14.1%20Ex%2017-21
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14.1%20Ex%2017-21

Course Number: MATH 32a, Winter 2008

College/University: UCLA

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S E C T I O N 14.1 y 15 10 5 -20 Vector-Valued Functions (ET Section 13.1) 455 x -5 -15 -20 20 The projection of the curve onto the x z-plane is traced by t cos t, 0, t , which is a wave with increasing amplitude moving in the z direction as shown in the following figure: z 20 15 10 5 -20 -15 -10 -5 -10 -15 -20 x 5 10 15 20 17. Find the points where the path r(t) = sin t, cos t, sin t cos 2t intersects the x...

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E S C T I O N 14.1 y 15 10 5 -20 Vector-Valued Functions (ET Section 13.1) 455 x -5 -15 -20 20 The projection of the curve onto the x z-plane is traced by t cos t, 0, t , which is a wave with increasing amplitude moving in the z direction as shown in the following figure: z 20 15 10 5 -20 -15 -10 -5 -10 -15 -20 x 5 10 15 20 17. Find the points where the path r(t) = sin t, cos t, sin t cos 2t intersects the x y-plane. SOLUTION The curve intersects the x y-plane at the points where z = 0. That is, sin t cos 2t = 0 and so either sin t = 0 or cos 2t = 0. The solutions are, thus: t = k or t = k + , 4 2 k = 0, 1, 2, . . . The values t = k yield the points: (sin k, cos k, 0) = 0, (-1)k , 0 . The values t = + k yield the points: 4 2 k = 0 : sin k = 1 : sin k = 2 : sin k = 3 : sin , cos , 0 = 4 4 1 1 , ,0 2 2 1 1 ,- ,0 2 2 3 3 , cos ,0 = 4 4 5 1 5 1 , cos ,0 = - ,- ,0 4 4 2 2 7 1 1 7 , cos ,0 = - , ,0 4 4 2 2 (Other values of k do not provide new points). We conclude that the curve intersects the x y-plane at the following points: 1 1 1 1 1 1 1 1 (0, 1, 0), (0, -1, 0), , , 0 , , - , 0 , - , - , 0 , - , , 0 2 2 2 2 2 2 2 2 18. Parametrize the intersection of the surfaces y 2 - z 2 = x - 2, y2 + z2 = 9 using t = y as the parameter (two vector functions are needed as in Example 2). 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 = 9 - y 2 . From the second equation we have: SOLUTION x = y 2 - z 2 + 2 = y 2 - 9 - y 2 + 2 = 2y 2 - 7 Taking t = y as a parameter, we have z = 9 - t 2 , x = 2t 2 - 7, yielding the following vector parametrization: r(t) = 2t 2 - 7, t, 9 - t 2 , for - 3 t 3. 19. Find a parametrization of the curve in Exercise 18 using trigonometric functions. 456 C H A P T E R 14 C A L C U L U S O F VE C T O R - VA L U E D F U N C T I O N S SOLUTION (ET CHAPTER 13) The curve in Exercise 18 is the intersection of the surfaces y 2 - z 2 = x - 2, y 2 + z 2 = 9. The circle y 2 + z 2 = 9 is parametrized by y = 3 cos t, z = 3 sin t. Substituting in the first equation and using the identity cos2 t - sin2 t = cos 2t, gives: x = 2 + y 2 - z 2 = 2 + (3 cos t)2 - (3 sin t)2 = 2 + 9 cos2 t - sin2 t = 2 + 9 cos 2t We obtain the following trigonometric parametrization: r(t) = 2 + 9 cos 2t, 3 cos t, 3 sin t 20. Viviani's Curve C is the intersection of the surfaces x 2 + y 2 = z 2 , y = z 2 (Figure 12). z x 2 + y2 = z 2 x y Viviani's curve (A) y = z2 (B) Viviani's curve viewed from the negative y-axis FIGURE 12 Viviani's curve is the intersection of the surfaces x 2 + y 2 = z 2 and y = z 2 . (a) Parametrize each of the two parts of C corresponding to x 0 and x 0 taking t = z as parameter. (b) Describe the projection of C onto the x y-plane. (c) Show that C lies on the sphere of radius 1 with center (0, 1, 0). This curve looks like a figure eight lying on a sphere [Figure 12(B)]. SOLUTION (a) We must solve for y and x in terms of z (which is a parameter). We get: y = z2 x 2 = z2 - y2 x = z2 - y2 = z2 - z4 Here, the from x = z 2 - z 4 represents the two parts of the parametrization: + for x 0, and - for x 0. Substituting the parameter z = t we get: y = t 2, We obtain the following parametrization: r(t) = t 1 - t 2 , t 