14.1%20Ex%2017-21

# 14.1%20Ex%2017-21 - S E C T I O N 14.1 Vector-Valued...

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Unformatted text preview: S E C T I O N 14.1 Vector-Valued Functions (ET Section 13.1) 455 20 x − 20 − 20 − 15 − 5 5 10 15 y The projection of the curve onto the xz-plane is traced by h t cos t , , t i , which is a wave with increasing amplitude moving in the z direction as shown in the following figure: 20 15 10 5 x − 20 − 15 − 10 − 5 − 20 − 15 − 10 5 10 15 20 z 17. Find the points where the path r ( t ) = h sin t , cos t , sin t cos 2 t i intersects the xy-plane. SOLUTION The curve intersects the xy-plane at the points where z = 0. That is, sin t cos 2 t = 0 and so either sin t = or cos 2 t = 0. The solutions are, thus: t = π k or t = π 4 + π k 2 , k = , ± 1 , ± 2 , . . . The values t = π k yield the points: ( sin π k , cos π k , ) = ³ , ( − 1 ) k , 0 . The values t = π 4 + π k 2 yield the points: k = : ³ sin π 4 , cos π 4 , = 1 √ 2 , 1 √ 2 , ¶ k = 1 : sin 3 π 4 , cos 3 π 4 , ¶ = 1 √ 2 , − 1 √ 2 , ¶ k = 2 : sin 5 π 4 , cos 5 π 4 , ¶ = − 1 √ 2 , − 1 √ 2 , ¶ k = 3 : sin 7 π 4 , cos 7 π 4 , ¶ = − 1 √ 2 , 1 √ 2 , ¶ (Other values of k do not provide new points). We conclude that the curve intersects the xy-plane at the following points: ( , 1 , ) , ( , − 1 , ) , ³ 1 √ 2 , 1 √ 2 , 0 , ³ 1 √ 2 , − 1 √ 2 , 0 , ³ − 1 √ 2 , − 1 √ 2 , 0 , ³ − 1 √ 2 , 1 √ 2 , 18. Parametrize the intersection of the surfaces y 2 − z 2 = x − 2 , y 2 + z 2 = 9 using t = y as the parameter (two vector functions are needed as in Example 2). SOLUTION We solve for z and x in terms of y . From the equation y 2 + z 2 = 9 we have z 2 = 9 − y 2 or z = ± p 9 − y 2 ....
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## This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

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14.1%20Ex%2017-21 - S E C T I O N 14.1 Vector-Valued...

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