Q1. f'(x) = 9xBQ3. yi= 72x5(5xo + 1 1)1'4q5. 16;1 = 12Q7. Yes (continuous)Q9. 1

Q2. dy/dx = -3 sin x04. s'= 0Q6. Limit = 1

Q8. f(x) = -cos X2

Q10. Graph

1. Ferris Wheel Problema. Graph

y(t)=C+AcosB(t-D)Vertical disPlacement = 25 = CAmplitude = 0.5(40) - 20 = APhase disp. (for cosine) = 3 = DPeriod = 60/3 = 20, so B = 2trJ2O = nl10(Note that B is the angular velocity in radians persecond.)

y(r) = 25 + 20 cos ft t, -.1

b.y'(t)=-2nsinftA-tl

c. y'(15) = -2rc sin ft ttu - 3) = 3.69316...

y(t) is increasing at about 3.7 fVsec.d. The fastest that y(t) changes is 2n, or 6.28... fVsec'

The seat is at y(t) = 25 ft above the ground then'

2. Pendulum Problema.y=C+AcosB(x -D).

B - 2ril6 = ru/3 rad/secD = phase displacement = 1.3 secA = 0.5(110 - 50) = 30 cm.C = 110 - 30 or 50 + 30, which equals 80 cm.

... d = 80 + eo cos $(t - 1.3)

g. 6' = -10n sin f;tt - r .s)

c. d'(5) = -1Orc sin {ts - 1.3) = 21.02195...

d'(11) = -10r sin flt t - 1.3) = 21.02135...

At both times the pendulum is moving away from thewall at about 21.0 cm/sec. The answers are thesame because the times are exactly one periodapart.

d. d'(20) = -10n sin $tzo - 1.3) = -21.o21s5-.

The pendulum is moving toward the wall. Since thederivative is negative, d is decreasing, which in thisproblem implies motion toward the wall.

e. The fastest is 10r = 31.4 cm/sec, when d = 80.

f. 0 =-10nsin|6- 1.3) + sin*(t- 1.3) = o +=f;tt-r.s)='sin-10 = 3c-is)=o+rn =t-1.3=3n + t=1.3+3n.The first positive time occurs when n = 0. When thevelocity is 0, the pendulum is at its maximum heightt = l*3_SCq.

3. Plavoround Problema. Curb has slope (3.25 - 0.75)144 = 2.5144.

.'. equation is f(x) = 0.75 + (2.5144\xb. Sinusoid has period 8 ft, so B - 2nl8 - rl4.

Amplitude = 0.5(0.75 - 0.25) = 0.25 ftLow end of ramp is a low point on the sinusoid..'. sinusoidal axis is at y = 0.25 when x = 0, and goesup with slope 2.5/44.Sinusoid is at a low point when x = 0. So phasedisplacement is 0 if the cosine is subtracted..'. equation isg(x) = 0.25 *fu. -0.25 cos f (x)

(There are other correct forms.)2.5 r. ft

c. S'(x) = 7a + 1f, sin f (x)

g'(s) = ff+ftsinf 1s1 = 0.1956... ft/ftGoing qp at about 0.2 vertical ft per horizontal ft

g'(r s) = ?f + ft sin f (r s) = -0.0820... fVft

Going down at about 0.08 vert. ft per horiz. ftNegative derivative implies g(x) is getting smallerand thus child is going down, and vice versa.

d. By tracing the g' graph, maximum value of g'(x) is0.2531... fUft (about 14.2' uP).Minimum is -0.1395... fUft (about 7.9" down).

4. Davlioht Problema. Let d = day number and L(d) = no. of minutes.

14 hours 3 minutes is 843 minutes.10 hours 15 minutes is 615 minutes..'. Amplitude = (1/2)(843 - 615) = 114 minutes.Sinusoidal axis is at L(d) = 615 + 114 =729 minAssuming a 365-day year, B =2n1365.Phase displacement = 172.

... L(d)= 729+1r+cosft $-fi21On August 7, d =219.L(219) = 729 + 1r+ cos ft eE - fi2)= 807.67...

or about 13 hours 28 minutes.

b. L'(d) =-'# sinfr @-nz)On Aug.7,d=219.

L'(21e) = -'# "in ft erc - fiz) = -1.42oos... .

Rate = -1.42 minlday.(Decreasing at about 1.42 min/day.)

c. The greatest rate occurs when the sine is 1 or -1 .

Rate is 228rd365 = 1.96 min/day.1/4 year is about 91 days. So greatest rate occurs atday 172 + 91, which is day 263 or day 81 (September20 or March 22).

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38 Colculus: Concepts ond Applicolions Problem Set 3'8