Fall 2009 Prelim 2 Solutions

# Fall 2009 Prelim 2 Solutions - Solutions to Prelim Exam 2 1...

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Solutions to Prelim Exam 2 1. Set x = t and y = 4 - t 2 . We get the parametrization r ( t ) = t i +(4 - t 2 ) j for - 1 t 2 . Hence d r dt = i - 2 t j and ± ± ± ± d r dt ± ± ± ± = 1 + 4 t 2 . (a) M = 2 - 1 (2 + t ) 1 + 4 t 2 dt (b) ¯ x = 1 M 2 - 1 t (2 + t ) 1 + 4 t 2 dt and ¯ y = 1 M 2 - 1 (4 - t 2 )(2 + t ) 1 + 4 t 2 dt . 2. (a) Sketch the circles: y x 0 r = 1 r = 2 cos( θ ) Since r = 1 and r = 2 cos( θ ), we get that the intersection points satisfy 1 = 2 cos( θ ). Hence θ = arccos( 1 2 ) = ± π 3 . The intersection points are (1 , π 3 ) and (1 , - π 3 ). (b) Area = π 3 - π 3 2 cos( θ ) 1 r dr dθ . (c) Area = 2 1 θ =arccos( r 2 ) θ = - arccos( r 2 ) r dθ dr. (d) Area = π 3 - π 3 r 2 2 ± ± ± ± 2 cos( θ ) 1 = π 3 - π 3 1 2 ( 4 cos 2 ( θ ) - 1 ) = π 3 - π 3 [ 2 (1 + cos(2 θ )) 2 - 1 2 ] = π 3 - π 3 ² cos(2 θ ) + 1 2 ³ = ² 1 2 sin(2 θ ) + 1 2 θ ³± ± ± ± π 3 - π 3 = 1 2 sin( 2 π 3 ) + π 3 ³ - ² sin( - 2 π 3 ) - π 3 ³] = 3

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## This note was uploaded on 06/11/2011 for the course MATH 1920 at Cornell.

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Fall 2009 Prelim 2 Solutions - Solutions to Prelim Exam 2 1...

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