This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Physics 105 Solutions to Problem Set 2 By Stephen Glassman, Fall 2007 2.8 As suggested, we begin by writing Newtons second law as mdv/F ( v ) = dt and integrate both sides from time t = 0 when v = v to some arbitrary time t = T when v = v ( T ). Z v ( T ) v mdv- cv 3 / 2 = Z T dt 2 m cv 1 / 2 v ( T ) v = T 2 m cv ( T ) 1 / 2- 2 m cv 1 / 2 = T On the last line, we readily see that v ( T ) = 0 = T = . Therefore, our mass never comes to rest. Finally, we solve for v ( T ) and obtain: v ( T ) = cT 2 m + 1 v 1 / 2 !- 2 2.12 Consider a function v ( x ( t )). Then by the chain rule, dv dt = dv dx dx dt = dv dx v . Also by the chain rule, d dx v 2 = 2 v dv dx . Therefore, F = p = m v = mv dv dx = m 1 2 dv 2 dx . We separate variables to obtain 2 m Fdx = dv 2 . If we integrate both sides from some initial state v , x to some final state v , x : Z x x 2 m F ( x ) dx = Z v v dv 2 = 2 m Z x x F ( x ) dx = v 2- v 2 For F ( x ) = F , we can write this expression as F x = 1 2 mv 2- 1 2 mv 2 = KE . 1 2.18 Unfortunately, Taylor used the variable x for the base point in his expression for the Taylor series. Im going to take it back and express Taylors theorem as: f ( a + ) = f ( a ) + f ( a ) + 1 2! f 00 ( a ) 2 + where f ( x ) is some sufficiently reasonable function, so long as the right hand side converges. It should be noted that an expansion for a particular f ( x ) around a particular base point a will generally only converge for some interval in . I will use the notation f ( n ) ( x ) to denote the n th derivative of f ( x ). Of particular interest is the special case a = 0 for which the Taylor series simplifies to: f ( ) = X n =0 f ( n ) (0) n ! n (a) Let f ( x ) = ln(1 + x ). Claim: f ( n ) ( x ) = (- 1) n- 1 ( n- 1)!(1 + x )- n n 1 Proof: We proceed using mathematical induction. For n = 1, the expression eval- uates to f ( x ) = (1 + x )- 1 which is precisely the derivative of ln(1 + x ). Now we assume the claim for n k and show it for n = k + 1. f ( k ) ( x ) = (- 1) k- 1 ( k- 1)!(1 + x )- k = f ( k +1) ( x ) = (- k )(- 1) k- 1 ( k- 1)!(1 + x )- k- 1 = (- 1) k k !(1 + x )- ( k +1) which is precisely the claim for n = k + 1....
View Full Document
This note was uploaded on 04/01/2008 for the course PHYSICS 105 taught by Professor Edgarknobloch during the Fall '07 term at University of California, Berkeley.
- Fall '07