SalasSV_09_06 - 552 CHAPTER 9 THE CONIC SECTIONS; POLAR...

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² 9.6 CURVES GIVEN PARAMETRICALLY So ±ar we have specifed curves by equations in rectangular coordinates or by equations in polar coordinates. Here we introduce a more general method. We begin with a pair o± ±unctions x = x ( t ), y = y ( t ) di±±erentiable on the interior o± an interval I . At the endpoints o± I (i± any) we require only continuity. For each number t in I we can interpret ( x ( t ), y ( t )) as the point with x -coordinate x ( t ) and y -coordinate y ( t ). Then, as t ranges over I , the point ( x ( t ), y ( t )) traces out a path in the xy -plane (Figure 9.6.1). We call such a path a parametrized curve and re±er to t as the parameter . Example 1 Identi±y the curve parametrized by the ±unctions x ( t ) = t + 1, y ( t ) = 2 t 5, t ( −∞ , ). SOLUTION We can express y ( t ) in terms o± x ( t ): y x ( x ( t ), y ( t )) Figure 9.6.1 y ( t ) = 2[ x ( t ) 1] 5 = 2 x ( t ) 7.
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9.6 CURVES GIVEN PARAMETRICALLY ± 553 The functions parametrize the line y = 2 x 7: as t ranges over the set of real numbers, the point ( x ( t ), y ( t )) traces out the line y = 2 x 7. ± Example 2 Identify the curve parametrized by the functions x ( t ) = 2 t , y ( t ) = t 2 , t [0, ). SOLUTION From the ±rst equation t = 1 2 x ( t ), and so y ( t ) = 1 4 [ x ( t )] 2 . The functions parametrize that part of the parabola y = 1 4 x 2 which lies in the right half plane: as t ranges over the interval [0, ), the point ( x ( t ), y ( t )) traces out the parabolic path y = 1 4 x 2 , x 0 (Figure 9.6.2). ± 1 y x 2 1 y = x 2 1 4 Figure 9.6.2 Example 3 Identify the curve parametrized by the functions x ( t ) = sin 2 t , y ( t ) = cos t , t [0, π ]. SOLUTION Note ±rst that x ( t ) = sin 2 t = 1 cos 2 t = 1 [ y ( t )] 2 . The points ( x ( t ), y ( t )) all lie on the parabola x = 1 y 2 . (Figure 9.6.3) y x y x = 1 – y 2 (0, 1); t = 0 (0, –1); t = π x ( t ) = sin 2 t , y ( t ) = cos t t [0, ] x Figure 9.6.3 At t = 0, x = 0 and y = 1; at t = π , x = 0 and y =− 1. As t ranges from 0 to π , the point ( x ( t ), y ( t )) traverses the parabolic arc x = 1 y 2 , 1 y 1 from the point (0, 1) to the point (0, 1). ± Remark Changing the domain in Example 3 to all real t does not give us any more of the parabola. For any given t we still have 0 x ( t ) 1 and 1 y ( t ) 1. As t ranges over the set of real numbers, the point ( x ( t ), y ( t )) traces out that same parabolic arc back and forth an in±nite number of times. ±
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554 ± CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS Straight Lines Given that ( x 0 , y 0 ) ±= ( x 1 , y 1 ), the functions (9.6.1) x ( t ) = x 0 + t ( x 1 x 0 ), y ( t ) = y 0 + t ( y 1 y 0 ), t ( −∞ , ) parametrize the line that passes through ( x 0 , y 0 ) and ( x 1 , y 1 ). PROOF If x 1 = x 0 , then we have x ( t ) = x 0 , y ( t ) = y 0 + t ( y 1 y 0 ). ( y 0 y 1 ) As t ranges over the set of real numbers, x ( t ) remains constantly x 0 and y ( t ) ranges over the set of real numbers. The functions parametrize the vertical line x = x 0 . Since x 1 = x 0 , both ( x 0 , y 0 ) and ( x 1 , y 1 ) lie on this vertical line. If x 1 x 0 , then we can solve the Frst equation for t : t = x ( t ) x 0 x 1 x 0 .
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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SalasSV_09_06 - 552 CHAPTER 9 THE CONIC SECTIONS; POLAR...

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