Math 1C Bria Osserma Practice Exam 1 (15 pts.) Determie the radius ad iterval of covergece of the power series (x ) +1. First we use the root test to determie for which values of x the series coverges absolutely. The absolute value series is a = x +1. lim a = lim x +1 x = lim x =. By the root test, the series a coverges if x < 1 or equivaletly x <, ad diverges if x > 1 or equivaletly x >. Thus the radius of covergece of the power series is. To d the iterval of covergece we have to test the edpoits, which are x = 1, 5. At x = 1, the series is ( ) +1 = ( 1), which diverges because the terms do't go to 0 as goes to iity. At x = 5, the series is = +1, which also diverges because the terms do't go to 0. covergece is ( 1, 5). So the iterval of
(0 pts.) (a) Fid the Taylor series for si x at x = π. What is the radius ad iterval of covergece? The derivatives of si x cycle betwee cos x, si x, cos x, ad si x. If they are ± cos x, they vaish at x = π. If they are ± si x, they are ±1 at x = π. So the Taylor series is 1 1 (x π ) + 1 4! (x π )4 + + ( 1) 1 ()! (x π ) +.... We use the ratio test to d for which x the series coverges absolutely. 1 a = ()! x π, ad so a +1 a = x π + ()! x π + ( + )! = x π ( + )( + 1), a +1 lim a = lim x π ( + )( + 1) = x π lim 1 ( + )( + 1) = 0. Thus, by the ratio test the series coverges absolutely for all x (the radius of covergece is iite). (b) If you approximate si x usig the Taylor polyomial of order 4 cetered at x = π, what is a (reasoable) boud for the error whe x π < 1? The fth derivative of si x is cos x, ad cos x 1 for all x. Thus the error of the approximatio at x = b is at most 1 5! b π 5. If we wat a error boud which holds for all x with x π < 1, we d the error is at most 1 5! = 1 10. Note: this ca be improved by observig that the Taylor polyomial of degree 4 for si x cetered at x = π is also the Taylor polyomial of degree 5, so we ca similarly obtai a sharper error boud of 1 6! x π 6 < 1. 6!
(15 pts.) Let u = 0, 1, 1, v = 1, 1, 0. (a) Usig projectio, write u as a sum of a vector parallel to v ad a vector orthogoal to v. Note that u = proj v u + (u proj v u), ad by deitio proj v u is parallel to v ad u proj v u is orthogoal to v. Now, proj v u = u v v v = 0 1 + 1 1 + 1 0 1 + 1 + 0 1, 1, 0 = 1 1, 1, 0 1 =, 1, 0. The u proj v u = 0, 1, 1 1, 1, 0 = 1, 1, 1, so we have 1 u =, 1, 0 + 1, 1, 1, where the rst is parallel to v ad the secod is orthogoal to it. (b) What is the agle betwee u ad v? The agle is cos 1 ( u v u v ) = cos 1 ( 1 0 +1 +1 ) = cos 1 ( 1 ) = 60.
4 (15 pts.) Let M be the plae passig through the three poits P = (, 0, 0), Q = (0,, 0), ad R = (0, 0, ). (a) What is the equatio for M? The vectors P Q =,, 0 ad P R =, 0, both lie o the plae, so we ca d a ormal vector by takig their cross product. We have i j k P Q P R = 0 = (4 0)i+(0 ( 4))j+(0 ( 4))k = 4, 4, 4. 0 Sice M cotais (, 0, 0), we ca write its equatio as 4(x ) + 4y + 4z = 0, or 4x + 4y + 4z = 8, or x + y + z =. (b) What is the distace from the origi to M? The distace from the origi O to M is the legth of the projectio of P O oto the ormal vector = P Q P R. This legth is P O = 4 + 0 4 + 0 4 4 + 4 + 4 = 8 4 =. 4
5 (10 pts.) Fid a parametric represetatio for the lie which is the itersectio of the two plaes x + z = 1 ad x + y z = 0. We rst d a poit which is i both plaes by solvig the equatios simultaeously. The rst equatio gives x = 1 z, so if we substitute ito the secod we get (1 z) + y z = 0, or y = z 1. If we put z = 0 (for example), we get x = 1 ad y = 1, so the poit (1, 1, 0) lies o both plaes. Next, the plaes have ormal vectors 1, 0, ad 1, 1, 1, so the lie of itersectio is perpedicular to both vectors, which meas it is parallel to i j k 1, 0, 1, 1, 1 = 1 0 1 1 1 = (0 ( 1) 1)i + ( 1 1 ( 1))j + (1 1 0 1)k = i + j + k. So a parametric represetatio for the lie is r(t) = 1, 1, 0 + t,, 1, or x = 1 t, y = 1 + t, z = t. 5
6 (10 pts.) A particle moves with positio at time t give by r(t) = t i + (cos t)j + (si t)k. Fid the velocity, speed, ad directio of the particle at all times t. The velocity v(t) is r (t) = ti (si t)j + (cos t)k. The speed is v(t) = 4t + 9 si t + 9 cos t = 4t + 9. The directio is v = v(t) t 4t +9 (si t) (cos t) i 4t j + k. +9 4t +9 7 (15 pts.) Two projectiles are red at the same time. Oe is red from (1, 0, 0) with iitial velocity 5, 10, 15, while the other is red from (0, 1, 0) with iitial velocity 10, 10, 10. At what time are the projectiles closest together? What is their closest distace? The motio equatios for the two projectiles are r 1 (t) = i + t(5i + 10j + 15k) 1 gt k = (1 + 5t)i + 10tj + (15t 1 gt )k ad r (t) = j + t(10i + 10j + 10k) 1 gt k = 10ti + (1 + 10t)j + (10t 1 gt )k, ad their distace at time t is r 1 r = (1 5t)i+( 1)j+(5t)k = (1 5t) + ( 1) + (5t) = 10t + 50t. This fuctio is miimized whe its square 10t + 50t is miimized, which (takig the derivative) is whe 10 + 100t = 0, or t = 1. At this 10 time, the distace betwee the projectiles is 10t + 50t = 10( 1 10 ) + 50( 1 100 ) = 1 + 1 =. 6