PHYSICS
105Sol2

# 105Sol2 - Physics 105 Solutions to Problem Set 2 By Stephen...

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Physics 105 Solutions to Problem Set 2 By Stephen Glassman, Fall 2007 2.8 As suggested, we begin by writing Newton’s second law as mdv/F ( v ) = dt and integrate both sides from time t = 0 when v = v 0 to some arbitrary time t = T when v = v ( T ). v ( T ) v 0 mdv - cv 3 / 2 = T 0 dt 2 m cv 1 / 2 v ( T ) v 0 = T 2 m cv ( T ) 1 / 2 - 2 m cv 1 / 2 0 = 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 0 - 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 0 , x 0 to some final state v , x : x x 0 2 m F ( x ) dx = v v 0 dv 2 = 2 m x x 0 F ( x ) dx = v 2 - v 2 0 For F ( x ) = F 0 , we can write this expression as F 0 Δ x = 1 2 mv 2 - 1 2 mv 2 0 = Δ KE . 1

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2.18 Unfortunately, Taylor used the variable x for the base point in his expression for the Taylor series. I’m going to take it back and express Taylor’s theorem as: f ( a + δ ) = f ( a ) + f ( a ) δ + 1 2! f ( a ) δ 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 δ . 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 ( δ ) = n =0 f ( n ) (0) n ! δ 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. Then f ( n ) (0) = ( - 1) n - 1 ( n - 1)! and: f ( δ ) = n =0 f ( n ) (0) n ! δ n = f (0) + n =1 f ( n ) (0) n ! δ n = 0 + n =1 ( - 1) n - 1 ( n - 1)! n ! δ n = - n =1 ( - 1) n n δ n by which I mean John Taylor, the author of the textbook. This is not to be confused with subsequent uses of “Taylor” all of which refer to Brook Taylor of Taylor series fame. Ug!
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