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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 = k or t = 4 + k 2 , k = , 1 , 2 , . . . The values t = k yield the points: ( sin k , cos k , ) = , ( 1 ) k , 0 . The values t = 4 + k 2 yield the points: k = : sin 4 , cos 4 , = 1 2 , 1 2 , k = 1 : sin 3 4 , cos 3 4 , = 1 2 , 1 2 , k = 2 : sin 5 4 , cos 5 4 , = 1 2 , 1 2 , k = 3 : sin 7 4 , cos 7 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 , ) , 1 2 , 1 2 , 0 , 1 2 , 1 2 , 0 , 1 2 , 1 2 , 0 , 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 .... View Full Document