Business Calc Homework w answers_Part_5

Business Calc Homework w answers_Part_5 - Section 1.4 (b) y...

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(b) y 5 t 5 ( 2 ˇ t w ) 2 5 x 2 The parametrized curve traces the left half of the parabola defined by y 5 x 2 (or all of the curve defined by x 52 ˇ y w ). 15. (a) [ 2 1, 5] by [ 2 1, 3] Initial point: (0, 0) Terminal point: None (b) y 5 ˇ t w 5 ˇ x w The parametrized curve traces all of the curve defined by y 5 ˇ x w (or the upper half of the parabola defined by x 5 y 2 ). 16. (a) [ 2 3, 9] by [ 2 4, 4] No initial or terminal point. (b) x 5 sec 2 t 2 1 5 tan 2 t 5 y 2 The parametrized curve traces all of the parabola defined by x 5 y 2 . 17. (a) [ 2 3, 3] by [ 2 2, 2] No initial or terminal point. Note that it may be necessary to use a t -interval such as [ 2 1.57, 1.57] or use dot mode in order to avoid “asymptotes” showing on the calculator screen. (b) x 2 2 y 2 5 sec 2 t 2 tan 2 t 5 1 The parametrized curve traces the left branch of the hyperbola defined by x 2 2 y 2 5 1 (or all of the curve defined by x 52 ˇ y 2 w 1 w 1 w ). 18. (a) [ 2 6, 6] by [ 2 5, 1] No initial or terminal point. Note that it may be necessary to use a t -interval such as [ 2 1.57, 1.57] or use dot mode in order to avoid “asymptotes” showing on the calculator screen. (b) 1 } 2 y } 2 2 2 x 2 5 sec 2 t 2 tan 2 t 5 1 The parametrized curve traces the lower branch of the hyperbola defined by 1 } 2 y } 2 2 2 x 2 5 1 (or all of the curve defined by y 52 2 ˇ x 2 w 1 w 1 w ). 19. (a) [ 2 9, 9] by [ 2 6, 6] No initial or terminal point. (b) y 5 4 t 2 7 5 2(2 t 2 5) 1 3 5 2 x 1 3 The parametrized curve traces all of the line defined by y 5 2 x 1 3. 20. (a) [ 2 6, 6] by [ 2 4, 4] No initial or terminal point. (b) y 5 1 1 t 5 2 2 (1 2 t ) 5 2 2 x 52 x 1 2 The parametrized curve traces all of the line defined by y 52 x 1 2. 21. (a) [ 2 3, 3] by [ 2 2, 2] Initial point: (0, 1) Terminal point: (1, 0) (b) y 5 1 2 t 5 1 2 x 52 x 1 1 The Cartesian equation is y 52 x 1 1. The portion traced by the parametrized curve is the segment from (0, 1) to (1, 0). 22. (a) [ 2 2, 4] by [ 2 1, 3] Initial point: (3, 0) Terminal point: (0, 2) (b) y 5 2 t 5 (2 t 2 2) 1 2 52} 2 3 } (3 2 3 t ) 1 2 52} 2 3 } x 1 2 The Cartesian equation is y 52} 2 3 } x 1 2. The portion traced by the curve is the segment from (3, 0) to (0, 2). 23. (a) [ 2 6, 6] by [ 2 2, 6] Initial point: (4, 0) Terminal point: None (b) y 5 ˇ t w 5 4 2 (4 2 ˇ t w ) 5 4 2 x 52 x 1 4 The parametrized curve traces the portion of the line defined by y 52 x 1 4 to the left of (4, 0), that is, for x # 4. Section 1.4 21
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24. (a) [ 2 1, 5] by [ 2 1, 3] Initial point: (0, 2) Terminal point: (4, 0) (b) y 5 ˇ 4 w 2 w t w 2 w 5 ˇ 4 w 2 w x w The parametrized curve traces the right portion of the curve defined by y 5 ˇ 4 w 2 w x w , that is, for x $ 0. 25. (a)
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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Business Calc Homework w answers_Part_5 - Section 1.4 (b) y...

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