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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online