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Chapter 3 17

# Chapter 3 17 - SECTION 3.3 RATES OF CHANGE IN THE NATURAL...

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Unformatted text preview: SECTION 3.3 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES 167 (b) 11(3) : ﬂax—3) = —15 x/gft/S (c)Itisatrestwhenv:0 c) t: lsor6s. (d) It moves in the positive direction when 1) > 0 4: (t — 1)(t — 6) > 0 4:) 0 S t < 1 or t > 6. (e) Distance in positive direction — |s(1) s(0)l : |s(8) 3(6)] — |58 — 0| + |4\/_ — (712x/6H :58+4\/2+12t/5z93.05ft Distance in negative direction 2 |s(6) — s(1)| = |—12 x/é — 58] = 58 +12 t/6 m 87.39 ft Total distance traveled = 58 + 4 ﬂ +12 s/é + 58 +12 s/6 = 116 + 4 \/§ + 24 t/é 5: 180.44 ft f 1:8, ( ) s: 4J2 z 5.6 t=6, 7 s=~12\/6{ z—29.4 t= , t=0. s=58 s=0 .— —> 0 S 7. s(t) — t3 4.5it2 7t 11(t) s’(t) 31:2 9t 7 ﬂ 5 4:» 3t2 — 9t — 12 z 0 4:) 3(t — 4)(t +1) : 0 4:) t: 4 or —1. Sincet 2 0, the particle reaches a velocity 0f5 m/s att = 4 s. 8. (a)s=5t+3t2 => v(t)= E :5+6t,sov(2)=5+6(2)=17m/s. dt (b)v(t)—35 —> 5l6t—35 6t—30 t45s. 2 dh 9. (a) h = 10t— 0.83t :> 11(15): E : 10 e 1.6615. so 11(3) = 10 —1.66(3): 5.02 m/s. 10::Vl7 (b)h—25 —> 1015 0.83t2_25 .> 0.83t2—10t+25:0 => kw The value t1 2 (10 — \/ 17) / 1.66 corresponds to the time it takes for the stone to rise 25 m and z 3.54 or 8.51. t2 : (10 + \/17)/1.66 corresponds to the time when the stone is 25 m high on the way down. Thus, v(t1) = 10 — 1.66[(10 — \/1_7)/1.66] : t/ﬁ as 4.12 m/s. 10. (a) At maximum height the velocity of the ball is 0 ft/s. v(t) — s’(t) _ 80 3215 — 0 <4 32t — 80 (I) t: go So the maximum height is 3(3) : 80(3) — 16(§)2 : 200 — 100 = 100 ft. (b) an) : 80t—16t2 ~ 96 16t2 8075 l 96 4 0 e, 16(t2 ~ 5t + 6) : 0 e» 16(t—3)(t~2) : 0. So the ball has a height of 96 ft on the way up at t : 2 and on the way down at t = 3. At these time velocities are 11(2) 2 80 — 32(2) 2 16 ft/s and 21(3) : 80 ~ 32(3) 2 —16 ft/s, respectively. 8 the 11. (a) A(\$) : .732 => A’(av) = 2.7:.A’(15): 30 mmZ/mm is the rate at which the area is increasing with respect to the side length as m reaches 15 mm. (b) The perimeter is P(m) : 43:, so A’(:c) : 2x = 5(4x) : %P(x). The ﬁgure suggests that if Act is small, then the change in the area of the square is approximately half of its perimeter (2 of the 4 sides) times Any, From the ﬁgure, AA = 2:0(Am) + (Ace)? If Am is small, then AA 2 293(Ax) and so AA/Am % 2x. ...
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