EE2090 Basic Engineering Mathematics. Assoc. Prof. M. Adams Room : S2.2-B2-23 Extn :

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1 EE2090 Basic Engineering Mathematics Assoc. Prof. M. Adams Room : S2.2-B2-23 Extn : eadams@ntu.edu.sg TOPICS 1. Multiple Integrals 2. Infinite Sequences & Series 3. Vectors 4. Introduction to Laplace Transforms EE2090 Basic Engineering Mathematics slide-1

2 Recommended Text Books Text: 1. Gerald L. Bradley & K.J. Smith, Calculus, 3rd Edition, Prentice Hall. 2. Erwin Kreyzig, Advanced Engineering Mathematics, 8th Edition, John Wiley, EE2090 Basic Engineering Mathematics slide-2

3 Course Contents Multiple Integrals 1. Double Integrals (a) Regions of Integration (b) Iterated Integrals (c) Double Integrals in Polar Coordinates 2. Triple Integrals Infinite Sequences & Series 1. Sequences and Series 2. Convergence and Tests for Convergence Vectors 1. Vectors in the Plane 2. Dot and Cross product 3. Lines and Planes in Space 4. Vector Methods for Measuring Distance in R 3 EE2090 Basic Engineering Mathematics slide-3

4 Course Contents Laplace Transforms 1. Definition (a) Linearity (b) Inverse Laplace Transform 2. Properties of Laplace Transforms (a) Differentiation (b) Integration (c) Shifting on the s-axis (d) Shifting on the t-axis EE2090 Basic Engineering Mathematics slide-4

5 MULTIPLE INTEGRALS We shall examine Double Integrals, and Triple Integrals First, the Double Integrals The Double Integral of f over the region R is 4

6 R f(x, y)da = lim m,n m i=1 n j=1 f(x ij, y ij) A if this limit exists. To understand this, let s review the familiar Definite Integral: 5

7 If f(x) is defined for a x b, then b a f(x)dx = lim n n i=1 f(x i ) x Divide interval [a, b] into n subintervals [x i 1, x i ]; x = (b a)/n; 6

8 x i are sample points in the subintervals; Riemann sum is n i=1 f(x i ) x Taking the limit of such sum as n to obtain the definite integral of f from a to b! The integral represents the area under the curve y = f(x) from a to b 7

9 Volumes and Double Integrals In a similar manner, we consider: 8

10 A function f of two variables defined on a closed rectangle R = [a, b] [c, d] = {(x, y) R 2 a x b, c y d} The graph of f is a surface, z = f(x, y); 9

11 Let S be the solid that lies above R and under the graph of f: S = {(x, y, z) R 3 0 z f(x, y), (x, y) R} Divide the rectangle R into m n subrectangles; 10

12 x = (b a)/m and y = (d c)/n; Area of subrectangle R ij is A = x y; Choose a sample point (x ij, y ij ) in each R ij; Volume of the thin column with base R ij and height f(x ij, y ij ) is f(x ij, y ij) A 11

13 Summing all such volumes, we obtain an approximation to the 12

14 total volume of S as m n V f(x ij, yij) A i=1 j=1 Note that sum; m i=1 n j=1 f(x ij, y ij ) A is called the double Riemann Finally, in the limits, as m, n, we have the double integral : m n f(x, y)da = f(x ij, yij) A R lim m,n i=1 j=1 13

15 Note, taking the sample point as (x i, y j ) gives R f(x, y)da = lim m,n m i=1 n j=1 f(x i, y j ) A A volume can thus be written as a double integral: 14

16 V = f(x, y)da R 15

17 Iterated Integrals It is usually very difficult to evaluate double integrals from first principle! A practical method for evaluating a double integral is by expressing it as an iterated integral. Suppose f is a function of two variables, continuous on the rectangle R = [a, b] [c, d] 16

18 Partial integrating with respect to y gives A(x) = d c f(x, y)dy Next integrating A(x) with respect to x gives b A(x)dx = b d a a c f(x, y)dy dx The integral on the right side of the equation above is called an iterated integral. 17

