16.6-bo'lim : Konservativ vektor maydonlari
9. \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) ni baholang, bu erda \(\vec F\left( {x,y} \right) = \left( {2y{{\bf{e}}^{x\,y}} + 2x{{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}}} \right)\,\vec i + \left( {2x{{\bf{e}}^{x\,y}} - 2y{{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}}} \right)\vec j\) va \(C\) quyida ko'rsatilgan egri chiziqdir.
Yechim
Ushbu muammoni hal qilishning ikki yo'li mavjud: qiyin va oson. Qiyin yo'l - egri chiziqli chiziqli integralni va berilgan vektor maydonini ko'rish va to'g'ridan-to'g'ri chiziqli integralni hisoblashni boshlash (ehtimol, bu holatda juda yoqimsiz). Eng oson yo'li - vektor maydonining konservativ yoki yo'qligini tekshirish va u potentsial funktsiyani topish va oldingi bo'limda ko'rgan chiziqli integrallar uchun asosiy teoremadan foydalanishdir.
Shunday qilib, keling, oson yo'ldan boramiz va vektor maydoni konservativ yoki yo'qligini tekshiramiz.
\[\begin{align*}P & = 2y{{\bf{e}}^{x\,y}} + 2x{{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}}&\hspace{0.25in}{P_y} & = 2{{\bf{e}}^{x\,y}} + 2xy{{\bf{e}}^{x\,y}} - 4xy{{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}}\\ Q & = 2x{{\bf{e}}^{x\,y}} - 2y{{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}} & \hspace{0.25in}{Q_x} & = 2{{\bf{e}}^{x\,y}} + 2xy{{\bf{e}}^{x\,y}} - 4xy{{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}}\end{align*}\]Shunday qilib, biz \({P_y} = {Q_x}\) va shuning uchun vektor maydoni konservativ ekanligini ko'rishimiz mumkin.
2-qadam
Endi biz faqat vektor maydoni uchun potentsial funktsiyani topishimiz kerak. Bu safar biz bu tafsilotlarni biroz tezroq va oldingi muammolarga qaraganda kamroq tushuntirish bilan ko'rib chiqamiz.
Birinchidan, \(Q\) ni \(y\) bilan birlashtiramiz.
\[\begin{align*}f\left( {x,y} \right) & = \int{{Q\,dy}}\\ & = \int{{2x{{\bf{e}}^{x\,y}} - 2y{{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}}\,dy}}\\ & = 2{{\bf{e}}^{x\,y}} + {{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}} + g\left( x \right)\end{align*}\]Endi \(x\) ga nisbatan farqlang va \(P\) ga tenglashtiring.
\[{f_x} = 2y{{\bf{e}}^{x\,y}} + 2x{{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}} + g'\left( x \right) = 2y{{\bf{e}}^{x\,y}} + 2x{{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}} = P\hspace{0.25in} \Rightarrow \,\,\,\,\,\,\,\,g'\left( x \right) = 0\]\(g\left( x \right)\) ni hal qilish \(g\left( x \right) = c\) ni beradi va shuning uchun bu vektor maydoni uchun potentsial funktsiya:
\[f\left( {x,y} \right) = 2{{\bf{e}}^{x\,y}} + {{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}} + c\]3-qadam
Endi biz potentsial funktsiyaga egamiz, biz oddiygina chiziqli integrallar uchun asosiy teoremadan foydalanishimiz mumkin:
\[ \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}} = f\left( {{\mbox{end point}}} \right) - f\left( {{\mbox{start point}}} \right)\]Masala bayonidagi grafikdan biz \(C\) ning boshlang’ich nuqtasi \(\left( {0,1} \right)\) va \(C\) ning tugash nuqtasi \(\left( {5,0} \right)\) ekanligini ko’rishimiz mumkin.
U holda integral,
\[\int\limits_{C}{{\vec F\centerdot d\vec r}} = f\left( {5,0} \right) - f\left( {0,1} \right) = \left( {2 + {{\bf{e}}^{25}} + c} \right) - \left( {2 + {{\bf{e}}^{ - 1}} + c} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{{{\bf{e}}^{25}} - {{\bf{e}}^{ - 1}}}}\]