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Assignment 8 Solutions

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

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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 -∞ 0 0 - 3 1 / 5 - 3 / 5 - 2 2 / 5 - 4 / 5 - 1 1 1 - 1 2 2 / 5 - 4 / 5 0 2 0 1 2 8 / 5 4 / 5 1 1 1 2 2 / 5 4 / 5 3 1 / 5 3 / 5 0 0 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.
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2: (a) Sketch the curve ( x, y ) = (cos t, sin 2 t ), showing all horizontal and vertical points. Solution: We have x ( t ) = - sin t and y ( t ) = 2 cos 2 t . The curve is horizontal when y ( t ) = 0, that is when t = π 4 + π 2 k with k Z , and the curve is vertical when x ( t ) = 0, that is when t = πk with k Z . We make a table of values and sketch the curve.
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