19 Using the iterated integral, we can now express a double integral as f(x, y)da = b d f(x, y)dydx = d b f(x, y)dxdy R a c c a Note that the order of integration does not matter. 18

20 Example: 1. Evaluate the iterated integral x 2 y dydx Solution: 19

21 First taking x as a constant, we obtain 2 1 x 2 y dy = ] y=2 [x 2y2 2 y=1 = 3 2 x2 Next, integrating the function of x from 0 to 3: x 2 y dydx = x2 dx = [ x 3 ] 3 =

22 2. Evaluate the double integral R (x 3y 2 ) da where R = {(x, y) 0 x 2, 1 y 2}. Solution: R (x 3y 2 )da = (x 3y 3 ) dy dx 21

23 = = [ xy y 3 ] y=2 y=1 dx ] 2 (x 7) dx = x2 2 7x 0 = Evaluate R y sin(xy) da, where R = [1, 2] [0, π]. 22

24 Solution: R y sin(xy) da = = = π 0 π 0 π y sin(xy) dxdy [ cos(xy)] x=2 x=1 dy ( cos 2y + cos y) dy 23

25 = 1 2 sin 2y + sin y ] π 0 = 0 4. Find the volume of the solid S that is bounded by the elliptic paraboloid x 2 + 2y 2 + z = 16, the planes x = 2 and y = 2, and the three coordinate planes. Solution: First observe that the solid S lies under the surface 24

26 z = 16 x 2 + 2y 2 and above the square R = [0, 2] [0, 2]. Thus: V = R (16 x 2 2y 2 ) da 25

27 2 2 = (16 x 2 2y 2 ) dxdy = = [ 16x 1 3 x3 2y 2 x ( y2 ) dy = ] x=2 x=0 dy [ 88 3 y 4 ] 2 3 y3 0 = 48 26

28 Double Integrals Over General Regions If f is continuous on a region D such that D = {(x, y) a x b, g 1 (x) y g 2 (x)} then D f(x, y)da = b a g2 (x) g 1 (x) f(x, y) dydx. D in this case is called a type I region 27

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30 Similarly, we may have D f(x, y)da = d c h2 (y) h 1 (y) f(x, y) dxdy. D in this case is called a type II region 29

31 30

32 Examples: 1. Evaluate D(x + 2y) da, where D is the region bounded by the parabolas y = 2x 2 and y = 1 + x 2. Solution: We need first to determine the boundaries of the region D. In this respect, it is always helpful to draw a sketch. 31

33 The parabolas intersect when 2x 2 = 1 + x 2, so that x 2 = 1, that is x = ±1. Referring to the figure, we note that the region is of type I. The lower boundary is y = 2x 2 and the upper boundary is 32

34 y = 1 + x 2. So 1 1+x 2 D (x + 2y) da = 1 2x 2 (x + 2y) dydx = = [ xy + y 2 ] 1+x 2 y=2x 2 dx ( 3x 4 x 3 + 2x 2 + x + 1) dx 33

35 = 3 x5 5 x x3 3 + x2 2 + x = ] Find the volume of the solid that lies under the paraboloid z = x 2 + y 2 and above the region D in the xy-plane bounded by the line y = 2x and the parabola y = x 2. 34

36 Solution: Figures show that it is possible to take the region D as either of type I or type II! Either case will yield the same result! Taking D as a type I region yields V = D (x 2 + y 2 ) da = 2 0 2x x 2 (x 2 + y 2 ) dydx 35

37 whereas, taking it as type II gives V = D (x 2 + y 2 ) da = 4 0 y (x 2 + y 2 ) dxdy 1 2 y and either should yield an answer of

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40 3. Evaluate xy da, where D is the region bounded by the line D y = x 1 and the parabola y 2 = 2x + 6. Solution: From the sketches,again D is both type I and type II. 39

