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Chapter 3 52

# Chapter 3 52 - 202 D CHAPTER3 DIFFERENTIATION RULES 40 f:c...

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Unformatted text preview: 202 D CHAPTER3 DIFFERENTIATION RULES 40. f(:c) = ace—I => f’(a:) = m(—e_z) + e” : (1 — we” => f”(\$)— — (1 - HEX—e”) + 8”(—1) = (x A 2)6"” => f’"(\$)= (90- 2)(- e )+ 6” = (3 - sole“ => f(4’(x )= (3 x)(— ix) + e‘w(-1)=(Uv - 4).:‘1 ﬁ => WWI): (1)"(96 - 706"”- So 01000;“: = (a: — 1000)e—I. 41. By measuring the slope of the graph of s 2 f(t) at t = 0, 1, 2, 3, 4, and 5, and using the method of Example 1 in Section 2.9, we plot the graph of the velocity function 12 : f’ (t) in the ﬁrst ﬁgure. The acceleration when t = 2 s is a = f"(2), the slope of the tangent line to the graph of f’ when t : 2. We estimate the slope of this tangent line to be a(2) : f”(2) : 1/(2) as 331 = 9 ft/sz. Similar measurements enable us to graph the acceleration function in the second ﬁgure. 50 25 0 10 20 t (b) Drawing a tangent line at t = 10 on the graph of a, (1 appears to decrease by 10 ft/s2 over a period of 20 s. So at t = 10 s, the jerk is approximately —10/20 = #06 (ft/\$2)/s or ft/s3. 43. (a) 5 = 2t3 — 1512 +361: + 2 9 11(1) # s'(t) _ 6152 301 1 36 —> a(t) — 1/(t) — 1:2: — 30 (b) (1(1) 2 12- 1 — 30 : —18 m/s2 (0) ya) : 6(t2 — 515 + 6): 6(t — 2)(t — 3) : 0 when t = 2 0r3 and (1(2) : 24 — 30 = —6 m/s2, a(3) = 36 — 30 : 6 m/s2. 44. (a) s g 2:3 3t2 I 12t > v(t) s'(t) 6t2 6t 12 —> a(t) - v’(t) # 12t — 6 (b) a(1):12-1—6:6m/s2 (c) 12(t) = 6(t2 — t — 2) = 605 + 1)(t — 2) = OWhent : —1 or2. Sincet 2 0,15 7a —1 and (1(2): 24 — 6 : 18 m/sz. 45. (a) s— — sin(% 75%)+cos( t,) 0 < t < 2.1)(15) — s '(t) # cos“: t) - % sith) - g — %[cos(%t) — sin(%t)] : a(t) = m) : gI—sin(%t) . g #cos(%t) . g] = #% [sin(%t) +cos(%t)] 2 (b) a(1) = —§—: [sin(% - 1) +cos(% . 1)] : —1- [- + 72—] = -% (1 + \/§) x —0.3745 m/s2 ...
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