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

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

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

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

b.d(0)

.'. 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'..

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.

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

+

C=0

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

v(t) c. 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.

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)

R2. a. f'(c)

=,1*9={d

b.f(x)=0'4x2-x+-5 f'(3)

o.4x2

- x+5-5.6

lim_ = x+3 x-3 ,,_ E-3X!.4I1 !2) =llm-.---...--.-.--X - it x-+3 (0.4x lim + 0.2\ - 1.4 = x+3 '

c. m(x) =

0.4x2-x+0.6

-3

.

Graph

=#;

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.

lhoblem Set 3-10

Solutions Monuol

4t