41 However, it is more complicated to deal with it as a type I region because the lower boundary consists of two parts! It is therefore preferable to express D as a type II region, giving the double 40

42 integral as : D xy da = = 1 2 = y y xy dxdy = 4 2 ( y y3 + 2y 2 8y [ y y4 + 2 y3 3 4y2 [ ] x 2 x=y+1 2 y ) ] 4 2 x= 1 2 y2 3 dy = 36 dy 41

43 Some Useful Properties of Double Integrals 1. D [f(x, y) + g(x, y)] da = f(x, y) da D + g(x, y) da D 2. D cf(x, y) da = c D f(x, y) da 42

44 3. f(x, y) da = f(x, y) da D D 1 + f(x, y) da D 2 4. Putting f(x, y) = 1 and integrate over a region D, we get the area of D: da = A(D) D 43

45 Example: Find the area of the region D between y = cos x and y = sin x over the interval 0 x π 4 Solution: From the graph, we find that 44

46 A = D da = π 4 0 cos x sin x dydx 45

47 = π 4 0 y] y=cos x y=sin x dx = π 4 0 [cos x sin x] =

48 Double Integrals in Polar Coordinates Description of the region of integration, complicated in terms of rectangular coordinates; R may be rather A change of coordinate system may greatly simply the problem; One such case is as shown in the figure and a change to polar coordinates greatly simplifies the problem. 47

49 Recall that the polar coordinates (r, θ) of a point are related to the rectangular coordinates (x, y) by r 2 = x 2 + y 2 x = r cos θ y = r sin θ 48

50 Double integral in polar coordinates is then given by the following: If f is continuous on a polar rectangle R given by 0 a r b, α θ β, where 0 β α 2π, then f(x, y)da = R α β b a f(r cos θ, r sin θ) r dr dθ. 49

51 50

52 51

53 Examples: 1. Evaluate R (3x + 4y2 ) da where R is the region in the upper half-plane bounded by the circles x 2 + y 2 = 1 and x 2 + y 2 = 4 Solution: Referring to the figure, the region can be described as R = { (x, y) y 0, 1 x 2 + y 2 4 } 52

54 In polar coordinates, 1 r 2, 0 θ π. Hence R (3x + 4y 2 ) da = = = = π 0 π 0 π 0 π (3r cos θ + 4r 2 sin 2 θ) r dr dθ (3r 2 cos θ + 4r 3 sin 2 θ) dr dθ [ r 3 cos θ + r 4 sin 2 θ ] r=2 [ r=1 dθ 7 cos θ + 15 ] (1 cos 2θ) 2 dθ 53

55 = 7 sin θ + 15θ sin 2θ ] π 0 = 15π 2 2. Find the volume of the solid bounded by the plane z = 0 and the paraboloid z = 1 x 2 y 2. Solution: Putting z = 0 we get x 2 + y 2 = 1. So, the plane intersects the paraboloid in the circle x 2 + y 2 = 1. The solid thus lies under the paraboloid and above the circular disk D given by x 2 +y

56 55

57 In polar coordinate, D is given by 0 r 1, 0 θ 2π. Noting that 1 x 2 y 2 = 1 r 2, the volume is then given by V = = D 2π 0 (1 x 2 y 2 ) da = dθ 1 0 2π 0 (r r 3 ) dr = 2π [ r r4 4 (1 r 2 ) r dr dθ ] 1 0 = π 2 3. Use a double integral to find the area enclosed by one loop of the four leaved rose r = cos 2θ. 56

58 Solution: From the sketch of the curve, we see that a loop is given by the region 57

59 D = {(r, θ) π/4 θ π/4, 0 r cos 2θ} 58

60 So the area is A(D) = = D π/4 π/4 π/4 da = [ 1 2 r2 π/4 π/4 ] cos 2θ 0 cos 2θ 0 dθ = 1 2 r dr dθ π/4 π/4 cos 2 2θ dθ [θ + 14 sin 4θ ] π/4 = 1 4 π/4 (1 + cos 4θ) dθ = 1 4 π/4 = π 8 59

