hw1sol - Math 53 Homework 1 Solutions (These solutions are...

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Unformatted text preview: Math 53 Homework 1 Solutions (These solutions are not necessarily complete; in particular some diagrams have been omitted for convenience and brevity.) 10.1 # 9: (a) x = t, y = 1- t (where t 0 because of the t ): for t = 0, ( x, y ) = (0 , 1); for t = 1, ( x, y ) = (1 , 0); for t = 2, ( x, y ) (1 . 414 ,- 1); and so on. 1 1- 1- 2- 3 b b b b b (0,1) t = 0 (1,0) t = 1 (2,-3) t = 4 (b) x = t , so t = x 2 , hence y = 1- t = 1- x 2 . ( x 0). # 11: (a) x = sin , y = cos , so x 2 + y 2 = sin 2 + cos 2 = 1. (b) right half of the unit circle, traced clockwise from (sin0 , cos 0) = (0 , 1) to (sin , cos ) = (0 ,- 1). # 13: (a) x = sin t , y = 1 / sin t , so xy = 1. (b) as t increases from 0 to / 2, x increases from 0 to 1; hence, we get the portion of the hyperbola y = 1 /x for 0 < x < 1, traced from left to right (downwards). # 31: (a) x = x 1 + ( x 2- x 1 ) t , y = y 1 + ( y 2- y 1 ) t , 0 t 1: the curve clearly passes through P 1 ( x 1 , y 1 ) at t = 0 and through P 2 ( x 2 , y 2 ) at t = 1. Since x and y each vary at a constant rate, the trajectory is along a straight line. (Or, more explicitly, we can eliminate t : t = x x 1 x 2 x 1 , so y = y 1 + y 2 y 1 x 2 x 1 ( x- x 1 ), which is indeed a line). Moreover, because the range of values of x for 0 t 1 is precisely from x 1 to x 2 (or similarly for y ), the trajectory is the line segment from P 1 to P 2 . (b) x =- 2 + (3- (- 2)) t =- 2 + 5 t and y = 7 + (- 1- 7) t = 7- 8 t . # 33: the circle of radius 2 centered at (0 , 1) can be parametrized by x = 2cos t , y = 1 + 2sin t where, as t varies from 0 to 2 , the trajectory goes around the circle counterclockwise, starting at (2 , 1), and hitting (0 , 3) at t = / 2. Hence: (a) to get a clockwise orientation, we should change...
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hw1sol - Math 53 Homework 1 Solutions (These solutions are...

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