Assignment 8 Solutions

Assignment 8 Solutions - MATH 138 Calculus 2, Solutions to...

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Unformatted text preview: MATH 138 Calculus 2, Solutions to Assignment 8 1: (a) Sketch the curve ( x, y ) = 2 1 + t 2 , 2 t 1 + t 2 , showing all horizontal and vertical points. Solution: We have x ( t ) =- 4 t (1 + t 2 ) 2 and y ( t ) = 2(1 + t 2 )- (2 t )(2 t ) (1 + t 2 ) 2 = 2- 2 t 2 (1 + t 2 ) 2 . The curve is horizontal when y ( t ) = 0, that is when t = 1, and vertical when x ( t ) = 0, that is at t = 0. We make a table of values and sketch the curve. t x y-- 3 1 / 5- 3 / 5- 2 2 / 5- 4 / 5- 1 1 1- 1 2 2 / 5- 4 / 5 2 1 2 8 / 5 4 / 5 1 1 1 2 2 / 5 4 / 5 3 1 / 5 3 / 5 y 1 x 1 2 (b) Find the Cartesian equation of this curve. Solution: We have x 2 = 4 (1 + t 2 ) 2 and y 2 = 4 t 2 (1 + t 2 ) 2 and so x 2 + y 2 = 4 + 4 t (1 + t 2 ) 2 = 4 1 + t 2 = 2 x . Thus the Cartesian equation is x 2 + y 2 = 2 x , or equivalently ( x- 1) 2 + y 2 = 1. 2: (a) Sketch the curve ( x, y ) = (cos t, sin 2 t ), showing all horizontal and vertical points....
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This note was uploaded on 09/18/2008 for the course MATH MAth 138 taught by Professor Marshman during the Spring '08 term at Waterloo.

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Assignment 8 Solutions - MATH 138 Calculus 2, Solutions to...

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