# 21CMidtermPrep2_OssermanSoshnikovKouba.pdf - Math 21C 1(15...

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This preview shows page 1 out of 19 pages. Unformatted text preview: Math 21C 1 (15 pts.) Brian Osserman Practice Exam 2 Determine the radius and interval of convergence of the power series ∞ X n(x − 2)n 3n+1 n=1 Solution: . First we use the root test to determine for which values of x the series converges absolutely. The absolute value series is ∞ X |an | = n=1 ∞ X n|x − 2|n n=1 3n+1 . r lim p n n→∞ n|x − 2|n n→∞ 3n+1 √ |x − 2| n n √ = lim n→∞ 3n3 |x − 2| = . 3 |an | = By the root test, the series P∞ |x − 2| < 3, and diverges if |x−2| 3 n=1 lim n |an | converges if |x−2| 3 < 1 or equivalently > 1 or equivalently |x − 2| > 3. Thus the radius of convergence of the power series is 3. To nd the interval of convergence we have to test the endpoints, which are x = −1, 5. At x = −1, the series is ∞ X n(−3)n n=1 3n+1 = ∞ X n(−1)n n=1 3 , which diverges because the terms don't go to 0 as n goes to innity. At x = 5, the series is ∞ ∞ X X n3n n = , n+1 3 3 n=1 n=1 which also diverges because the terms don't go to 0. So the interval of convergence is (−1, 5). 2 (20 pts.) (a) Find the Taylor series for sin x at x = π 2 . What is the radius and interval of convergence? Solution: The derivatives of sin x cycle between cos x, − sin x, − cos x, and sin x. If they are ± cos x, they vanish at x = π2 . If they are ± sin x, they are ±1 at x = π2 . So the Taylor series is 1 π 1 π 1 π 1 − (x − )2 + (x − )4 + · · · + (−1)n (x − )2n + . . . . 2 2 4! 2 (2n)! 2 We use the ratio test to nd for which x the series converges absolutely. ∞ X |an | = n=1 and ∞ X n=1 1 π |x − |2n , (2n)! 2 |x − π2 |2n+2 (2n)! |x − π2 |2 |an+1 | = = , |an | |x − π2 |2n+2 (2n + 2)! (2n + 2)(2n + 1) so |x − π2 |2 1 |an+1 | π = lim = |x− |2 lim = 0. n→∞ (2n + 2)(2n + 1) n→∞ |an | 2 n→∞ (2n + 2)(2n + 1) lim Thus, by the ratio test the series converges absolutely for all x (the radius of convergence is innite). (b) If you approximate sin x using the Taylor polynomial of order 4 centered at x = π 2 , what is a (reasonable) bound for the error when |x − π2 | < 1? Solution: The fth derivative of sin x is cos x, and | cos x| ≤ 1 for all x. Thus the error of the approximation at x = b is at most 1 |b 5! − π2 |5 . If we want an error bound which holds for all x with |x − π2 | < 1, we nd the error is at most 1 5! = 1 120 . Note: this can be improved by observing that the Taylor polynomial of degree 4 for sin x centered at x = π 2 is also the Taylor polynomial of degree 5, so we can similarly obtain a sharper error bound of 6!1 |x − π 6 | 2 < 1 6! . 2 3 (15 pts.) Let u = h0, 1, 1i, v = h1, 1, 0i. (a) Using projection, write u as a sum of a vector parallel to v and a vector orthogonal to v. Solution: Note that u = projv u + (u − projv u), and by denition projv u is parallel to v and u − projv u is orthogonal to v. Now, u·v v |v|2 0·1+1·1+1·0 = h1, 1, 0i 12 + 12 + 02 1 h1, 1, 0i = 2   1 1 = , ,0 . 