13.6-bo'lim : Zanjir qoidasi
6. Quyidagi ma'lumotlarni hisobga olgan holda, \(\displaystyle \frac{{\partial w}}{{\partial t}}\) va \(\displaystyle \frac{{\partial w}}{{\partial s}}\) ni aniqlash uchun zanjir qoidasidan foydalaning.
\[w = \sqrt {{x^2} + {y^2}} + \frac{{6z}}{y}\,\hspace{0.5in}x = \sin \left( p \right),\,\,\,\,y = p + 3t - 4s,\,\,\,\,z = \frac{{{t^3}}}{{{s^2}}},\,\,\,\,p = 1 - 2t\]Yechim
Xo'sh, bizda buning uchun eslatmalardan formula yo'q, shuning uchun biz birinchi navbatda xulosa qilishimiz kerak. Buning uchun bizga quyidagi daraxt diagrammasi kerak bo'ladi.
Ushbu daraxt diagrammalarining ba'zilari juda chalkash bo'lishi mumkin. Biz har bir lotin uchun amal qilishimiz kerak bo'lgan novdalarni biroz aniqroq qilish uchun bizni qiziqtirgan o'zgaruvchilarni rangladik.
2-qadam
Mana bizdan topishimiz so'ralayotgan ikkita hosila uchun formulalar.
\[\frac{{\partial w}}{{\partial t}} = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{dp}}\frac{{dp}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{\partial y}}{{\partial p}}\frac{{dp}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{\partial y}}{{\partial t}} + \frac{{\partial w}}{{\partial z}}\frac{{\partial z}}{{\partial t}}\hspace{0.5in}\frac{{\partial w}}{{\partial s}} = \frac{{\partial w}}{{\partial y}}\frac{{\partial y}}{{\partial s}} + \frac{{\partial w}}{{\partial z}}\frac{{\partial z}}{{\partial s}}\]3-qadam
Mana bu muammo uchun ish.
\[\begin{align*}\frac{{\partial w}}{{\partial t}} & = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{dp}}\frac{{dp}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{\partial y}}{{\partial p}}\frac{{dp}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{\partial y}}{{\partial t}} + \frac{{\partial w}}{{\partial z}}\frac{{\partial z}}{{\partial t}}\\ & = \left[ {\frac{x}{{\sqrt {{x^2} + {y^2}} }}} \right]\left[ {\cos \left( p \right)} \right]\left[ { - 2} \right] + \left[ {\frac{y}{{\sqrt {{x^2} + {y^2}} }} - \frac{{6z}}{{{y^2}}}} \right]\left[ 1 \right]\left[ { - 2} \right] + \\ & \hspace{0.5in}\hspace{0.5in}\hspace{0.25in}\,\,\,\,\,\,\, + \left[ {\frac{y}{{\sqrt {{x^2} + {y^2}} }} - \frac{{6z}}{{{y^2}}}} \right]\left[ 3 \right] + \left[ {\frac{6}{y}} \right]\left[ {\frac{{3{t^2}}}{{{s^2}}}} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{ - 2x\cos \left( p \right)}}{{\sqrt {{x^2} + {y^2}} }} + \frac{y}{{\sqrt {{x^2} + {y^2}} }} - \frac{{6z}}{{{y^2}}} + \frac{{18{t^2}}}{{y{s^2}}}}}\end{align*}\] \[\begin{align*}\frac{{\partial w}}{{\partial s}} & = \frac{{\partial w}}{{\partial y}}\frac{{\partial y}}{{\partial s}} + \frac{{\partial w}}{{\partial z}}\frac{{\partial z}}{{\partial s}}\\ & = \left[ {\frac{y}{{\sqrt {{x^2} + {y^2}} }} - \frac{{6z}}{{{y^2}}}} \right]\left[ { - 4} \right] + \left[ {\frac{6}{y}} \right]\left[ { - \frac{{2{t^3}}}{{{s^3}}}} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{ - 4y}}{{\sqrt {{x^2} + {y^2}} }} + \frac{{24z}}{{{y^2}}} - \frac{{12{t^3}}}{{y{s^3}}}}}\end{align*}\]Har bir lotinning ikkinchi bosqichida biz “formula”da qaysi atama qaysi hosiladan kelganini va umuman kerak emasligini aniqlash uchun qavs qo‘shdik. Uchinchi bosqichga o'tish uchun kerak bo'lganda biroz soddalashtirdik.
Bundan tashqari, biz barcha orqa almashtirishni tugatganimizdan so'ng, har bir hosiladan kelib chiqadigan tartibsizlik tufayli bu lotinlarda hech qanday "orqa almashtirish" qilmadik.