61 We shall now examine the triple integrals The Triple Integral of the function f is defined in a similar manner as that for the double integrals; The function f is defined on a rectangular box B = {(x, y, z) a x b, c y d, r z s} Using the iterated integrals, we can express a triple integral as: 60

62 If f is continuous on the rectangular box B = [a, b] [c, d] [r, s], then s f(x, y, z) dv = B r d b c a f(x, y, z) dx dy dz. In the case of a triple integral over a general bounded region E, we have 61

63 E = {(x, y, z) a x b, g 1 (x) y g 2 (x), u 1 (x, y) z u 2 (x, y)} then E f(x, y, z) dv = s r d c b a f(x, y, z) dx dy dz. Just as in the case of the double integral, setting f(x, y, z) = 1 62

64 for all points in E, the triple integral represents the volume of E: V (E) = dv E 63

65 Examples: 1. Evaluate the triple integral rectangular box given by B xyz 2 dv, where B is the B = {(x, y, z) 0 x 1, 1 y 2, 0 z 3} Solution: B xyz 2 dv = xyz 2 dx dy dz 64

66 = = [ y 2 z 2 4 [ x 2 yz 2 2 ] y=2 y= 1 ] x=1 x=0 dy dz dz = z3 4 ] 3 0 = 27 4 note that the order of integration does not matter 65

67 2. Use triple integral to find the volume of the tetrahedron T bounded by the planes x + 2y + z = 2, x = 2y, x = 0, and z = 0. Solution: The tetrahedron T and its projection D on the xy-plane are shown in figures. 66

68 67

69 The lower boundary of T is the plane z = 0 and the upper boundary is the plane x + 2y + z = 2, that is, z = 2 x 2y. Therefore: V (T ) = = dv = T 1 1 x/2 0 x/ x/2 x/2 2 x 2y (2 x 2y) dy dx = dz dy dx 68

70 SEQUENCES A sequence is a succession of numbers that are listed according to a given prescription or rule. We can write the sequence explicitly as: a 1, a 2,..., a n,... where n is a positive integer, or simply as {a n } For example 1, 1 2, 1 3,, 1 n, is a sequence and a n = 1 n. 69

71 a n is called the general term a n+1 is the successor a n 1 is the predecessor We can also write out the rule as a general term and generate the sequence subsequently. For instance: a n = n n+1 generates a 1 = 1 2, a 2 = 2 3, a 3 = 3 4, Formally, we can state: 70

72 A sequence {a n } is a function whose domain is a set of nonnegative integers and whose range is a subset of the real numbers. The functional values a 1, a 2, a 3,... are called the terms of the sequence, and a n, is called the nth term, or the general term, of the sequence. 71

73 LIMITS OF SEQUENCES How does a given sequence {a n } behave as n gets arbitrarily large? Consider the sequence whose a n = This generates: a 1 = 1 2, a 2 = 2 3, a 3 = 3 4, n n+1 It appears that the terms of the sequence are approaching 1 as n increases!! 72

74 In general, if the terms of the sequence approach the number L as n increases without bound, we say that the sequence converges to the limit L and write L = lim n a n In the above example we would expect L = lim n n n + 1 = 1 We may define formally the above results as follows: 73

75 The sequence {a n } converges to the number L, and we write L = lim n a n if for every ɛ > 0, there is an integer N such that a n L < ɛ whenever n > N. Otherwise, the sequence diverges. Note that the notation L = lim n a n means that eventually the terms of the sequence {a n }, can be made as close to L as may be desired by taking n sufficiently large. 74

76 The following results should be useful: Theorem 1. [Limit theorem for sequences] If lim n a n = L and lim n b n = M, then Linearity rule for sequences lim (ra n + sb n ) = rl + sm n Product rule for sequences lim (a nb n ) = LM n 75

77 Quotient rule for sequences lim n a n bn = L M provided M 0 Root rule for sequences lim n exists. m a n = m L provided m a n is defined for all n and m L 76

