9. Velocitv from Disolacement Problemt

d'(1)=18-9.8=8.2d'(3)= 18-9.8'3 =-11.4d' is called velocity in physics.

b. At t = 1 the football is going up at 8.2 m/sec.At t = 3 the football is going down at 11.4 m/sec.The ball is going up when the derivative is positiveand coming down when lhe derivative is negative.The ball is going up when the graph slopes up andcoming down when the graph slopes down.

c. d'(4) - -21 .2, which suggests that the ball is goingdown at 41 .2 nlsec. However, d(4) = -6.4, *6;"6reveals that the ball has gone underground. Thefunction gives meaningful answers in the real worldonly if the domain of t is restricted to values thatmake d(t) nonnegative.

10. Disolacement from Velocity Problemv(t) = 1510'oSince v(t) = x'(t), x(t) must have had t1'6 in it. Thederivative of ti'6 can be assumed to be 1.6t0'6. So thecoefficient of t1'6 must be 15/1 .6, or 9.375. But x(0)was 50. Thus, x(t) = 9.375t1'6 + 50.The derivative x'(t) really does equal v(t).Using this equation,x(10) = 9.375(101 6) + 50 = 423.225... .

So the distance traveled is 423.225... - 50 =373.225..., or about 373 ft.The concept is antiderivative. (See Problem 39 inProblem Set 3-4.)By the trapezoidal rule, with n = 100 increments,distance = 373j62... = 373 it.Concept: definite integral.

1 1. Average Versus lnstantaneous Velocity ProblemThe average rate is defined to be the change in thedependent variable divided by the change in theindependent variable (such as total distance dividedby total time). Thus, the difference quotient is anaverage rate.The instantaneous rate is the limit of this average rateas the change in the independent variable approacheszero.

12. a. Graph ol velocity is the dashed line.

Problem Set 3-6. poge 106 lntroduction to Sine. Cosine. ond Composite Functions

1. Graph. f(x) = si1'1 1 2. f'(x) = cos x, so f(2) = 6ss 2 = -0.416... .

y2 - (cos 2)x + (sin 2- 2 cos 2) = -0.41...x + 1.74...Graph. The line appears to be tangent to the sine graphal x = 2. The sine function and the linear function havethe same slope at x = 2, which means they have thesame rate of change. The derivative is the rate ofchange of the function.

b. Y is a relative maximum when t = 0, 4,8,... sec.Y is a relative minimum when t = 2, 6, 10, ... sec.

c. The velocity is a relative maximum when is t = 3 secor 7 sec. The displacement graph at these timesappears to be increasing the fastest.

d. The equation used in the text isy=2+0.85tcosft

The student could observe that the period is 4,leading to the coefficienl ol n/2. The amplitudedecreases in a way that suggests an exponentialfunction with base close to, but less than, 1. Theadditive 2 raises the graph up two units, as can beascertained by the fact that the graph seems toconverge to 2 as t gets larger. The graph shownabove agrees with the numerical derivative. Notethat the actual maximum and minimum values occurslightly before the values of t read from the graph inpart (b). For instance, the maximum near t = 4 isaclually at t = 3.9346... .

Problem Set 3-6 Solufions Monuol 35