Example 42 if dz xdx ydy is dz exact here m z x

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Example 4.2 If dz = xdx + ydy , is dz exact? Here M = ∂z ∂x vextendsingle vextendsingle vextendsingle vextendsingle y = x. (4.9) Thus z = 1 2 x 2 + f ( y ) . (4.10) Thus ∂z ∂y vextendsingle vextendsingle vextendsingle vextendsingle x = df dy = y. (4.11) Thus f ( y ) = 1 2 y 2 + C, (4.12) and z ( x, y ) = 1 2 ( x 2 + y 2 ) + C. (4.13) Examine a difference in z : z 2 z 1 = 1 2 ( x 2 2 x 2 1 + y 2 2 y 2 1 ) . (4.14) Note the constant C drops out and that the difference in z only depends on the end points and not the path between points 1 and 2. Here dz is exact. Note also that ∂M ∂y vextendsingle vextendsingle vextendsingle vextendsingle x = 0 , ∂N ∂x vextendsingle vextendsingle vextendsingle vextendsingle y = 0 , (4.15) thus, our condition for exactness, Eq. (4.8), is satisfied. Example 4.3 If dz = ydx xdy , is dz exact? Here M = ∂z ∂x vextendsingle vextendsingle vextendsingle vextendsingle y = y. (4.16) CC BY-NC-ND. 2011, J. M. Powers.
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80 CHAPTER 4. WORK AND HEAT Thus z = yx + f ( y ) . (4.17) So ∂z ∂y vextendsingle vextendsingle vextendsingle vextendsingle x = x + df dy = x. (4.18) Thus df dy = 2 x. (4.19) But functions of y cannot be functions of x . Therefore, we have generated nonsense. We cannot find z ( x, y )! Thus z is not a state variable, and dz is inexact. We give a new notation for such differentials: δz. If we choose a path, we can still find differences in z . Let us examine the integral of z along two paths, illustrated in Fig. 4.2. x y 1 2 P a th A Path B, C 1 Path B, C 2 (x,y)=(1,1) (x,y)=(0,1) (x,y)=(0,0) Figure 4.2: Sketch of two paths in the x y plane. Path A : Integrate from ( x, y ) = (0 , 0) to ( x, y ) = (1 , 1) along the path x = y and find z 2 z 1 . On path A, x = y and dx = dy . So eliminate y to get δz = xdx xdx = 0 . (4.20) Thus, δz = 0, and z = C after integrations. Thus z 2 = z 1 = C , and z 2 z 1 = 0 . Path B : Integrate from ( x, y ) = (0 , 0) to ( x, y ) = (1 , 1) along the path given first by the y -axis from 0 to 1, then by the line y = 1 from x = 0 to x = 1 and find z 2 z 1 . We have integraldisplay 2 1 δz = integraldisplay C 1 ydx xdy + integraldisplay C 2 ydx xdy (4.21) CC BY-NC-ND. 2011, J. M. Powers.
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4.2. WORK 81 On C 1 , we have x = 0, and dx = 0. On C 2 , we have y = 1 and dy = 0. So we get integraldisplay 2 1 δz = integraldisplay C 1 y (0) (0) dy + integraldisplay C 2 (1) dx x (0) = integraldisplay 1 0 dx. (4.22) Thus z 2 z 1 = 1 . (4.23) We chose a different path, and found a different difference in z . Note also that ∂M ∂y vextendsingle vextendsingle vextendsingle vextendsingle x = 1 , ∂N ∂x vextendsingle vextendsingle vextendsingle vextendsingle y = 1; (4.24) thus, our condition for exactness, Eq. (4.8), is not satisfied. 4.2 Work 4.2.1 Definitions From Newtonian mechanics, we know going from state 1 to state 2, that the work 1 W 2 is done by a force moving through a distance. It is defined as 1 W 2 = integraldisplay 2 1 F · d x . (4.25) In differential form, we have δW = F · d x . (4.26) In one-dimensional systems, we have 1 W 2 = integraldisplay 2 1 Fdx, (4.27) δW = Fdx. (4.28) Note that we have anticipated that the work differential is inexact. This is an important point, as work integrals will be path-dependent, and work will not be a state variable for a system. Here F is a three-dimensional force vector, x is a three-dimensional distance vector, and · is the dot product operator. Recall that the dot product of two vectors yields a scalar.
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