Pre-Calc Homework Solutions 162

# Pre-Calc Homework - 162 Section 4.4 di dt 35 Since 2 sin t 2 cos t the largest magnitude of the 2 sin t 2 cos t 0 or cos2 t and cos2 t 1 2 38 No

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35. Since } d d i t }52 2 sin t 1 2 cos t , the largest magnitude of the current occurs when 2 2 sin t 1 2 cos t 5 0, or sin t 5 cos t . Squaring both sides gives sin 2 t 5 cos 2 t , and we know that sin 2 t 1 cos 2 t 5 1, so sin 2 t 5 cos 2 t 5 } 1 2 } . Thus the possible values of t are } p 4 } , } 3 4 p } , } 5 4 p } , and so on. Eliminating extraneous solutions, the solutions of sin t 5 cos t are t 5 } p 4 } 1 k p for integers k , and at these times ) i ) 5 ) 2 cos t 1 2 sin t ) 5 2 ˇ 2 w . The peak current is 2 ˇ 2 w amps. 36. The square of the distance is D ( x ) 5 1 x 2 } 3 2 } 2 2 1 ( ˇ x w 1 0) 2 5 x 2 2 2 x 1 } 9 4 } , so D 9 ( x ) 5 2 x 2 2 and the critical point occurs at x 5 1. Since D 9 ( x ) , 0 for x , 1 and D 9 ( x ) . 0 for x . 1, the critical point corresponds to the minimum distance. The minimum distance is ˇ D w (1 w ) w 5 } ˇ 2 5 w } . 37. Calculus method: The square of the distance from the point (1, ˇ 3 w ) to ( x , ˇ 1 w 6 w 2 w x w 2 w ) is given by D
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## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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