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(a) Here n (b) Here l 36 and C 4p and n 2pr 45p. Then l n 2pr 360 36 9 45p p. 360 2 40 2pr, and r 360

CHAP. 19]

Evaluate the following quantities without explicitly finding X(eJo):

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10.12 Area of a sector (a) Find the area K of a 300 sector of a circle whose radius is 12. (b) Find the measure of the central angle of a sector whose area is 6p if the area of the circle is 9p. (c) Find the radius of a circle if an arc of length 2p has a sector of area 10p.

( b ) 9Mw)

10.13 Area of a segment of a circle (a) Find the area of a segment if its central angle measures 60 and the radius of the circle is 12. (b) Find the area of a segment if its central angle measures 90 and the radius of the circle is 8. (c) Find each segment formed by an inscribed equilateral triangle if the radius of the circle is 8.

x ( e j W )=

1 2 4 s 23

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which is simply the sum of the values of x ( n ) . Therefore, for the given sequence it follows that

Fig. 20-2

(b) To evaluate the phase, note that because x ( n ) is real and even, X ( e J W is real and even and, therefore, the phase ) is equal zero or n for all o.

n pr2 360 1 4 (144) 23

( c ) From the inverse DTFT,

60 144p 360 36 23.

note that when n = 0 :

24p.

Therefore, d(u2 ) dt du 2u dt du 2u dt du u dt d(x 2 + y2 ) dt d(x 2 ) d(y2 ) = + [by the power chain rule] dt dt dx dy = 2x + 2y [by the power chain rule] dt dt dx dy =x +y = 300x + 400y dt dt =

x(0) =

1 2 bh 1 2 (8)(8)

Therefore,

32. 48 23.

lr(n)12 = 3871

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Now we must nd x, y, and u after 2 hours. Since x is increasing at the constant rate of 300 kilometers per hour and t is measured from the beginning of the ight, x = 300t (distance = speed time, when speed is constant). Similarly, y = 400t. Hence, at t = 2, x = 300(2) = 600 y = 400(2) = 800

and is used as a measure of the time delay of a sequence. Find an expression for c in terms of the DTFT of x ( n ) , and find the value of c for the sequence x ( n ) that has a DTFT as shown in the figure below.

1 (c) R 8. Since s R 23 8 23, the area of ^ABC is 4s2 23 Also, area of circle O pR2 64p. 1 Hence, area of segment BDC 3(64p 48 23).

To find the value of c in terms of X ( e J W ) first note that the denominator is simply the value of X ( e J o ) evaluated at ,

Fig. 10-11

o=O:

10.14 Area of a segment formed by an inscribed regular polygon Find the area of each segment formed by an inscribed regular polygon of 12 sides (dodecagon) if the radius of the circle is 12. (See Fig. 10-12.)

nx(n)

Substituting in (4), 1000 du = 300(600) + 400(800) = 180 000 + 320 000 = 500 000 dt du 500 000 = = 500 kilometers per hour dt 1000

Fig. 10-12

Therefore, and c may be evaluated in terms of X ( e J m )as follows:

Therefore,

n 30 pr2 144p 12p. 360 360 To find the area ^OAB, we draw altitude AD to base OB. Since m/AOB Area of sector OAB Then the area of ^OAB is

23 .7

1 2 bh 1 2 (12)

20.1 Air is leaking out of a spherical balloon at the rate of 3 cubic inches per minute. When the radius is 5 inches, how fast is the radius decreasing

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