Final Exam Solution

Then f t f f f x t y t z t f c t x y z then b c f

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Unformatted text preview: R RR R These show that c(t) is a flow line for F and p(t) is a flow line for G. (d) (2 points) Explain how it is that only one of these vector fields is irrotational but both have flow lines that circle around the z-axis. Solution: A vector field being “irrotational” is a local property. That means G has no vortices. But that does not mean its trajectories cannot rotate around the z-axis. Math 3010 Final Exam - Page 6 of 14 Dec. 8, 2012 3. Line integrals (a) (5 points) Let C be the curve parametrized by a differentiable path c : [a, b] → R3 . Let f : R3 → R be of class C 1 . Prove that C ∇f · ds = f (c(b)) − f (c(a)) . Solution: Let c(t) = (x(t), y (t), z (t)) and F (t) = f (c (t)). Then F ′ (t) = ∂f ′ ∂f ′ ∂f ′ x (t) + y (t) + z (t) = ∇f · c′ (t). ∂x ∂y ∂z Then, b C ∇f · ds = ∇f (c(t)) · c′ (t)dt a b F ′ (c(t)) dt = a = F (c(b)) − F (a) , by FTC = f (c(b)) − f (c(a)) . (b) (5 points) Let C be the straight line segment starting from (3, 0, 2) and ending at 2, π , 1 . Evaluate C F · ds where 4 F (x, y, z ) = (y + yz ) cos(xy ), (x + xz ) cos(xy ) + 2yz, sin(xy ) + y 2 + z 2 . Solution: Notice that 1 F (x, y, z ) = ∇ (1 + z ) sin(xy ) + y 2 z + z 3 3 Thus, π F · ds = f 2, , 1 − f (3, 0, 2) , 4 C π2 1 π + = (1 + 1) sin + 2 16 3 π2 1 −. = 16 3 − (1 + 2) sin (0) + 0 + 8 3 , Math 3010 Final Exam - Page 7 of 14 Dec. 8, 2012 4. (10 points) Let f (x, y, z ) = xyz and S be the triangle with vertices (1, 0, 0), (0, 2, 0)...
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This note was uploaded on 01/19/2014 for the course MATH 3010 taught by Professor Magpantay during the Winter '13 term at York University.

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