2 2 Then u − projv u = h0, 1, 1i − 12 , 21 , 0 = − 21 , 21 , 1 , so we have     1 1 1 1 u= , ,0 + − , ,1 , 2 2 2 2 projv u = where the rst is parallel to v and the second is orthogonal to it. (b) What is the angle between u and v? Solution: u·v ) = cos−1 ( √02 +112 +12 √2 ) = cos−1 ( 12 ) = The angle is cos−1 ( |u||v| 60◦ . 3 4 (15 pts.) Let M be the plane passing through the three points P = (2, 0, 0), Q = (0, 2, 0), and R = (0, 0, 2). (a) What is the equation for M ? Solution: −→ −→ The vectors P Q = h−2, 2, 0i and P R = h−2, 0, 2i both lie on the plane, so we can nd a normal vector by taking their cross product. We have i j k −→ −→ P Q× P R = −2 2 0 = (4−0)i+(0−(−4))j+(0−(−4))k = h4, 4, 4i . −2 0 2 Since M contains (2, 0, 0), we can write its equation as 4(x − 2) + 4y + 4z = 0, or 4x + 4y + 4z = 8, or x + y + z = 2. (b) What is the distance from the origin to M ? Solution: The distance from the origin O to M is the length of the −→ −→ −→ projection of P O onto the normal vector n = P Q × P R. This length is −→ | − 2 · 4 + 0 · 4 + 0 · 4| 8 2 |P O · n| √ = = √ =√ . |n| 42 + 42 + 42 4 3 3 4 5 (10 pts.) Find a parametric representation for the line which is the intersection of the two planes x + 2z = 1 and x + y − z = 0. Solution: We rst nd a point which is in both planes by solving the equations simultaneously. The rst equation gives x = 1 − 2z , so if we substitute into the second we get (1 − 2z) + y − z = 0, or y = 3z − 1. If we put z = 0 (for example), we get x = 1 and y = −1, so the point (1, −1, 0) lies on both planes. Next, the planes have normal vectors h1, 0, 2i and h1, 1, −1i, so the line of intersection is perpendicular to both vectors, which means it is parallel to i j k h1, 0, 2i × h1, 1, −1i = 1 0 2 1 1 −1 = (0 · (−1) − 2 · 1)i + (2 · 1 − 1 · (−1))j + (1 · 1 − 0 · 1)k = −2i + 3j + k. So a parametric representation for the line is r(t) = h1, −1, 0i + t h−2, 3, 1i , or x = 1 − 2t, y = −1 + 3t, z = t. 5 6 (10 pts.) A particle moves with position at time t given by r(t) = t2 i + (cos 3t)j + (sin 3t)k. Find the velocity, speed, and direction of the particle at all times t. Solution: The velocity v(t) is r0 (t) = 2ti − 3(sin 3t)j + 3(cos 3t)k. The speed is |v(t)| = v |v(t)| 7 (15 pts.) = √ 2t i 4t2 +9 − √ 4t2 + 9 sin2 3t + 9 cos2 3t = 3(sin 3t) √ j 4t2 +9 + √ 4t2 + 9. The direction is 3(cos 3t) √ k. 4t2 +9 Two projectiles are red at the same time. One is red from (1, 0, 0) with initial velocity h5, 10, 15i, while the other is red from (0, 1, 0) with initial velocity h10, 10, 10i. At what time are the projectiles closest together? What is their closest distance? Solution: The motion equations for the two projectiles are 1 1 r1 (t) = i + t(5i + 10j + 15k) − gt2 k = (1 + 5t)i + 10tj + (15t − gt2 )k 2 2 and 1 1 r2 (t) = j + t(10i + 10j + 10k) − gt2 k = 10ti + (1 + 10t)j + (10t − gt2 )k, 2 2 and their distance at time t is |r1 −r2 | = |(1−5t)i+(−1)j+(5t)k| = p (1 − 5t)2 + (−1)2 + (5t)2 = √ 2 − 10t + 50t2 . This function is minimized when its square 2 − 10t + 50t2 is minimized, which (taking the derivative) is when −10 + 100t = 0, or t = 1 10 . At this time, the distance between the projectiles is √ r 2 − 10t + 50t2 = 1 1 2 − 10( ) + 50( )= 10 100 6 r 1 2−1+ = 2 r 3 . 2 ...
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