# qr code vb.net open source y 6 4 2 x 6 4 2 2 4 6 2 4 6 in Visual Studio .NET Generate Code 39 Extended in Visual Studio .NET y 6 4 2 x 6 4 2 2 4 6 2 4 6

y 6 4 2 x 6 4 2 2 4 6 2 4 6
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Figure B-37
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Illustration for the solution to Problem 9 in Chap 18
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572 Worked-Out Solutions to Exercises: 11-19
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The object described by these equations is a circular helix having a radius of 2 units, and centered on the y axis Here are some values of x, y, and z that we can calculate as t varies, causing a point on the helix to complete a single revolution in a moving plane perpendicular to the y axis: When t = 0, we have x = 2, y = 0, and z = 0 When t = p /2, we have x = 0, y = 1/4, and z = 2 When t = p, we have x = 2, y = 1/2, and z = 0 When t = 3p /2, we have x = 0, y = 3/4, and z = 2 When t = 2p, we have x = 2, y = 1, and z = 0
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Every time t increases by 2p, our point makes exactly one revolution in a moving plane that s always perpendicular to the y axis In addition, we can see that every time t increases by 2p, our point gets 1 unit more distant from the xz plane in the +y direction The pitch of the helix is therefore 1 linear unit per revolution The sense of rotation is rather tricky to describe Suppose that we re somewhere on the y axis, and we direct our gaze in the y direction As the value of the parameter t increases, causing a point on the helix to move generally toward us, the point appears to revolve counterclockwise around the y axis The helix s sense of rotation is therefore counterclockwise in the +y direction Figure B-38 is a perspective drawing of the graph of this object in Cartesian xyz space
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Helix is centered on the y axis
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+y Helix turns counterclockwise as we move in the +y direction
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Pitch of helix =1
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Each axis division is 1 unit
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Radius of helix =2
Figure B-38
Illustration for the solution to Problem 10 in Chap 18 Each axis division represents 1 unit
19
19
1 Here s the sequence S again, for reference: S = 5, 3, 1, 1, 3, 5, We can read the value of the first term directly as s0 = 5 After that, the terms proceed as follows: The second term is 3, which is 2 less than the first term The third term is 1, which is 2 less than the second term The fourth term is 1, which is 2 less than the third term The fifth term is 3, which is 2 less than the fourth term The sixth term is 5, which is 2 less than the fifth term
We re told that S is an arithmetic sequence, so we can be sure that the terms always differ by the same constant c The general form is S = s0, (s0 + c), (s0 + 2c), (s0 + 3c), where s0 is the initial term and c is the constant In this case, c = 2, giving us S = , [5 + ( 2)], [5 + 2 ( 2)], [5 + 3 ( 2)], = 5, (5 2), (5 4), (5 6), (5 8), (5 10), = 5, 3, 1, 1, 3, 5, The brackets in the first line and the parentheses in the second line aren t technically necessary, but they serve to visually isolate the terms In general, the nth term of the series is what we get when we multiply the constant c by n 1 and then add it to s0 The hundredth term in S is therefore s0 + (n 1)c = 5 + (100 1) ( 2) = 5 + 99 ( 2) = 5 + ( 198) = 5 198 = 193 2 The values of the terms in the sequence S become large negatively without bound We can choose an integer that s as large negatively as we want, and we ll always be able to find an element in S that s still more negative Therefore, S does not converge 3 We re told that t0 = 2 and k = 4 Let s plug these values into the general formula for an infinite geometric sequence T = t0, t0k, t0k 2, t0k 3, t0k4,
574 Worked-Out Solutions to Exercises: 11-19
When we do that for the first seven terms, we get t0 = 2 t1 = 2 ( 4) = 8 t2 = 2 ( 4)2 = 2 16 = 32 t3 = 2 ( 4)3 = 2 ( 64) = 128 t4 = 2 ( 4)4 = 2 256 = 512 t5 = 2 ( 4)5 = 2 ( 1024) = 2048 t6 = 2 ( 4)6 = 2 4096 = 8192 This series does not converge The terms actual values alternate between positive and negative The absolute values increase by a factor of 4 with each succeeding term In an intuitive sense, T approaches both positive infinity and negative infinity 4 This time, we re given the values t0 = 2 and k = 1/4 We can plug these values into the general formula for an infinite geometric sequence T = t0, t0k, t0k2, t0k3, t0k4, Calculating the first seven terms, we obtain t0 = 2 t1 = 2 ( 1/4) = 1/2 t2 = 2 ( 1/4)2 = 2 1/16 = 1/8 t3 = 2 ( 4)3 = 2 ( 1/64) = 1/32 t4 = 2 ( 4)4 = 2 1/256 = 1/128 t5 = 2 ( 4)5 = 2 ( 1/1024) = 1/512 t6 = 2 ( 4)6 = 2 1/4096 = 1/2048 This series converges on 0 The terms actual values alternate between positive and negative The absolute values decrease by a factor of 4 with each succeeding term In an intuitive sense, T approaches 0 from both sides 5 Here s the sequence B again, for reference: B = 0/1, 1/2, 2 /3, 3/4, 4/5, , (n 1)/n, When we add the elements of B, we obtain the series B+ = 0/1 + 1/2 + 2 /3 + 3/4 + 4/5 + + (n 1)/n + In summation notation, we can write B+ =