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

lhoblem Set 3-10 Solutions Monuol 4t