<p> Summary 5 Independence of Path – Exact Differentials.</p><p>Independence of path: conservative fields and exact differentials.  Simple closed curve C: x=x(t), y=y(t), a< t <b. A curve represented by x=x(t), y=y(t), a≤≤ t b, is said to be a simple closed curve if (x(a), y(a)) =(x(b), y(b)) but they are not equal for any other values of t. C is simple closed if the points on the curve are the same points only at t=a and t=b, that is, (x(a), y(a))=(x(b), y(b)). It is closed because the end points coincide; it is simple because it only intersects at the end points.  Positive direction. A simple closed curve is traversed in the positive sense if the inside of C is to the left as the curve is traversed in the direction of increasing t. Pdx  Qdy The symbol  denotes a line integral along or around C in the positive C direction.  Smooth on a set S. A function f is smooth on a set S if its first partial derivatives are continuous in S  Open set. A set S is said to be open if for each point (x, y) in S there is a neighborhood of (x, y) contained in S.  Connected set. A set S is said to be connected if each two points in S can be connected by a broken-line path lying entirely in S.  Region: An open connected set is called a region  Simply connected: A set is simply conned if the inside of each simple closed curve in S also lies in S; that is, S has no holes in it. The interior of a simple closed curve is itself simply connected.</p><p>F Tds Definition . The line integral C is said to be independent of the path in the region D provided that, given any two points A and B of D, the integral has the same value along every piecewise smooth curve, or path, in D from A to B. In this case we may write B F  T d s  F  T d s C  A In this case, the value of the integral depends only on the end points and not on the path between B the end points. Therefore, Pdx  Qdy  Rdz  Pdx  Qdy  Rdz or C  A B Pdx  Qdy  Pdx  Qdy C  A Note: The fundamental theorem of calculus tells us that if g(x) is continuous on an interval b [a,b], then g ( x ) d x  g ( b )  g ( x ) . Thus, the value of the integral depends only on the a values of the anti-derivative g(x) at the ends points of [a,b].</p><p>Can we find something similar to an anti-derivative?</p><p>Definition: suppose there is a scalar function f(x,y) such that f x  P a n d f y  Q . T h e n f  P i  Q j  F ( x , y ) . In this case the vector field F is said to be conservative. The function f is called a potential for the vector field F. Similarly, if there is a scalar function f(x, y ,z)) such that f x  P, f y  Q and f z  R. Then f  Pi  Qj  Rk  F(x, y, z). In this case the vector field F is said to be conservative. The function f is called a potential for the vector field F.</p><p>Exact Differential: the differential expression Pdx+Qdy is called an exact differential (or simply exact) if there is a function f(x, y) such that df=Pdx + Qdy.</p><p>Example: The vector field F(x, y)  2xy  3y2 i  (x2  6xy) j is conservative because it has a potential f (x, y)  x2 y  3xy2 . Also, the differential df  2xy  3y2 dx  (x2  6xy) dy is called an exact differential. Example: The vector field F(x, y, z)  yzi  (xz) j  (xy  2z)k is conservative because it has a potential f (x, y, z)  xyz  z 2 . Also, the differential df  yz dx  xz dy  (xy  2z)k is called an exact differential.