2 , t for - 1 t 1 (1) x = t 2 - t 4 = t 1 - t 2 . (b) The projection of the curve onto the x y-plane is the curve on the x y-plane obtained by setting the z-coordinate of r(t) equal to zero. We obtain the following curve: t 1 - t 2 , t 2 , 0 , -1 t 1 We also note that since x = t 1 - t 2 , then x 2 = t 2 - (1 t 2 ), but also y = t 2 , so that gives us the equation x 2 = y(1 - y) for the projection onto the x y plane. We rewrite this as follows. x 2 = y(1 - y) x 2 + y2 - y = 0 x 2 + y 2 - y + 1/4 = 1/4 x 2 + (y - 1/2)2 = (1/2)2 We can now identify this projection as a circle in the x y plane, with radius 1/2, centered at the x y point (0, 1/2). S E C T I O N 14.1 Vector-Valued Functions (ET Section 13.1) 457 (c) The equation of the sphere of radius 1 with center (0, 1, 0) is: x 2 + (y - 1)2 + z 2 = 1 (2) To show that C lies on this sphere, we show that the coordinates of the points on C (given in (1)) satisfy the equation of the sphere. Substituting the coordinates from (1) into the left side of (2) gives: x 2 + (y - 1)2 + z 2 = t 1 - t 2 2 + (t 2 - 1)2 + t 2 = t 2 (1 - t 2 ) + (t 2 - 1)2 + t 2 = (t 2 - 1)(t 2 - 1 - t 2 ) + t 2 = 1 We conclude that the curve C lies on the sphere of radius 1 with center (0, 1, 0). 21. Show that any point on x 2 + y 2 = z 2 can be written in the form (z cos , z sin , z) for some . Use this to find a parametrization of Viviani's curve (Exercise 20) with as parameter. SOLUTION We first verify that x = z cos , y = z sin , and z = z satisfy the equation of the surface: x 2 + y 2 = z 2 cos2 + z 2 sin2 = z 2 cos2 + sin2 = z 2 We now show that if (x, y, z) satisfies x 2 + y 2 = z 2 , then there exists a value of such that x = z cos , y = z sin . Since x 2 + y 2 = z 2 , we have |x| |z| and |y| |z|. If z = 0, then also x = y = 0 and any value of is adequate. If z = 0 then x 1 and y 1, hence there exists 0 such that x = cos 0 . Hence, z z z z2 - x 2 y = = 1- z z2 x 2 = 1 - cos2 0 = sin 0 z If x and y are both positive, we choose 0 such that 0 < 0 < . If x > 0 and y < 0 we choose 0 such that z z z z 2 3 < < 2 . If x < 0 and y < 0 we choose such that < < 3 , and if x < 0 and y > 0 we choose such 0 0 0 0 z z z z 2 2 that < 0 < . In either case we can represent the points on the surface as required. Viviani's curve is the intersection 2 of the surfaces x 2 + y 2 = z 2 and x = z 2 . The points on these surfaces are of the form: x 2 + y2 = z2: x = z2: (z cos , z sin , z) (z 2 , y, z) (1) The points (x, y, z) on the intersection curve must satisfy the following equations: z 2 = z cos y = z sin The first equation implies that z = 0 or z = cos . The second equation implies that y = 0 or y = cos sin = 1 sin 2 . 2 The x coordinate is obtained by substituting z = cos in x = z cos (or in x = z 2 ). That is, x = cos2 . We obtain the following vector parametrization of the curve: r(t) = cos2 , 1 sin 2 , cos 2 22. Use sine and cosine to parametrize the intersection of the cylinders x 2 + y 2 = 1 and x 2 + z 2 = 1 (use two vector-valued functions). Then describe the projections of this curve on the three coordinate planes. The circle x 2 + z 2 = 1 in the x z-plane is parametrized by x = cos t, z = sin t, and the circle x 2 + y 2 = 1 in the x y-plane is parametrized by x = cos s, y = sin s. Hence, the points on the cylinders can be written in the form: SOLUTION x 2 + z 2 = 1: x 2 + y 2 = 1: (cos t, y, sin t) (cos s, sin s, z) The points (x, y, z) on the intersection of the two cylinders must satisfy the following equations: cos t = cos s y = sin s z = sin t The first equation implies that s = t + 2 k. Substituting in the second equation gives y = sin (t + 2 k) = sin (t) = sin t. Hence, x = cos t, y = sin t, z = sin t. We obtain the following vector parametrization of the intersection: r(t) = cos t, sin t, sin t

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