260S12Ex3solns

# For the next 3 problems γ t x t y t a t b is a

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For the next 3 problems, γ ( t ) = ( x ( t ) , y ( t )), a t b , is a smooth curve in the plane and we consider the line integral J := γ p ( x, y ) dx + q ( x, y ) dy . Give a proof or counterexample for each of the following. A–2. If γ ( t ) is a horizontal line segment and p ( x, y ) = 0 on this segment, then J = 0. Solution: Parametrize the horizontal line segment, say a x b as γ ( t ) = ( t, c ), where a t b . Then dx/dt = 1 and dy/dt = 0. Thus J := γ p ( x, y ) dx dt + q ( x, y ) dy dt dt = b a 0 dx dt + q ( x, y )0 dt = 0 . A–3. If γ ( t ) is a vertical line segment and p ( x, y ) = 0 on this segment, then J = 0. Solution: If α y δ and x = c , write the curve as γ ( t ) = ( c, t ), with , α t δ . Then dx/dt = 0 and J := γ p ( x, y ) dx dt + q ( x, y ) dy dt dt = β α [0 + q ( c, t ) t ] dt. It is easy to pick q so that this is not zero – and we get a counterexample. For instance use q ( x, y ) 1. 1

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A–4. If p ( x, y ) 0 and q ( x, y ) 0 on γ , and if in defining γ we know that dx/dt > 0 and dy/dt > 0, then J 0. Solution: Just as above, J := γ p ( x, y ) dx dt + q ( x, y ) dy dt dt but now all the terms in the integrand are non-negative. Hence J 0. Part B : Four traditional problems (15 points each, so 60 points). B–1. Let F = y i +(3+2 x ) j +2 k , and γ ( t ) be the straight line from(0 , 0 , 0) to (1 , 2 , - 3). Compute γ F · d s . Solution: First note γ F · d s = γ F ( γ ( t )) · dt dt.
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