Log in
HomeStudy Guides420,000US230,000

MATH 2214 Midterm: MATH 2214 Virginia Tech 4.2 Exam

39 views3 pages
turquoisegnu128
31 Jan 2019
School
Virginia Tech
Department
Mathematics
Course
MATH 2214
Professor
All

For unlimited access to Study Guides, a Grade+ subscription is required.

Document Summary

Solutions 3. 10: a) the characteristic equation is r2 + 2. 0 = 0 with roots r = i 0, so: (i) we have yc = c1 cos( 0t) + c2 sin( 0t). yp = a cos( t) + b sin( t), p = a 2 cos( t) b 2 sin( t), y . 0 2) sin( t) = f cos( t), p + 2 y . 0yp = a( 2 hence a = f/( 2 (ii) we have. 0 2), b = 0. yp = at cos( 0t) + bt sin( 0t), y p = 0at sin( 0t) + 0bt cos( 0t) + a cos( 0t) + b sin( 0t), 0at cos( 0t) 2 p = 2 y p + 2 y . 0bt sin( 0t) 2a 0 sin( 0t) + 2b cos( 0t), 0y = 2a 0 sin( 0t) + 2b cos( 0t) = f cos( 0t), hence a = 0, b = f/(2 0): the spring constant is.

Get access

Grade+
$40 USD/m
Billed monthly
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
10 Verified Answers

Related textbook solutions

Calculus
4 Edition,
Rogawski
ISBN: 9781319050733
Single Variable Calculus: Early Transcendentals
4th Edition, 2018
Stewart
ISBN: 9781337687805
CALCULUS:EARLY TRANSCENDENTALS
4 Edition,
Rogawski
ISBN: 9781319050740

Related Questions

[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

question 23 please

[8] 9. Either prove the statement is true or show that it is false. (a) sin (sin(2)) = 27. Solution: It is FALSE. The range of sin-' r is (-1,1], hence sin (sin(27)) = sin (0) = 0 (b) If f is continuous at a, then f is continuous at a. Solution: It is TRUE. Let g(x) = \f()]. If lim f(x) = f(a), then we have since the absolute value function is continuous lim g(x) = lim f() = lim f(x) = f(a) = g(a) So, g = \f is continuous at a. (c) If [g] is continuous at a, then g is continuous at a. -1 if r <0 Solution: It is FALSE. Let g(x) = 3 · Then, g = 1, so g is continuous at 0 i if >0 by the Continuity Theorems, but lim g(t) = -1 + lim g(1) 240- 1- 0+ So, g(x) is not continuous at a. (d) If f is continuous at a, then f is differentiable at a. Solution: It is FALSE. We know that f(1) = 2 is continuous at 0, but is not differentiable at 0.