</p><p>Condition for exactness Theorem: suppose that the functions P, and Q are continuous and have continuous first-order derivatives in a simply connected region D, then the vector field F=Pi+Qj is conservative in R, P Q and hence has a potential function f(x,y), if and only if  at each point in D. y x Similarly, a differential Pdx + Qdy + Rdz is exact if the curlF=0 where F(x, y, z)=Pi + Qj + Rk Note: Let F(x, y, z) =Pi + Qj +Rk be vector field. i j k      R Q   R P   Q P   CurlF  0   0  i    j    k    0 x y z  y z   x z   x y  P Q R R Q R P Q P    0 and   0 and   0 y z x z x y R Q R P Q P   and  and  y z x z x y Note: the following theorem is the analog of the fundamental theorem of calculus for line integrals. F  T d s  P d x  Q d y  R d z Theorem: The line integral C ç is independent of path in the plane region D if and only if there exist a scalar function f (a potential) in D such that F  dr  F Tds  Pdx  Qdy  Rdz  C ç C</p><p>Fundamental Theorem of line integrals. Suppose Pdx + Qdy +Rdz is exact, that is, the function F(x, y, z) =Pi +Qj +Rz has a potential f(x, y, z) in a region D such that f  F . Let C: x=x(t), y=y(t), z=z(t), a < t < b be a curve in D from the point A=(x(a), y(a), z(a)) to B=(x(b), y(b), z(b)). Then  Pdx  Qdy  Rdz  F  dr  f (B)  f (A) C C Proof: b  Pdx  Qdy  Rdz P(x(t), y(t), z(t))x(t)  Q(x(t), y(t), z(t))y(t)  R(x(t), y(t), z(t))z(t)dt C a b b f dx f dy f dz    dt  f x(t), y(t), z(t)dt  x dt y dt z dt   a   a  f x(b), y(b), z(b) f x(a), y(a), z(a)  f x(t), y(t), z(t) b  f (b)  f (a)   a</p><p>Corollary 1. If Pdx + Qdy + Rdz is exact, then the line integral is independent of the path in R and depends only on the values of a potential f at the end points.</p><p>Corollary 2. If Pdx + Qdy + Rdz is exact and C is a simple closed curve, then,  Pdx  Qdy  Rdz  0. C Proof: f(b)=f(a) because the curve is simple closed. Pdx  Qdy  Rdz  f (b)  f (a)  0 Therefore,  C Note: The following statements are equivalent:  F(x, y) =Pi+Qj is conservative  Pdx +Qdy is exact  F has a potential f. P Q   y x P d x  Q d y  The line integral C is independent of path P d x  Q d y  If C is a closed region, then C = 0 Problem 1. Use the curl to find out if the vector field F(x, y)  y2  3x2 i  2xyj is conservative. If it is conservative, find a potential Solution: i j k i j k       F(x, y)  y2  3x2 i  2xyj  CurlF   x y z x y z P Q R y2  3x2 2xy 0   i0  0)  j(0  0 k(2y  2y)  0, it is conservative</p><p>f P  y2  3x2   y2  3x2  f (x, y)  xy2  x3  c(y) x f Q   2xy  c(y)  2xy  c(y)  2xy  c(y)  0  c(y) is a cons tan t, let cy  0 y Therefore f (x, y)  xy2  x3 is a potential</p><p>Problem 2. 2 2 Use the result of problem 1 to evaluate the line integral  y  3x dx  2xydy where C is the C x2 y 2 arc on the ellipse   1 from A=(2, 0) to ((0, -1) 4 1 Solution: F(x, y)  y2  3x2 i  2xyj is conservative implies that the differential df  y2  3x2 dx  2xydy is exact and the field is conservative. Therefore the value of the integral is independent of the path, it only depends on A and B Therefore, B (0,1) y 2  3x2 dx  2xydy  y 2  3x2 dx  2xydy  f (B)  f (A)  xy 2  x3  0  8  8   (2,0) C A Problem 3. If a vector field F(x, y, z) = PI +Qj +Rk is conservative, then we say that the differential expression Pdx + Qdy + Rdz is