Zoezi la thermodynamiki lililofanyiwa kazi

Kukokotoa integrali ya umbo tofauti kwenye njia

Zoezi 6 · Somo 4 — Kanuni ya Kwanza ya Thermodynamiki

umbo tofauti
diferenshali kamili
kigezo cha Schwarz
integrali ya mstari

Toleo la kazi — zoezi hili bado liko chini ya ukaguzi.

Swali

Fikiria maumbo mawili tofauti

\omega_1 = y\,\mathrm{d}x + x\,\mathrm{d}y, \qquad \omega_2 = y\,\mathrm{d}x,

na njia tatu zinazounganisha asili $O(0,0)$ na nukta $M(1,1)$ : njia $\gamma_1$ ni kipande cha mlalo $O\to(1,0)$ kikifuatwa na kipande cha wima $(1,0)\to M$ ; njia $\gamma_2$ ni kipande cha wima $O\to(0,1)$ kikifuatwa na kipande cha mlalo $(0,1)\to M$ ; na njia $\gamma_3$ ni kipande cha mstari kinachounganisha $O$ moja kwa moja na $M$ .

Onyesha kuwa $\omega_1$ ni kamili kwa kupata funksioni $f$ ambayo $\omega_1=\mathrm{d}f$ .
Kokotoa $\int_{\gamma_1}\omega_1$ na $\int_{\gamma_2}\omega_1$ kwa kuweka parameta kwa kila kipande. Pata tena jibu bila kukokotoa.
Tumia kigezo cha Schwarz kwa $\omega_2$ . Tunaweza kuhitimisha nini?
Kokotoa $\int_{\gamma_1}\omega_2$ , $\int_{\gamma_2}\omega_2$ na $\int_{\gamma_3}\omega_2$ . Toa maoni.

Kidokezo

Kwenye kipande cha mlalo

\mathrm{d}y=0

, na kwenye kipande cha wima

\mathrm{d}x=0

: kila integrali hupunguzwa kuwa integrali rahisi.

Jibu la kina

Suluhisho

Swali la 1.

f(x,y)=xy

inafaa, kwa sababu

\partial f/\partial x=y

\partial f/\partial y=x

Swali la 2. Kwa mkunjo uliowekewa parameta kwa $t\mapsto\bigl(x(t),y(t)\bigr)$ , tuna

\int_\gamma\omega_1 = \int \left[y(t)x'(t)+x(t)y'(t)\right]\,\mathrm{d}t.

Njia $\gamma_1$ ni muungano wa vipande viwili. Kwenye cha kwanza, $x(t)=t$ na $y(t)=0$ , huku $t\in[0,1]$ ; hivyo $\mathrm{d}x=\mathrm{d}t$ na $\mathrm{d}y=0$ . Kwenye cha pili, $x(t)=1$ na $y(t)=t$ , bado $t\in[0,1]$ ; hivyo $\mathrm{d}x=0$ na $\mathrm{d}y=\mathrm{d}t$ . Kwa hiyo,

\begin{aligned} I_1 = \int_{\gamma_1}\omega_1 &= \int_0^1\left(0\times 1+t\times 0\right)\,\mathrm{d}t + \int_0^1\left(t\times 0+1\times 1\right)\,\mathrm{d}t \\ &= 0+\int_0^1\mathrm{d}t = 1. \end{aligned}

Kwa $\gamma_2$ , kipande cha kwanza kina $x(t)=0$ , $y(t)=t$ , kisha cha pili kina $x(t)=t$ , $y(t)=1$ , huku $t\in[0,1]$ katika hali zote mbili. Hivyo

\begin{aligned} I_2 = \int_{\gamma_2}\omega_1 &= \int_0^1\left(t\times 0+0\times 1\right)\,\mathrm{d}t + \int_0^1\left(1\times 1+t\times 0\right)\,\mathrm{d}t \\ &= 0+\int_0^1\mathrm{d}t = 1. \end{aligned}

Thamani hizi mbili ni sawa. Tungeweza kutabiri bila hesabu: kwa kuwa $\omega_1=\mathrm{d}f$ na $f(x,y)=xy$ , integrali yake hutegemea ncha pekee,

\int_\gamma\omega_1=f(M)-f(O)=f(1,1)-f(0,0)=1.

Swali la 3. Tuandike $\omega_2=A(x,y)\,\mathrm{d}x+B(x,y)\,\mathrm{d}y$ . Hapa $A(x,y)=y$ na $B(x,y)=0$ . Kama umbo lingekuwa kamili, kigezo cha Schwarz kingetoa

\frac{\partial A}{\partial y}=\frac{\partial B}{\partial x}.

Lakini $\partial A/\partial y=1$ ilhali $\partial B/\partial x=0$ . Derivativi mseto ni tofauti: kwa hiyo $\omega_2$ si kamili. Integrali yake inaweza kutegemea njia kati ya $O$ na $M$ , kama hesabu inayofuata inavyoonyesha.

Swali la 4. Kwa kuwa $\omega_2=y\,\mathrm{d}x$ , ni badiliko la $x$ linalofanyika wakati $y$ si sifuri pekee linaloweza kuchangia integrali. Kwa kutumia tena uparametishaji wa awali, kwenye $\gamma_1$ tunapata

\begin{aligned} J_1 =\int_{\gamma_1}\omega_2 &=\int_0^1 0\times 1\,\mathrm{d}t +\int_0^1 t\times 0\,\mathrm{d}t \\ &=0. \end{aligned}

Kwenye $\gamma_2$ , kipande cha kwanza kina $x(t)=0$ , hivyo $\mathrm{d}x=0$ ; kwenye cha pili, $x(t)=t$ , $y(t)=1$ na $\mathrm{d}x=\mathrm{d}t$ . Basi

\begin{aligned} J_2 =\int_{\gamma_2}\omega_2 &=\int_0^1 t\times 0\,\mathrm{d}t +\int_0^1 1\times 1\,\mathrm{d}t \\ &=1. \end{aligned}

Hatimaye, kipande cha ulalo $\gamma_3$ kinawekwa parameta waziwazi kwa

x(t)=t,\qquad y(t)=t,\qquad t\in[0,1],

hivyo $\mathrm{d}x=\mathrm{d}t$ na $\mathrm{d}y=\mathrm{d}t$ . Tunapata

J_3 =\int_{\gamma_3}\omega_2 =\int_0^1 y(t)x'(t)\,\mathrm{d}t =\int_0^1 t\,\mathrm{d}t =\frac12.

Njia zote tatu zina ncha zilezile lakini zinatoa thamani tatu tofauti: $J_1=0$ , $J_2=1$ na $J_3=1/2$ . Huu ndio utegemezi wa njia unaotambulisha umbo tofauti lisilo kamili.