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Pre-Calc Homework Solutions 23

Pre-Calc Homework Solutions 23 - Section 1.5 39 Possible...

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39. Possible answers: (a) x 5 a cos t , y 5 2 a sin t , 0 # t # 2 p (b) x 5 a cos t , y 5 a sin t , 0 # t # 2 p (c) x 5 a cos t , y 5 2 a sin t , 0 # t # 4 p (d) x 5 a cos t , y 5 a sin t , 0 # t # 4 p 40. Possible answers: (a) x 5 2 a cos t , y 5 b sin t , 0 # t # 2 p (b) x 5 2 a cos t , y 5 2 b sin t , 0 # t # 2 p (c) x 5 2 a cos t , y 5 b sin t , 0 # t # 4 p (d) x 5 2 a cos t , y 5 2 b sin t , 0 # t # 4 p 41. Note that m / OAQ 5 t , since alternate interior angles formed by a transversal of parallel lines are congruent. Therefore, tan t 5 } O AQ Q } 5 } 2 x } , so x 5 } ta 2 n t } 5 2 cot t . Now, by equation (iii), we know that AB 5 } ( A A Q O ) 2 } 5 1 } A A Q O } 2 ( AQ ) 5 (cos t )( x ) 5 (cos t )(2 cot t ) 5 } 2 s c i o n s t 2 t } . Then equation (ii) gives y 5 2 2 AB sin t 5 2 2 } 2 s c i o n s t 2 t } ? sin t 5 2 2 2 cos 2 t 5 2 sin 2 t . The parametric equations are: x 5 2 cot t , y 5 2 sin 2 t , 0 , t , p Note: Equation (iii) may not be immediately obvious, but it may be justified as follows. Sketch segment QB . Then / OBQ is a right angle, so n ABQ , n AQO , which gives } A A Q B } 5 } A A Q O } .
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