78 Examples of Convergent sequences Find the limit of each of these convergent sequences: { } 100 n { 2n 2 } + 5n 7 n 3 { 3n 4 } + n 1 5n 4 + 2n

79 Solutions 1. As n grows arbitrarily large, 100/n gets smaller and smaller. Thus, lim n 100 n = 0 2. We cannot use the quotient rule of Theorem 1 because neither 78

80 the limit in the numerator nor the one in the denominator exists. However, 2n 2 + 5n 7 n 3 = 2 n + 5 n 2 7 n 3 and by using the linearity rule, we then have lim x 2n 2 + 5n 7 n 3 = lim n 7 lim n = n + 5 lim n 1 n 3 1 n 2 79

81 = 0 3. Divide the numerator and denominator by n 4, the highest power of n that occurs in the expression, to obtain lim n 3n 4 + n 1 5n 4 + 2n = lim n n 3 1 n n n 4 =

82 Examples of Divergent sequences Show that the following sequences diverge: 1. {( 1) n } 2. { n 5 + n 3 } + 2 7n 4 + n Solution: 1. The sequence defined by {( 1) n } is 1, 1, 1, 1,... and this 81

83 sequence diverges by oscillation because the nth term is always either 1 or 1. Thus a n cannot approach one specific number L as n grows large. n 5 + n lim n 7n 4 + n = lim n 2 n 5 n 7 n n 3 n 5 The numerator tends toward 1 as n, and the denominator approaches 0. Hence the quotient increases without bound, and the sequence must diverge. The following theorem relates how the limit of a sequence can be 82

84 deduced from the limit of a continuous function: Theorem 2. Given the sequence {a n }, let f be a continuous function such that a n = f(n) for n = 1, 2,.... If lim f(x) exists x and lim f(x) = L, the sequence {a n} converges and lim a n = L. x n Before proceeding further, the following result is always helpful: 83

85 l Hospital s Rule Let f and g be differentiable functions with g (x) 0 on an open interval containing c (except possibly at c itself). Suppose f(x) lim x g(x) produces an indeterminate form 0 0 lim x f (x) g (x) +, or. Then lim x or and that = L where L is either a finite number, f(x) g(x) = L 84

86 Examples 1. Calculate lim n 2. Evaluate lim n ln n n n 2 1 e n 85

87 Solution 1. Note that both numerator and denominator approach as n. Also note that l Hospital s Rule cannot be applied directly to sequences but to functions of real variables. So, we should first apply the Rule to related function f(x) = ln x/x 86

88 and obtain lim x ln x x = lim x 1/x 1 = 0 By Theorem 2, we have lim n ln n n = 0 87

89 2. By a similar approach, we have lim x x 2 1 e x = lim x = lim x = 0 2x e x 2 e x Hence, lim n n 2 1 e n = 0 Note that l Hospital s Rule is being applied twice! 88

90 Infinite Series An infinite series is an expression of the form a 1 + a 2 + a 3 + = k=1 a k and the nth partial sum of the series is n S n = a 1 + a a n = k=1 a k The series is said to converge with sum S if the sequence of 89

91 partial sums {Sn} converges to S. In this case, we write a k = lim S n = S n k=1 If the sequence {S n } does not converge, the series k=1 a k diverges and has no sum. Note that we can also consider certain series in which the starting point is not 1; for example, the series can be denoted 90

92 by k=3 1 k or k=2 1 k+1 Some Examples 1. Show that the series Solution k=1 1 2 k converges. This series has the following partial sums: S 1 = 1 2 =

93 S 2 = = = 3 4 S 3 = = = 7 8 S n = n. The sequence of partial sums is 1 2, 3 4, 7 8, 15 16, 31 32, 63 64, and in general 92

94 (by mathematical induction) S n = n Because lim (1 1 n 2 n ) = 1, we conclude that the series converges and its sum is S = 1 2. Show that the series Solution k=1 ( 1) k diverges. 93

