^3},\displaystyle \frac{{\sin \left( {3t - 3} \right)}}{{t - 1}},{{\bf{e}}^{2t}}} \right\rangle \) bu yerda \(\mathop {\lim }\limits_{t \to 1} \vec r\left( t \right)\) ni hisoblang.
Yechim

Bu erda, albatta, unchalik ko'p narsa yo'q.

\[\begin{align*}\mathop {\lim }\limits_{t \to 1} \vec r\left( t \right) & = \left\langle {\mathop {\lim }\limits_{t \to 1} {t^3},\mathop {\lim }\limits_{t \to 1} \frac{{\sin \left( {3t - 3} \right)}}{{t - 1}},\mathop {\lim }\limits_{t \to 1} {{\bf{e}}^{2t}}} \right\rangle \\ & = \left\langle {\mathop {\lim }\limits_{t \to 1} {t^3},\mathop {\lim }\limits_{t \to 1} \frac{{3\cos \left( {3t - 3} \right)}}{1},\mathop {\lim }\limits_{t \to 1} {{\bf{e}}^{2t}}} \right\rangle \\ & = \left\langle {1,3,{{\bf{e}}^2}} \right\rangle \end{align*}\]

E'tibor bering, biz foydalanishimiz kerak edi L'Hospital qoidasi \(y\) komponentida.

Endi hosilalar haqida g'amxo'rlik qilaylik va cheklovlar qanday ishlashini ko'rganimizdan so'ng, hosilalar uchun bizda quyidagilar borligi ajablanarli emas.

Vektor funksiyaning hosilasi

\[\vec r'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle = f'\left( t \right)\vec i + g'\left( t \right)\vec j + h'\left( t \right)\vec k\]
2-misol \(\vec r\left( t \right) = {t^6}\,\vec i + \sin \left( {2t} \right)\vec j - \ln \left( {t + 1} \right)\vec k\) uchun \(\vec r'\left( t \right)\) ni hisoblang.
Yechim

Darhaqiqat, lotinlarni olishdan tashqari, bu muammo uchun juda ko'p narsa yo'q.

\[\vec r'\left( t \right) = 6{t^5}\,\vec i + 2\cos \left( {2t} \right)\vec j - \frac{1}{{t + 1}}\vec k\]

Holbuki, biz bilgan asosiy faktlarning aksariyati hanuzgacha saqlanib qolgan, ammo bu erda vektor funktsiyalarining hosilalari haqidagi ba'zi faktlar aniq bo'lishi mumkin.

Vektor funktsiyalari uchun differensiallash formulalari

\[\begin{align*}& \frac{d}{{dt}}\left( {\vec u + \vec v} \right) = \vec u' + \vec v'\\ & {\left( {c\vec u} \right)^\prime } = c\,\vec u'\\ & \frac{d}{{dt}}\left( {f\left( t \right)\vec u\left( t \right)} \right) = f'\left( t \right)\vec u\left( t \right) + f\left( t \right)\vec u'\left( t \right)\\ & \frac{d}{{dt}}\left( {\vec u\centerdot \vec v} \right) = \vec u'\centerdot \vec v + \vec u\centerdot \vec v'\\ & \frac{d}{{dt}}\left( {\vec u \times \vec v} \right) = \vec u' \times \vec v + \vec u \times \vec v'\\ & \frac{d}{{dt}}\left( {\vec u\left( {f\left( t \right)} \right)} \right) = f'\left( t \right)\vec u'\left( {f\left( t \right)} \right)\end{align*}\]

Bundan tashqari, biz kerak bo'lganda foydalanishimiz uchun yo'ldan chiqib ketishimiz kerak bo'lgan bitta tezkor ta'rif mavjud.

A silliq egri - bu \(\vec r'\left( t \right)\) uzluksiz va \(\vec r'\left( t \right) \ne 0\) har qanday \(t\) uchun har qanday egri chiziq, ehtimol oxirgi nuqtalardan tashqari. Masalan, spiral silliq egri chiziqdir.

Nihoyat, vektor funktsiyalarining integrallarini muhokama qilishimiz kerak. Ham chegaralarni, ham hosilalarni qo'llanma sifatida ishlatganda, bizda noaniq integrallar uchun integratsiya uchun quyidagilar ham borligi ajablanarli emas.

