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19.2 The volume V of a sphere of radius r is given by the formula V = 4 r 3 /3. (a) How fast is the volume changing relative to the radius when the radius is 10 millimeters (b) What is the change in volume when the radius changes from 10 to 10.1 millimeters

+ I)) =

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A segment of a circle is a part of a circle bounded by a chord and its arc. A minor segment of a circle is the smaller of the two segments thus formed. Thus in Fig. 10-10, the shaded section of circle Q is minor segment ACB.

+ 1)e-In" =

To evaluate this sum, a "trick" is to use the identity 1This allows us to write the DTFT as follows: DTFT(x(2n

Fig. 10-9

+ 1)) =

PRINCIPLE 1:

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dV 4 3 = Dr r dr 3 When r = 10, dV = 4 (10)2 = 400 400(3.14) = 1256 dr (b) 4 4000 (10)3 = 3 3 4 4121.204 4 V (10.1) = (10.1)3 = (1030.301) = 3 3 3 V (10) = V = V (10.1) V (10) = 4121.240 4000 3 3 3.14 = (4121.204 4000) = (121.204) (121.204) 3 3 3 = 126.86 cubic millimeters The change predicted from (19.1) and part (a) is V dV dr r = 1256(0.1) = 125.6 cubic millimeters = 4 (3r 2 ) = 4 r 2 3

x(n)e-jn" =

Fig. 10-10

~ ( e j " )= xR(eJ")

PRINCIPLE 2:

~(eJo) ,

19.3 An oil tank is being lled. The oil volume V , in gallons, after t minutes is given by V = 1.5t 2 + 2t How fast is the volume increasing when there is 10 gallons of oil in the tank [Hint: To answer the question, how fast , you must always nd the derivative with respect to time.]

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n In a circle of radius r, the length l of an arc of measure n equals of the circumference 360 pnr n of the circle, or l . 2pr 360 180 n In a circle of radius r, the area K of a sector of measure n equals of the area of the 360 n 2. circle, or K pr 360 Area of a sector of n Area of the circle length of an arc of measure n circumference of the circle n 360

where x R ( e j w )and x l ( e j w ) are the real part and the imaginary part of x ( e j W ) ,respectively. Find the sequence y(n) that has a DTFT given by

The key to solving this problem is to recall that if x(n) is real, and if X(eJW) written in terms of its real and is imaginary parts, XR(ejw) the DTFT of the even part of x(n), and Xl(eJW) the DTFT of the oddpart: is is x,(n> = [x(n)

PRINCIPLE 3: PRINCIPLE 4:

x&) = f [x(n) - x(-n)]

The area of a minor segment of a circle equals the area of its sector less the area of the triangle formed by its radii and chord. If a regular polygon is inscribed in a circle, each segment cut off by the polygon has area equal to the difference between the area of the circle and the area of the polygon divided by the number of sides.

IXl(eJw)

When there are 10 gallons in the tank, 1.5t 2 + 2t = 10 Solving by the quadratic formula, t= 2 4 4(1.5)( 10) 2 4 + 60 2 64 2 8 = = = =2 2(1.5) 3 3 3 or 10 3 or 1.5t 2 + 2t 10 = 0

PRINCIPLE 5:

Therefore, the DTFT of -j x , ( n ) is X l ( e i " ) ,

10.11 Length of an arc (a) Find the length of a 36 arc in a circle whose circumference is 45p. (b) Find the radius of a circle if a 40 arc has a length of 4p.

+ 2 ) - jx&)

it follows that y ( n ) , which is formed from these two sequences, is as shown below:

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