95 The series can be expanded (written out) as ( 1) k = k=1 and we see that the nth partial sum is 1 if n is odd S n = 0 if n is even Because the sequence {S n } has no limit, the given series must diverge. 94

96 Another Example: Show that the series converges and find its sum. k=1 1 k 2 +k Solution Using partial fractions, we find that 1 k 2 + k = 1 k(k + 1) = 1 k + 1 k + 1 Thus, the nth partial sum of the given series can be represented as 95

97 follows: S n = = = n k=1 n k=1 1 k 2 + k [ 1 k 1 ( ] k + 1 ) ( ) + 3 ) ( 1 n 1 n + 1 ( )

98 ( = ( + 1 n + 1 ) n ) EE /05 = 1 1 n ( ) 3 1 n The limit of the sequence of partial sums is [ lim S n = lim 1 1 ] = 1 n n n + 1 so the series converges, with sum S = 1. 97

99 The above series is called a telescoping series or ( collapsing series) as there is internal cancellation in the partial sum! A little note: When the starting point of a series is not important, we may denote the series by writing a k instead of a k k=1 98

100 GENERAL PROPERTIES OF INFINITE SERIES Theorem 3. [Linearity of infinite series] If a k and b k are convergent series, then so is (ca k + db k ) for constants c, d, and (cak + db k ) = c a k + d b k 99

101 Example Show that the series k=1 [ ] 4 k 2 +k 6 converges, and find its sum. 2 k Solution Earlier examples show that 1 k 2 +k and 1 both converge with 2 k=1 k=1 k sum 1, the linearity property allows us to write the given series as 4 k=1 1 k 2 + k 6 k=1 1 = 4(1) 6(1) = 2 2k 100

102 What about the case when one series is convergent while the other diverges? Theorem 4. If either a k or b k diverges and the other converges, then the series (a k + b k ) must diverge. 101

103 GEOMETRIC SERIES A geometric series is an infinite series in which the ratio of successive terms in the series is constant. If this constant ratio is r, then the series has the form ar k = a + ar + ar 2 + k=0 ar ar n + for a 0 For example, is a geometric series because each 102

104 term is 1 2 the preceding term. The ratio of a geometric series may be positive or negative. For example, 2 ( 3) k = k=0 is a geometric series with r = 1 3. The following theorem tells us how to determine whether a given geometric series converges or diverges and, if it does converge, what its sum must be. 103

105 Theorem 5. [Geometric series theorem] The geometric series ar k with a 0 diverges if r 1 and k=0 converges if r < 1 with sum ar k = a 1 r k=0 Example Determine whether each of the following geometric series converges or diverges. If the series converges, find its sum. 104

106 1. k=0 ( 1 3 ) k k=2 3 ( 1 5 ) k Solution 1. r = 3 2 satisfies r 1, the series diverges. 105

107 2. We have r = 1 5 and r < 1, and the geometric series converges. Note that the first value of k is 2 (not 0), so the value a (the first value) is a = 3( 1 5 )2 = 3 25 and 3 k=2 ( 1 ) k = a r = 25 1 ( 1 5 ) =

108 Testing for Convergence or Divergence of a Series The convergence or divergence of an infinite series is determined by the behavior of its nth partial sum, S n, as n. We have used algebraic methods to find formulae for the nth partial sum of a series. It is often difficult or even impossible to find a usable formula for the nth partial sum of a series, and other techniques must be 107

109 used to determine convergence or diverge. Theorem 6. [The integral test] If a k = f(k) for k = 1, 2,..., where f is a positive, continuous, and decreasing function of x for x 1, then and f(x)dx k=1 a k either both converge or both diverge

110 Note the following: It is NOT necessary to start the series or the integral at n = 1. It is NOT necessary that f be always decreasing. What is important is that f be ultimately decreasing, that is, decreasing for x larger than some number! 109

