# vcprac01s - The University of Sydney School of Mathematics...

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Unformatted text preview: The University of Sydney School of Mathematics and Statistics Solutions to Practice Session 1 MATH2061: Vector Calculus Summer School 2012 1. Evaluate integraldisplay C φds if φ = xy + z and C is the straight line segment from (1 , 1 , 1) to ( − 2 , 2 , 5). Solution: C has parametric equations x = 1 − 3 t , y = 1 + t , z = 1 + 4 t , for t : 0 → 1. So integraldisplay C φds = integraldisplay 1 ((1 − 3 t )(1 + t ) + (1 + 4 t )) √ 9 + 1 + 16 dt = √ 26 integraldisplay 1 − 3 t 2 + 2 t + 2 dt = 2 √ 26 . 2. Describe geometrically the vector fields determined by each of the following vector functions: (a) F = 3 i + 2 j ; (b) F = − x i − y j . Solution: (a) The vector at ( x,y ) has tail at ( x,y ) and head at ( x + 3 ,y + 2). (b) The vector at ( x,y ) has tail at ( x,y ) and head at the origin. (a) (b) 3. Evaluate contintegraldisplay (2 x − y + 4) dx + (5 y + 3 x − 6) dy around a triangle in the xy-plane with vertices at (0 , 0) , (3 , 0) , (3 , 2), traversed in the anti-clockwise direction. Solution: Let A and B be the points (3 , 0) and (3 , 2) respectively. Then contintegraldisplay (2 x − y + 4) dx + (5 y + 3 x − 6) dy = integraldisplay OA (2 x − y + 4) dx + (5 y + 3 x − 6) dy + integraldisplay AB (2 x − y + 4) dx + (5 y + 3 x − 6) dy + integraldisplay BO (2 x − y + 4) dx + (5 y + 3 x − 6) dy. O A B Copyright c circlecopyrt 2012 The University of Sydney 1 Along OA, y = 0 , x = t, t : 0 → 3 , dx = dt, dy = 0 ....
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## This note was uploaded on 02/06/2012 for the course MATH 2061 taught by Professor Notsure during the Three '09 term at University of Sydney.

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vcprac01s - The University of Sydney School of Mathematics...

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