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23. Disolacement Problem a. d'(t) = 70 - 9.8t d(t)=70t-4'912+C b.d(0) =6 =+ C=6 .'. d(t) = 70t - 4.9t2 + 6 c. d(5) = 233.5 m d(6) = 249.6 m d(9) = 239.1 m These three numbers show that d(t) has a high point somewhere between 5 and 9. d. The high point occurs when d'(t) = 0. 70 -9.8t = 0 + I =70/9.8 =7.1428'.. d(7.1428...) = 256 The arrow was highest, 256 m, at about t = 7.1 sec. 24. Acceleration Problem a. v'(t) = 18 sin 3t v(t) = -6 cos 3t + C b.v(0)=-6 =+ -6=-6cos0+C + C=0 "' v(t) = -6 cos 3t This negative velocity means that Calvin was swinging backwards at t = 0. c. t v(t) 0-6 0.2 -5 o.4 -2.2 0.6 1.4 0.8 4.4 1 5.9 1.2 5.4 1.4 2.9 1.6 -0.5 1.8 -3.8 2 -5.8 d. The maximum velocity is 6 fVsec at t = n/3 and other times. The acceleration at this time is zero. 25. Derivative and Antiderivative Problem a. g'(x) = 0.6x 9(x)=0'3x2+C b. i. g(x) = 0.3x2 ii. g(x) = 0.3x2 + 3 iii. g(x) = 0.3x2 + 5 c. Graph. Allthe graphs are obtained by shifting the solution function vertically through some displacement C. So all the graphs are "related" and thus can be called a family. Review Problems RO. Students should realize that the most important idea in Chapter 3 is the ability to lind exactly,by algebra, the derivatives they had been able to find only approximately before. They should realize that the concept of limit has been applied to do this. Consequently, they can put a "check" in the "apply limits" box in the 4-by-4 matrix of concepts and techniques. By working the real-world problems that have appeared in most sections, students should realize that the algebraic techniques just provide another way to carry out the applications of derivatives they were doing graphically and numerically in the first two chapters. Rl. a. the numerical derivative l(2) = 12. b. m(x) =#; m(2) takes the indeterminate form 0/0. By tracing the graph or constructing a table, the limit of m(x) seems to be 12 as x approaches 2. c. m(x) =x2 + 2x+ 4. lim m(x) -22 +2.2+4=12. x+2 d. The answer to part c. is exactly 12. The answer to part a. is approximately 12. R2. a. f'(c) =,1*9={d b.f(x)=0'4x2-x+-5 f'(3) = lim_ x+3 ,,_ E-3X!.4I1 !2) =llm-.---...--.-.--- x-+3 X - it = lim (0.4x + 0.2\ - 1.4 x+3 ' 0.4x2-x+0.6 c. m(x) = -3 o.4x2 - x+5-5.6 x-3 . Graph lhoblem Set 3-10 Solutions Monuol 4t

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d(7.1428...) = 256 The arrow was highest, 256 m, at about b.d(0) =6 =+ C=6 .'. d(t) = 70t - 4.9t2 + 6 solution function vertically through some displacement C. So all the graphs are "related" and thus can be called a family. 24. Acceleration Problem a. v'(t) = 18 sin 3t = lim (0.4x + 0.2\ - 1.4 d(t)=70t-4'912+C m(2) takes the indeterminate form 0/0. By tracing the graph or constructing a table, the limit of m(x) seems to be 12 as x approaches 2. c. m(x) =x2 + 2x+ 4. v(t) = -6 cos 3t + C -5.8

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23. Disolacement Problema. d'(t) = 70 - 9.8t

d(t)=70t-4'912+Cb.d(0) =6 =+ C=6

.'. d(t) = 70t - 4.9t2 + 6c. d(5) = 233.5 m

d(6) = 249.6 md(9) = 239.1 mThese three numbers show that d(t) has a high pointsomewhere between 5 and 9.

d. The high point occurs when d'(t) = 0.70 -9.8t = 0 + I =70/9.8 =7.1428'..d(7.1428...) = 256The arrow was highest, 256 m, at aboutt = 7.1 sec.

24. Acceleration Problema. v'(t) = 18 sin 3t

v(t) = -6 cos 3t + Cb.v(0)=-6 =+ -6=-6cos0+C + C=0

"' v(t) = -6 cos 3t

This negative velocity means that Calvin wasswinging backwards at t = 0.

c. t v(t)0-60.2 -5o.4 -2.20.6 1.40.8 4.41 5.91.2 5.41.4 2.91.6 -0.51.8 -3.82 -5.8

d. The maximum velocity is 6 fVsec at t = n/3 and othertimes. The acceleration at this time is zero.

25. Derivative and Antiderivative Problema. g'(x) = 0.6x

9(x)=0'3x2+Cb. i. g(x) = 0.3x2

ii. g(x) = 0.3x2 + 3iii. g(x) = 0.3x2 + 5

c. Graph. Allthe graphs are obtained by shifting thesolution function vertically through somedisplacement C. So all the graphs are "related" andthus can be called a family.

Review ProblemsRO. Students should realize that the most important idea

in Chapter 3 is the ability to lind exactly,by algebra,the derivatives they had been able to find onlyapproximately before. They should realize that theconcept of limit has been applied to do this.Consequently, they can put a "check" in the "applylimits" box in the 4-by-4 matrix of concepts andtechniques. By working the real-world problems thathave appeared in most sections, students shouldrealize that the algebraic techniques just provideanother way to carry out the applications ofderivatives they were doing graphically andnumerically in the first two chapters.

Rl. a. the numerical derivative l(2) = 12.

b. m(x) =#;m(2) takes the indeterminate form 0/0.By tracing the graph or constructing a table, the limitof m(x) seems to be 12 as x approaches 2.

c. m(x) =x2 + 2x+ 4.lim m(x) -22 +2.2+4=12.

x+2d. The answer to part c. is exactly 12.

The answer to part a. is approximately 12.

R2. a. f'(c) =,1*9={db.f(x)=0'4x2-x+-5

f'(3) = lim_x+3

,,_ E-3X!.4I1 !2)=llm-.---...--.-.---x-+3 X - it

= lim (0.4x + 0.2\ - 1.4x+3 '

0.4x2-x+0.6c. m(x) = -3

o.4x2 - x+5-5.6x-3

. Graph