Vektor funksiyaning noaniq integrali

\[\begin{align*}\int{{\vec r\left( t \right)\,dt}} & = \left\langle {\int{{f\left( t \right)\,dt}},\int{{g\left( t \right)\,dt}},\int{{h\left( t \right)\,dt}}} \right\rangle + \vec c\\ \int{{\vec r\left( t \right)\,dt}} & = \int{{f\left( t \right)\,dt}}\,\,\vec i + \int{{g\left( t \right)\,dt}}\,\,\vec j + \int{{h\left( t \right)\,dt}}\,\,\vec k + \vec c\end{align*}\]

va aniq integrallar uchun quyidagi.

Vektor funksiyaning aniq integrali

\[\begin{align*}\int_{{\,a}}^{{\,b}}{{\vec r\left( t \right)\,dt}} & = \left\langle {\int_{{\,a}}^{{\,b}}{{f\left( t \right)\,dt}},\int_{{\,a}}^{{\,b}}{{g\left( t \right)\,dt}},\int_{{\,a}}^{{\,b}}{{h\left( t \right)\,dt}}} \right\rangle \\ \int_{{\,a}}^{{\,b}}{{\vec r\left( t \right)\,dt}} & = \int_{{\,a}}^{{\,b}}{{f\left( t \right)\,dt}}\,\,\vec i + \int_{{\,a}}^{{\,b}}{{g\left( t \right)\,dt}}\,\,\vec j + \int_{{\,a}}^{{\,b}}{{h\left( t \right)\,dt}}\,\,\vec k\end{align*}\]

Noaniq integrallar bilan biz integratsiya doimiysini qo'ydik, bu holda doimiy doimiy doimiy emas, balki vektor bo'lishi kerakligi aniq edi.

Bundan tashqari, aniq integrallar uchun biz uni ba'zan quyidagicha yozamiz:

\[\begin{align*}\int_{{\,a}}^{{\,b}}{{\vec r\left( t \right)\,dt}} & = \left. {\left( {\left\langle {\int{{f\left( t \right)\,dt}},\int{{g\left( t \right)\,dt}},\int{{h\left( t \right)\,dt}}} \right\rangle } \right)} \right|_a^b\\ \int_{{\,a}}^{{\,b}}{{\vec r\left( t \right)\,dt}} & = \left. {\left( {\int{{f\left( t \right)\,dt}}\,\,\vec i + \int{{g\left( t \right)\,dt}}\,\,\vec j + \int{{h\left( t \right)\,dt}}\,\,\vec k} \right)} \right|_a^b\end{align*}\]

Boshqacha qilib aytganda, biz noaniq integralni bajaramiz, so'ngra vektorni komponentlar bo'yicha emas, balki butun sifatida baholaymiz.

3-misol \(\vec r\left( t \right) = \left\langle {\sin \left( t \right),6,4t} \right\rangle \) uchun \(\int{{\vec r\left( t \right)\,dt}}\) ni hisoblang.
Yechim

Biz qilishimiz kerak bo'lgan narsa - har bir komponentni birlashtirish va u bilan ishlash.

\[\int{{\vec r\left( t \right)\,dt}} = \left\langle { - \cos \left( t \right),6t,2{t^2}} \right\rangle + \vec c\]
4-misol \(\vec r\left( t \right) = \left\langle {\sin \left( t \right),6,4t} \right\rangle \) uchun \(\int_{{\,0}}^{{\,1}}{{\vec r\left( t \right)\,dt}}\) ni hisoblang.
Yechim

Bunday holda, biz qilishimiz kerak bo'lgan narsa avvalgi misoldagi natijani qayta ishlatish va keyin baholashdir.

\[\begin{align*}\int_{{\,0}}^{{\,1}}{{\vec r\left( t \right)\,dt}} & = \left( {\left\langle { - \cos \left( t \right),6t,2{t^2}} \right\rangle } \right)_0^1\\ & = \left\langle { - \cos \left( 1 \right),6,2} \right\rangle - \left\langle { - 1,0,0} \right\rangle \\ & = \left\langle {1 - \cos \left( 1 \right),6,2} \right\rangle \end{align*}\]