Calculus

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  • Topic: IPv4, Classless Inter-Domain Routing, 1981
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SOLUTIONS TO SUGGESTED PROBLEMS FROM THE TEXT PART 2
3.5 2 3 4 6 15 18 28 34 36 42 43 44 48 49 3.6 1 2 6 12 17 19 23 30 31 34 38 40 43a 45 51 52 1 4 7 8 10 14 17 19 20 21 22 26 r’(θ) = cosθ – sinθ 2 2 cos θ – sin θ = cos2θ z’= -4sin(4θ) -3cos(2 – 3x) 2 cos(tanθ)/cos θ f’(x) = [-sin(sinx)](cosx) -sinθ w’ = (-cosθ)e y’ = cos(cosx + sinx)(cosx – sinx) 2 T’(θ) = -1 / sin θ x q(x) = e / sin x F(x) = -(1/4)cos(4x) (a) dy/dt = -(4.9π/6)sin(πt/6) (b) indicates the change in depth of water (a) Graph at end (b) Max on 1 July; 4500; yes; 1 Jan (c) pos 1 April; neg 1 Oct (d) 0 2 2 2 (a) a cosθ + √l – a sin θ (b) i: -2a cm/sec 2 2 2 ii: -a√2 – a / (√l – a /2 cm/sec

28 36 37 42 52a 52b 1 2 4b 5 8 13 17 26a 29 39 41 1 2 3 17 22 29 36 44a 46 49 2 5 8 10 14 16b 21 25 26 27 5.2 1 6 8 10 14

Sketch at end Sketch at the end
x = 0: not max/min x = 3/7: local max x = 1: local min

4.2

-1/3 g decreasing near x = x0 g has local min at x1 Sketch at end Sketch at end x = 4; y = 57 Max: 20 at x = 1 Min: -2 at x = -1; x = 8 Max: 2 at x = 0; x = 3 Min: 16 at x = -1; x = 2 (a) f(1) local min; f(0), f(2) local max (b) f(1) global min; f(2) global max

Global min = 2 at x = 1, No global max D=C r = 3B/(2A) Sketch at end Sketch at end. x = L/2 x = 2a Min: -2amps; Max: 2 amps (a) xy + πy /8 (b) x + y + πy/2 (c) x = 100/(4 + π); y = 200/(4 + π) 2

2t / (t + 1) 1 / (x – 1) cosα/sinα (lnx) + 1 e . 1/x 1 -sin (lnt) / t 2 2 / (√1 – 4t ) 1 / t lnt 2 1 / (1 + 2u + 2u ) 0.8 -1 ‹ x ‹ 1 1 / ((ln 10) x) y=x-1 (a) f’(x) = 0 (b) f is constant 225 -0.12 Critical points: x = -3and x = 2 F(-3) local max; f(2) local min Critical points: x = 0 and x = 2 No local max; f(2) local min

2

4.4

5.1

r = (4/π) ; h = 2(4/π) (±1,0), (±1, 1/2 ) (0.59, 0.35) (a + b)/2 x = 4/5 x = 1, P = (1,1) Upper = 408; Lower = 390 (a) Always same direction: speed up, then slow down. (b) Over = 15, Under = 6

1/3

1/3

4.1

82 meters 250 meters Total change = 15 cm to the right 14.5 miles 8 cm to the right 20 minutes (a) Car A Largest Max velocity; (b) Car A stops first (c) Car B travel farther (a) A: 8hrs; B: 4 hrs (b) 100 km/h (c) A: 400 km; B: 100 km

Local max x = 1 Local max x = -1 Sketch at end
(a) Inc x › 0; Dec x ‹ 0 (b) No global max; local & global min f(0) (a) Inc -1 ‹ x ‹ 0 and x › 1; Dec x ‹ -1 and 0 ‹ x ‹1 (b) local max f(0); local min f(-1) & f(1)

Sketch at end Sketch at end Sketch at end
Set derivative = 0. No critical point Second derivative -, graph concave down

(a) Left smaller (b) 0, 2, 6, 1/3 Best estimate = 93.47 543 Sum written out at the end (a) 13 (b) 1

Sketch at the end

5.2

5.3

24b 24c 31 33 35 36 1 2 4 6 9 10 14 15a 15b 16 25 29 30 31 35 43 2 3 12 13 14 17 21 23 24 28 29 30 34 35a 39a 48 50 1 6 9 14 18 25 28 36

46 118 (a) 13 (b) -2 (c) 11 (d) 15 (a) -2 (b) –A/2 Sketch at the end A = 2; b = 6; f(x) = 2 + 3(4/3) Dollars Meter per second Integral represent the change in position from t = 1 and t = 3: meter Integral represent the change in world’s population between 2000 and 2004: billions of people

40 43 50 54 61 64 71 76 79 6.3 1 2 3 5 7

x /4 – 2x + C 3 2 x /3 + 5x /2 + 8x + C 12 πx + x /12 + C 1.0986 1 21 e - 1 – sin 1 c=6 Explanation at end x /4 + 5x +C 4x + ln|x| + C 3/2 8t /3 + C Set y(π) = 0 in equation. Varify 10e + 15 2z – cos z + 6 Set y(0) = 2 in equation. Verify (a) x + x + C (b) Sketch at end 2 (c) x + x + 3

2

4

4

2 2202.5

2

t

Sketch at end
(a) estimate (b) 177.270 (c) Lower

9 10 11 17a 18 21 2 3 4 8 13 20

$300 000 12 newton.meter
Ave f = ¼; Ave g = ¼; Ave f.g = 0 Statement not true

h(t) = -16t + 40t + 30 2 19.55 fet/sec 5/6 miles Explanation at back t^2 (1/2)e + C 3x (1/3)e + C -(1/2)cos(2x) + C (1/33)(t – 3) + C -2√4 − + C x^3 + 1 (1/3)e +C 4 (sin α)/4 + C 3 (1/3)(ln z) + C 2 (1/6)ln(1 + 3t ) + C sin +C sin x e +C 2 (1/2)ln(x + 1) + C (1/6)tan 2x + C 1 – (1/e) 3 e(e – 1) 2/5 14/3 y = 3x 3 11

2

5.4

(a) 8.5 (b) 1.7 (b) 0.64 Description at end Table at end...
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