# Substituting in 1 we obtain the following

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Chapter 8 / Exercise 8
Mathematical Practices, Mathematics for Teachers: Activities, Models, and Real-Life Examples
Larson
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Substituting in (1) we obtain the following parametrization: c (θ) = ( cos + π), sin + π)) 6. Find a path c(t) that traces the parabolic arc y = x 2 from ( 0 , 0 ) to ( 3 , 9 ) for 0 t 1. solution The second coordinates of the points on the parabolic arc are the square of the first coordinates. Therefore the points on the arc have the form: c(t) = (αt,α 2 t 2 ) (1) We need that c( 1 ) = ( 3 , 9 ) . That is, c( 1 ) = (α,α 2 ) = ( 3 , 9 ) α = 3 Substituting in (1) gives the following parametrization: c(t) = ( 3 t, 9 t 2 ) 7. Find a path c(t) that traces the line y = 2 x + 1 from ( 1 , 3 ) to ( 3 , 7 ) for 0 t 1. solution Solution 1: By one of the examples in section 12.1, the line through P = ( 1 , 3 ) with slope 2 has the parametrization c(t) = ( 1 + t, 3 + 2 t) But this parametrization does not satisfy c( 1 ) = ( 3 , 7 ) . We replace the parameter t by a parameter s so that t = αs + β . We get c (s) = ( 1 + αs + β, 3 + 2 (αs + β)) = (αs + β + 1 , 2 αs + 2 β + 3 ) We need that c ( 0 ) = ( 1 , 3 ) and c ( 1 ) = ( 3 , 7 ) . Hence, c ( 0 ) = ( 1 + β, 3 + 2 β) = ( 1 , 3 ) c ( 1 ) = + β + 1 , 2 α + 2 β + 3 ) = ( 3 , 7 ) We obtain the equations 1 + β = 1 3 + 2 β = 3 α + β + 1 = 3 2 α + 2 β + 3 = 7 β = 0 = 2 Substituting in (1) gives c (s) = ( 2 s + 1 , 4 s + 3 )
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Chapter 8 / Exercise 8
Mathematical Practices, Mathematics for Teachers: Activities, Models, and Real-Life Examples
Larson
Expert Verified
1628 C H A P T E R 11 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS (LT CHAPTER 12) Solution 2: The segment from ( 1 , 3 ) to ( 3 , 7 ) has the following vector parametrization: ( 1 t) 1 , 3 + t 3 , 7 = 1 t + 3 t, 3 ( 1 t) + 7 t = 1 + 2 t, 3 + 4 t The parametrization is thus c(t) = ( 1 + 2 t, 3 + 4 t) 8. Sketch the graph c(t) = ( 1 + cos t, sin 2 t) for 0 t 2 π and draw arrows specifying the direction of motion. solution From x = 1 + cos t we have x 1 = cos t . We substitute this in the y coordinate to obtain y = sin 2 t = 2 sin t cos t = ± 2 sin 2 t cos t = ± 2 1 cos 2 t cos t = ± 2 1 (x 1 ) 2 (x 1 ) We can see that the graph is symmetric with respect to the x -axis, hence we plot the function y = 2 1 (x 1 ) 2 (x 1 ) and reflect it with respect to the x -axis. When t = 0 we have c( 0 ) = ( 2 , 0 ) . when t increases near 0, cos t is decreasing and sin 2 t is increasing, hence the general direction at the point ( 2 , 0 ) is upwards and left. As t approaches π/ 2, the x -coordinate decreases to 1 and the y -coordinate to 0. Likewise, as t moves from π/ 2 to π , the x -coordinate moves to 0 while the y -coordinate falls to 1 and then rises to 0. The resulting graph is seen here in the corresponding figure. x y 1 −1 2 1 Plot of Exercise 8 In Exercises 9–12, express the parametric curve in the form y = f(x) . 9. c(t) = ( 4 t 3 , 10 t) solution We use the given equation to express t in terms of x . x = 4 t 3 4 t = x + 3 t = x + 3 4 Substituting in the equation of y yields y = 10 t = 10 x + 3 4 = − x 4 + 37 4 That is, y = − x 4 + 37 4 10. c(t) = (t 3 + 1 ,t 2 4 ) solution The parametric equations are x = t 3 + 1 and y = t 2 4. We express t in terms of x : x = t 3 + 1 t 3 = x 1 t = (x 1 ) 1 / 3 Substituting in the equation of y yields y = t 2 4 = (x 1 ) 2 / 3 4
Chapter Review Exercises 1629 That is, y = (x 1 ) 2 / 3 4 11. c(t) = 3 2 t ,t 3 + 1 t solution We use the given equation to express t in terms of x : x = 3 2 t 2 t = 3 x t = 2 3 x Substituting in the equation of y yields y = 2 3 x 3 + 1 2 /( 3 x) = 8 ( 3 x) 3 + 3