20 Find the average value of the sine function from 0 to . p Solution: The average value of the function uses the definition of the average

Example 9-20 Find the average value of the sine function from 0 to . p Solution: The average value of the function uses the definition of the average

value of a function over a range (see Chapter 7, Integration). The integration is over the first half-cycle of the sine function as shown in Fig. 8-22. The average value of the function x ⫽ sin u from 0 to is p

x avg ⫽ 1 Z

sin udu

p⫺ 03 0

CALCULUS FOR THE UTTERLY CONFUSED

Carry out the integration to find the average value of the sine function over one half-cycle.

x Z avg ⫽1 ⫺ ⫽1 pS ⫺ cos u T

0 p[ cos p ⫺ (⫺cos 0)]

p[ (⫺1) ⫺ (⫺1)] ⫽ ⫽ p 0.64

Figure 8-23 shows the rectangle with

height 0.64 and base p with area

equal to the area under the first half- cycle of sin u .

pq Example 8-21 The power delivered

by a loudspeaker is P⫽P o sin 2 v t

Fig. 8-23

where P o is the peak power and is v

a constant with the units of recipro- cal time. What is the average power in terms of the peak power?

Solution: Start by graphing sin vt (Fig. 8-24). The v t is not important to the graph. When v t has gone from 0 to 2 , the sine p

function has gone through one cycle.

P 2 P o sin wt

2π w t

Fig. 8-24

Now graph the sin 2 curve. Look first at the range from 0 to . The p sin 2 curve starts at zero when v t⫽ 0 , and goes to 1 when v t⫽p /2 , and then back to zero again when v t⫽p .

Trigonometric Functions

The sin 2 curve, however, has a different shape from the sin curve. The unique shape of the sin 2 curve is due to the fact that when a number less than 1 is squared, the result is smaller [ 0.5 2 ⫽ 0.25 ]. The smaller the number, the smaller

the result on squaring [ 0.9 2 ⫽ 0.81 but 0.3 2 ⫽ 0.09 ]. When the sin curve is negative, the sin 2 curve is positive (see Fig. 8-24). The

sin 2 curve is periodic in so the average value of the p sin 2 curve is the average value between 0 and . p

The average value of this sin 2 -type function follows the definition of the aver- age value of the function.

avg ⫽

sin 2 v p⫺ td (vt)

The

1 sin 2 u du integral is (from the Mathematical Tables) 1 u⫺ 1 sin 2u so

The average power for the loudspeaker is one-half the peak power.

Example 8-22 What is the area bounded by y 1 ⫽ cos x , y 2 ⫽ sin x , and x⫽ 0 ? Solution: The sine and cosine functions are

shown in Fig. 8-25. The integral of the area

between the curves is in the x-direction and y has form

2 = sin x 1( cos x ⫺ sin x)dx

. The integral is

from x⫽ 0 to the intersection point of the

two curves. At this point y = cos x 1 or sin x ⫽ cos x sin x ⫽ cos x 1 or tan x ⫽ 1 x

From the graph of tan x , tan x ⫽ 1 when

x⫽p /4 . Check the number in your hand cal- culator. Take the inverse tangent of 1. Table 8-1

Fig. 8-25

also shows sin x equal to cos x at x⫽p /4 so the complete integral, complete with limits, is

p /4

A⫽ 3 ( cos x ⫺ sin x)dx

CALCULUS FOR THE UTTERLY CONFUSED

Example 8-23

A certain company is selling golf shirts with the college logo through college bookstores. Because of an expanding market overall sales are growing at 2t thousands of dollars per month, t. Superimposed over this linear growth is a sinusoidal variation following the academic year and described by

(0.1) cos 2p 12 in thousands of dollars and with t⫽ 0 corresponding to January. Determine the rate of sales at the end of March (t ⫽ 3) .

Solution: The equation describing sales is

S⫽ 2t ⫹ 0.1 cos 2p

The general expression for the change in sales is

(0.1)p dS ⫽

2p t

d 2p t

2dt ⫺ (0.1) sin

12 Q2 12 R or dS ⫽ dt 2⫺ sin

6 12 The rate of change of sales at the end of March (t ⫽ 3) is obtained by evaluat-

At the end of March sales are at the linear rate of 2 thousand dollars per month. Example 8-24 For the previous problem find the total sales for the 3 months

from January through March. Solution: Sales follow S⫽

2t ⫹ 0.1 cos 2p t . The total sales over any period

is the integral of S, with appropriate limits. The limits for this January to March period are t ⫽ 0 and t ⫽ 3.

Trigonometric Functions

S 0S3 ⫽ 3 [2t ⫹ (0.1) cos pt/6]dt

S 0S3 ⫽

3 0.1 2tdt ⫹ 3 p cos d

Total sales for the January to March period were 9.2 thousands of dollars. ✔ Solve right and non-right triangles

✔ Know how to derive simple identities ✔ Differentiate trigonometric functions ✔ Integrate trigonometric functions ✔ Apply differentiation and integration to practical problems ✔ Use tables for differentiation of trigonometric functions ✔ Use tables for integration of trigonometric functions