exact, that is, there is a function f, such that  1   1 2 2  df=Pdx + Qdy +Rdz. Test the differential   2xydx    x  3y dy for exactness and  x   y  find the potential f if it exists. Solution:</p><p> i j k i j k       F(x, y)  y 2  3x2 i  2xyj  CurlF   x y z x y z P Q R  1   1 2 2    2xy   x  3y  0  x   y    i0  0)  j(0  0 k(2x  2x)  0, it is conservative  1  f  1  P    2xy     2xy  f (x, y)  ln x  x2 y  c(y)  x  x  x  f 1 1 Q   x2  c(y)  x2  c(y)   x2  3y 2  c(y)   3y 2 y y y  c(y)  ln y  y3  c (let c  0) Therefore f (x, y)  ln x  x2 y  ln y  y3  ln(xy)  x2 y  y3 is a potential</p><p>(1,1) y 2 y Problem 4: Evaluate  2xe dx  x e dy (0,0) Test for exactness: i j k    F(x, y, z)  2xe yi  x2e y j  0k  CurlF  x y z 2xe y x2e y o   i0  0 j0  0 2xe y  2xe y  0  exact, there is a potential f f P   2xe y  f (x, y)  x2e y  C(y)   x2e y  C(y)  Q  x2e y  x y C(y)  0  C(y) is a cons tant. Let C(y)  0 Therefore, f (x, y)  x2e y is a potential Check : f  2xe yi  x2e y j Therefore, (1,1) (1,1) 1 2xe ydx  x2e ydy  x2e y  e1   (0,0) e (0,0) Problem 5. Show that 2xy  z 2 dx  x2  6yz2 dy  (2xz  6y2 z)dz is exact and find a potential. i j k    F(x, y, z)  2xy  z 2 i  x2  6yz 2 j  2xz  6y2 zk  CurlF  x y z P Q R</p><p> i j k       i12yz 12yz)  j(2z  2z k(2x  2x)  0 x y z 2xy  z 2  x2  6yz 2  2xz  6y2 z Therefore the differential is exact.</p><p>f P  2xy  z 2   2xy  z 2  f (x, y, z)  x2 y  xz 2  c(y, z) x f Q   x2  c(y, z)  x2  c(y, z)  x2  6yz 2  c(y, z)  6yz 2  c(y, z)  3y2 z 2  c(z) y  f (x, y, z)  x2 y  xz 2  3y2 z 2  c(z) f R   2xz  6y2 z  c(z)  2xz  6y2 z  c(z)  2xz  6y2 z  c(z)  0  c(z) is a cons tan t z Let c(z)  0,therefore f (x, y, z)  x2 y  xz 2  3y2 z 2 is a potential</p><p>Problem 6. (1,2) Evaluate  ( y 2  2xy)dx  (x2  2xy)dy (0, 0) i j k i j k       CurlF   x y z x y z P Q R y2  2xy x2  2xy 0   i0  0)  j(0  0  k(2x  2y  2y  2x)  0 Therefore the differential is exact.</p><p>Find a potential f f P   y2  2xy  f (x, y)  xy2  x2 y  C(y) x f  Q   2xy  x2  C(y)  2xy  x2  C(y)  x2  2xy  C(y)  0  C(y) is cons tan t y Let C(y)  0. Therefore, f (x, y)  xy2  x2 y is a potential (1,2) (1,2) Evaluate (y 2  2xy)dx  (x2  2xy)dy  xy2  x2 y  4  2  0  6  (0,0) (0, 0)</p><p>Problem 7. ( , ) Evaluate  (sin y  y cos x)dx  (sin x  x cos y)dy ( / 2, / 2) i j k i j k       CurlF   x y z x y z P Q R sin y  y cos x sin x  xcos y 0   i0  0)  j(0  0 k(cos x  cos y  cos y  cos x)  0 Therefore the differential is exact. f P   sin y  y cos x  f (x, y)  xsin y  y sin x  c(y) x f Q   xcos y  sin x  c(y)  sin x  x cos y  c(y)  0  c(y)  c y Therefore, f (x, y)  xsin y  y sin x is a potentia ( , )         (sin y  y cos x)dx  (sin x  x cos y)dy   sin   sin  sin   sin     2 2 2 2 ( / 2, / 2)    </p><p>Problem 8. (2,3,1 2 Evaluate  (2x  2zdx  2yzdy  (2x  y )dz (1,1,1) i j k    CurlF  x y z P Q R</p><p> i j k      i2y  2y)  j(2  2  k(2y  2y)  0 x y z 2x  2z 2yz 2x  y2  Therefore the differential is exact. f P  2x  2z   2x  2z  f (x, y, z)  x2  2xz  c(y, z) x f Q   0  0  c(y, z)  c(y, z)  2yz  c(y, z)  y2 z  c(z) y  f (x, y, z)  x2  2xz  y 2 z  c(z) f R   2x  y2  c(z)  2x  y2  c(z)  2x  y2  c(z)  0  c(z) is a cons tan t z Let c(z)  0,therefore f (x, y, z)  x2  2xz  y2 z is a potential</p><p>(2,3,1) (2,3,1) (2x  2zdx  2yzdy  (2x  y 2 )dz  x2  2xz  y 2   9  4  13  (1,1,1) (1,1,1)</p>