HS-MATH-CSSM5_--_Chapter_3-_Differentiation_Rules.pdf

A 2 kt 25 a c k 1 at dc akt1a 2 k a 2 ktak a 2 kakt1

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a 2 kt 25. (a) [C] = -k 1 ::::} at+ . d[C] (akt+1)(a 2 k)-(a 2 kt)(ak) a 2 k(akt+1-akt) a 2 k rate of reaction= dt = (akt + 1 )2 = (akt + 1 )2 = (akt + 1 )2 a 2 kt a 2 kt +a- a 2 kt __ a_ (b) If x = [ C], then a - x = a - akt + 1 = akt + 1 - akt + 1 · ( a ) 2 _ a 2 k = d [C] [from part (a)] So k(a- x) 2 = k akt + 1 - (akt + 1)2 dt dx dt' a2kt (a2kt)jt - a2k ---+ a2k =amoles/L. (c) As t---+ oo, [C] = akt + 1 = (akt + 1)/t - ak + (1/t) ak d[C] a 2 k (d) As t---+ oo, dt = (akt + 1)2 ---+ 0. (e) As t increases, nearly all of the reactants A and Bare converted into product C. In practical terms, the reaction virtually stops. 27. (a) Using v = pl ( R 2 - r 2 ) with R = 0.01, l = 3, P = 3000, and"' = 0.027, we have v as a function of r: 4ry 3000 ( 2 2) - - v(r) = 4 ( 0 . 027 ) 3 0.01 - r . v(O) = 0.925 cmjs, v(0.005) = 0.694 cmjs, v(0.01) = 0. P P Pr (b) v(r) = -l (R 2 - r 2 ) ::::} v'(r) = -l ( -2r) = --l. When l = 3, P = 3000, and"'= 0.027, we have 4ry 4ry 2ry 3000r - , - v'(r) =- 2 ( 0 . 027 ) 3 . v'(O) = 0, v'(0.005) = -92.592 (cmjs)jcm, and v (0.01) = -185 .185 (cm/s)/cm. (c) The velocity is greatest where r = 0 (at the center) and the velocity is changing most where r = R = 0.01 em (at the edge). 29. (a) C(x) = 2000 + 3x + O.Olx 2 + 0.0002x 3 ::::} C'(x) = 3 + 0.02x + 0.0006x 2 (b) C' (100) = 3 + 0.02(100) + 0.0006(10,000) = 3 + 2 + 6 = $11/pair. C' (100) is the rate at which the cost is increasing as the 100th pair of jeans is produced. It predicts the cost of the 101st pair. (c) The cost of manufacturing the 101st pair of jeans is C(101) - C(100) = (2000 + 303 + 102.01 + 206.0602) - (2000 + 300 + 100 + 200) = 11.0702 ~ $11.07 31. (a) A(x) = p(x) ::::} A'(x) = xp'(x)- p(x) · 1 = xp'(x)- p(x). A'(x) > 0 ::::} A(x) is increasing; that x x 2 x 2 is, the average productivity increases as the size of the workforce increases. (b) p' ( x) is greater than the average productivity ::::} p' ( x) > A( x) ::::} p' ( x) > p( x) ::::} X xp'(x) > p(x) ::::} xp'(x)- p(x) > 0 ::::} xp'(x)_-; p(x) > 0 ::::} A'(x) > 0.
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88 D CHAPTER 3 DIFFERENTIATION RULES PV PV 1 . 33. PV = nRT =* T = nR = (l0)(0.0 821 ) = 0. 821 (PV). Usmg the Product Rule, we have ~~ = 0.~ 21 [P(t)V 1 (t) + V(t)P 1 (t)] = 0.~ 21 [(8)(-0.15) + (10)(0.10)] ~ -0.2436 K/min. 35. (a) If the populations are stable, then the growth rates are neither positive nor negative; that is, dC dW dt = 0 and dt = 0. (b) "The caribou go extinct" means that the population is zero, or mathematically, C = 0. . dC dW (c) We have the equatiOns dt = aC- bCW and dt = -cW + dCW. Let dCjdt = dW/dt = 0, a= 0.05, b = 0.001, c = 0.05, and d = 0.0001 to obtain 0.05C - 0.001CW = 0 (1) and -0.05W + O.OOOlCW = 0 (2). Adding 10 times (2) to (1) eliminates the CW-terms and gives us 0.05C- 0.5W = 0 =* C = lOW. Substituting C = lOW into (1) results in 0.05(10W) - 0.001(10W)W = 0 {::} 0.5W- 0.01 W 2 = 0 {::} 50W- W 2 = 0 {::} W(50- W) = 0 {::} W = 0 or 50. Since C =lOW, C = 0 or 500. Thus, the population pairs (C, W) that lead to stable populations are (0, 0) and (500, 50). So it is possible for the two species to live in harmony. 3.4 Derivatives of Trigonometric Functions 1. j (X) = X - 3 sin X =* !' (X) = 1 - 3 COS X 3. y = sinx + lOtanx =* y 1 = cosx + 10sec 2 x 5. g( t) = t 3 cost =* g 1 ( t) = t 3 (- sin t) + (cost) · 3t 2 = 3t 2 cost - t 3 sin t or t 2 ( 3 cost - t sin t) 7. h(B) =esc(}+ e 9 cot(} =* h 1 ( 8) = -esc(} cot(} + e 9 (- csc 2 8) + (cot 8)e 9 = -esc(} cot(} + e 9 (cot(} - csc 2 8) 9 _ X 1 _ (COS X) ( 1) - (X) (- sin X) _ COS X + X sin X .y---=*Y- ------ cosx (cosx) 2 cos 2 x 11. f(B) = sec(} =* 1 +sec(} f'(B) = (1 + sec8)(sec8tan8)- (sec8)(sec8tan8) = (sec8tan8)[(1 +sec B)- sec8] = sec8tan8 (1 +sec 8) 2 (1 +sec 8) 2 (1 +sec 8) 2 13. y = sinx x2 1 x 2 cosx- (sinx)(2x) x(xcosx- 2sinx) xcosx- 2sinx =* y = = = -----,,---- (x2)2 x4 x3 15. y =sec(} tan(} =* y 1 =sec(} (sec 2 8) +tan(} (sec(} tan 8) =sec(} (sec 2 (} + tan 2 8) Using the identity 1 + tan 2 (} = sec 2 8, we can write alternative forms of the answer as sec (} ( 1 + 2 tan 2 (}) or sec (} ( 2 sec 2 (} - 1) 17 .
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