14.2Ex18 - r 1 t =-8 t 4 − t 3 ® r 2 t = e t − 6 ® 19 d dt r 1 t r 2 t SOLUTION By the Product Rule for dot products we have d dt r 1 r 2

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SECTION 14.2 Calculus of Vector-Valued Functions (ET Section 13.2) 469 x y t = 0 t = 1 t = − 1 r ' (1) = ⟨− 2, 1 17. Sketch the curve r 1 ( t ) = - t , t 2 ® together with its tangent vector at t = 1. Then do the same for r 2 ( t ) = - t 3 , t 6 ® . SOLUTION Note that r 1 0 ( t ) = h 1 , 2 t i and so r 1 0 ( 1 ) = h 1 , 2 i . The graph of r 1 ( t ) satisFes y = x 2 . Likewise, r 2 0 ( t ) = - 3 t 2 , 6 t 5 ® and so r 2 0 ( 1 ) = h 3 , 6 i . The graph of r 2 ( t ) also satisFes y = x 2 . Both graphs and tangent vectors are given here. 2 r 2 ( t ) 1 r 1 ( t ) 18. Sketch the cycloid r ( t ) = - t sin t , 1 cos t ® together with its tangent vectors at t = π 3 and 3 4 . SOLUTION The tangent vector r 0 ( t ) is the following vector: r 0 ( t ) = d dt h t sin t , 1 cos t i = h 1 cos t , sin t i Substituting the given values gives the following vectors: r 0 ³ 3 ± = D 1 cos 3 , sin 3 E = * 1 2 , 3 2 + r 0 ² 3 4 = ¿ 1 cos 3 4 , sin 3 4 À = * 1 + 2 2 , 2 2 + The cycloid r ( t ) = h t sin t , 1 cos t i and the two tangent vectors are shown in the following Fgure: t = 0 y = 2 t = 2 p x y 1 2 3 2 , t = p 3 t = 3 p 4 2 2 2 2 1 + , In Exercises 19–22, use the appropriate Product Rule to evaluate the derivative, where
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Unformatted text preview: r 1 ( t ) =-8 t , 4 , − t 3 ® , r 2 ( t ) =-, e t , − 6 ® 19. d dt ( r 1 ( t ) · r 2 ( t ) ) SOLUTION By the Product Rule for dot products we have: d dt r 1 · r 2 = r 1 · r 2 + r 1 · r 2 We compute the derivatives of r 1 and r 2 : r 1 = d dt-8 t , 4 , − t 3 ® =-8 , , − 3 t 2 ® r 2 = d dt-, e t , − 6 ® =-, e t , ® By (1) we have: d dt r 1 ( t ) · r 2 ( t ) =-8 t , 4 , − t 3 ® ·-, e t , ® +-8 , , − 3 t 2 ® ·-, e t , − 6 ® = 4 e t + 18 t 2...
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This note was uploaded on 02/13/2010 for the course MATH MATH 32A taught by Professor Park during the Fall '09 term at UCLA.

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