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20esolns4

# 20esolns4 - Homework 4 Solutions MTH 20E 4.3.4 4.3.8 4.3.12...

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Homework 4 Solutions MTH 20E 4.3.4 4.3.8 4.3.12 Graphics package trouble - check back later! 4.3.18 In order to show V ( c ( t )) is decreasing, it is su ffi cient to show that its derivative is non-positive. By the chain rule, ( V ( c ( t ))) = V ( c ( t ))˙ c ( t ) = - ( c ( t ))˙ c ( t ) 0 . The second equality comes by virtue of c being a flow line for -∇ V . 7.1.2a Since c ( t ) = (cos t, - sin t, 1), c ( t ) = 2, and c f ds = 2 2 π 0 cos t dt = 0 . 7.1.6 (a)This follows from the standard path integral formula, with the consid- eration that c ( t ) = dx d θ 2 + dy d θ 2 = d d θ ( r cos θ ) 2 + d d θ ( r sin θ ) 2 = dr d θ cos θ - r sin θ 2 + dr d θ sin θ + r cos θ 2 ( usingProductRule ) = r 2 + dr d θ 2 . (b) L ( r ) = 2 π 0 (1 + cos θ ) 2 + ( - sin θ ) 2 d θ = 2 π 0 2 + 2 cos θ d θ = 2 π 0 2 cos ( θ 2 ) d θ ( Half - AngleFormula ) 1

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= 8 π 2 0 cos u du = 8 . 7.2.12 (a)This results stems directly from the definitions of a path integral and length of graph, with the understand that arc length is a path integral over the function f 1. (b) L ( f ) = 2 1 1 + ( 1 x ) 2 dx = 2 1 x 2 + 1 x dx = x 2 + 1 - ln 1 + x 2 + 1 x 2 1 ( TableofIntegration 56) = 5 - ln 1 + 5 2 - 2 + ln (1 + 2) . 7.2.1a c F · d s = 1 0 ( t, t, t ) · (1 , 1 , 1) dt = 1 0 3 t dt = 3 2 . 7.2.2a c x dy - y dx = 2 π 0 (cos t )(cos t ) - (sin t )( - sin t ) dt = 2 π 0 1 dt = 2 π . 7.2.15 It is possible to compute this line integral directly; any simple curve
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