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line_integrl_sol

# line_integrl_sol - MIT OpenCourseWare http/ocw.mit.edu...

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MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 200 7 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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4. Line Integrals in the Plane 4A. Plane Vector Fields 4A-1 a) All vectors in the field are identical; continuously differentiable everywhere. b) The vector at P has its tail at P and head at the origin; field is cont. diff. everywhere. c) All vectors have unit length and point radially outwards; cont. diff. except at (0,O). d) Vector at P has unit length, and the clockwise direction perpendicular to OP. 4B. Line Integrals in the Plane l 2 a) On Cl: y = 0, dy = 0; therefore (x2 - y) dx + 2% dy = x2 dx = = - 3 -1 3' b) C: use the parametrization x = cos t, y = sin t; then dx = - sin t dt, dy = cost dt c) L C = C l + C 2 + C 3 ; L1o+1 C 1 : x = d x = O ; C2: y = 1 - x ; C 3 : y = d y = 0 1 ydx-xdy = (1-x)~x-x(-dx)+ LO = Jdldx = 1. d) C : x = 2 c o s t , y=sint; dx=-2sintdt L y d x = J d Z T - 2 ~ i ~ ~ t d t = - 2 ~ e) L C : x = t 2 , y = t 3 ; 1 dx=2tdt, dy=3t2dt + 2 + 1 2 2 6y dx x dy = 6t3(2t dt) t2(3t2 dt) = (15t4) dt = 3t5] = 3 . 31. 1 4B-2 a) The field F points radially outward, the unit tangent t to the circle is always perpendicular to the radius; therefore F . t = 0 and Jc F . dr = Jc F . t ds = 0 The field F is always tangent to the circle of radius a, in the clockwise direction, and b) of magnitude a. Therefore F = -at, so that Jc F . dr = Jc F . t ds = - Jc ads = -2aa2.
2 E. 18.02 EXERCISES i + j 4B-3 a) maximum if C is in the direction of the field: C = - JZ i + j b) minimum if C is in the opposite direction to the field: C = -- . . 4 i - j c) zero if C is perpendicular to the field: C = f - JZ d) max = 4, min = -4 by (a) and (b), fir the max or min F and C have respectively the same or opposite constant direction, so Jc F . dr = f IF1 . ICI = f a . 4C. Gradient Fields and Exact Differentials 4C-1 a) F = Vf = 3x2y i + (x3 + 3y2) j b) (i) Using y as parameter, C1 is: x = Y2, y = y; thus dx = 2y dy, and b) (ii) Using y as parameter, C2 is: x = 1, y = y; thus dx = 0, and b) (iii) By the Fundamental Theorem of Calculus for line integrals, Vf . dl- = f (B) - f (A). P 4C-2 a) F = V f = (xyexY + exY) i + (x2exy)j . b) (i) Using x as parameter, lo C is: x = x, y = l/x, so dy = -dx/x2, and so 0 F . dr = (e + e) dx + (x2e) (-dx/x2) = (2ex - ex)] = -e. b) (ii) Using the F.T.C. for line integrals, F . dr = f (1,l) - f (0, co) = 0 - e = -e. Ic 4C-3 a) F = Vf = (cosxcosy)i - (sinxsiny)j. b) Since / F . dr is path-independent, for any C connecting A : (xo, yo) to B : (xi, yi), J c we have by the F.T.C. for line b integrals, F . dr = sin XI cos yl - sin xo cos yo This difference on the right-hand side is maximized if sinxl cos yl is maximized, and sin xo cos yo is minimized. Since I sin x cos yl = I sin X I I cos yl 5 1, the difference on the right

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line_integrl_sol - MIT OpenCourseWare http/ocw.mit.edu...

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