lec_week13

lec_week13 - MIT OpenCourseWare http://ocw.mit.edu 18.02...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 18.02 Lecture 30. – Tue, Nov 27, 2007 Handouts: practice exams 4A and 4B. Clarification from end of last lecture: we derived the diffusion equation from 2 inputs. u = concentration, F = ﬂow, satisfy: 1) from physics: F = − k u , 2) from divergence theorem: ∂u/∂t = − div F . Combining, we get the diffusion equation: ∂u/∂t = − div F = + k div ( u ) = k 2 u . Line integrals in space. Force field F = P, Q, R , curve C in space, dr = dx, dy, dz Work = C F dr = C P dx + Q dy + R dz . ⇒ · Example: F = yz, xz, xy . C : x = t 3 , y = t 2 , z = t . 0 ≤ t ≤ 1. Then dx = 3 t 2 dt , dy = 2 tdt , dz = dt and substitute: 1 F dr = yzdx + xzdy + xydz = 6 t 5 dt = 1 · C C (In general, express ( x, y, z ) in terms of a single parameter: 1 degree of freedom) Same F , curve C = segments from (0 , , 0) to (1 , , 0) to (1 , 1 , 0) to (1 , 1 , 1). In the xy-plane, z = 0 = F = xy k ˆ , so F dr = 0, no work. For the last segment, x = y = 1, dx = dy = 0, so ⇒ · 1 F = z, z, 1 and dr = , , dz . We get 1 dz = 1 . Both give the same answer because F is conservative, in fact F = ( xyz ). Recall the fundamental theorem of calculus for line integrals: P 1 f dr = f ( P 1 ) − f ( P ) . · P Gradient fields. F = P, Q, R = f x , f y , f z ? Then f xy = f yx , f xz = f zx , f yz = f zy , so P y = Q x , P z = R x , Q z = R y . Theorem: F is a gradient field if and only if these equalities hold (assuming defined in whole space or simply connected region) Example: for which a, b is axy ˆ ı + ( x 2 + z 3 )ˆ j + ( byz 2 − 4 z 3 ) k ˆ a gradient field? P y = ax = 2 x = Q x so a = 2; P z = 0 = 0 = R x ; Q z = 3 z 2 = bz 2 = R y so b = 3. Systematic method to find a potential: (carried out on above example) f x = 2 xy , f y = x 2 + z 3 , f z = 3 yz 2 − 4 z 3 : f x = 2 xy f ( x, y, z ) = x 2 y + g ( y, z ). ⇒ f y = x 2 + g y = x 2 + z 3 g y = z 3 g ( y, z ) = yz 3 + h ( z ), and f = x 2 y + yz 3 + h ( z ). ⇒ ⇒ f z = 3 yz 2 + h ( z ) = 3 yz 2 − 4 z 3 ⇒ h ( z ) = − P 1 4 z 3 ⇒ h ( z ) = − z...
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This note was uploaded on 05/04/2011 for the course MATH 18.02 taught by Professor Auroux during the Spring '08 term at MIT.

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lec_week13 - MIT OpenCourseWare http://ocw.mit.edu 18.02...

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