111 Example - Harmonic series diverges Test the series (called the harmonic series) 1 k = k=1 for convergence. Solution Because f(x) = 1 x is positive, continuous, and decreasing for x 1, 110

112 the conditions of the integral test are satisfied. 1 1 dx = lim x b b 1 1 dx = lim [ln b ln 1] = x b The integral diverges, so the harmonic series diverges. Another example Test the series k for convergence. e k/5 k=1 Solution The function f(x) = x e x/5 = xe x/5 is positive and continuous for 111

113 all x > 0. Is it decreasing? We find that f (x) = x ( 15 ) ( e x/5 + e x/5 = 1 x ) 5 e x/5 The critical number is found when f (x) = 0, so we solve ( 1 x 5) e x/5 = 0 1 x 5 = 0 x = 5 We see that f (x) < 0 for x > 5, so it follows that f is decreasing for x > 5. The conditions for the integral test have been established! Computing the improper integral, we have: 112

114 5 xe x/5 dx = lim [ = lim b b b 5 5xe x/5 b 5 xe x/5 dx b [ = lim 5xe x/5 25e x/5] b 5 b = 5 lim b b + 5 e b/5 = 5 lim 1 b ( lim b 50e 1 )eb/5 + 50e 1 ( 5e x/5 )dx ] 113

115 = 50e 1 Thus, the improper integral converges, which in turn assures the convergence of given series. Note, however, that the series converges because the integral converges to a finite number 50e 1, but this does NOT mean that the series k converges to the same number 50e 1!! e k/5 k=1 114

116 Theorem 7. [The Ratio Test] Given the series a k with a k > 0, suppose that a k+1 lim = L k a k The ratio test states the following: If L < 1, then a k converges. If L > 1 or if L is infinite, then a k diverges. If L = 1, the test is in-conclusive. 115

117 Example - Convergence using the ratio test Test the series for convergence. k=1 2 k k! Solution Let a k = 2k k! and note that L = lim k a k+1 a k = lim k 2 k+1 (k+1)! 2 k k! 116

118 = lim k k!2 k+1 (k + 1)!2 k = lim k 2 k + 1 = 0 Thus L < 1, and the ratio test tells us that the given series converges. Another example Divergence using the ratio test Test the series k k k! for convergence. k=1 Solution Let a k = kk k! and note that 117

119 L = lim k = lim k = lim k a k+1 a k = lim k (k+1) k+1 (k+1)! k k k! k!(k + 1) k+1 k k (k + 1)! ( 1 + k) 1 k = e > 1 = lim k (k + 1) k k k Because L > 1, the given series diverges. 118

120 Theorem 8. [The root test] Given the series a k with a k 0, suppose that lim k k ak = L. The root test states the following: If L < 1, then a k converges. If L > 1 or if L is infinite, then a k diverges. If L = 1, the root test is in-conclusive. 119

121 Example - Convergence with the root test Test the series 1 for convergence. (ln k) k k=2 Solution Let a k = 1 (ln k) k L = lim k = lim k and note that k (ln k) k k ak = lim k 1 ln k = 0 Because L < 1, the root test tells us that the given series converges. 120

122 Another example - Divergence with the root test Test the series ( k 2 k) for convergence. k=1 Solution L = lim k k2 ( 1 + k) 1 ( = lim ) k k k 1/k 121

123 = e > 1 Since L > 1, the series diverges. Remark The ratio test is especially useful for series a k for which the general term a k involves factorials or powers, and the root test applies naturally if a k involves a power of k. 122

124 Vectors in the Plane and in Space A vector is a quantity that has both magnitude and direction. velocity, force A vector is sometimes represented by a directed line, segment,pq, with initial point P and terminal point Q. We may write PQ as P Q. The magnitude of PQ is denoted as PQ 178

125 Two vectors are equal if they have the same magnitude and the same direction, even if they are in different locations. A vector with magnitude 0 is called a null vector or zero vector. The 0 vector has no specific direction. A scalar multiple sv is parallel to v with magnitude s v and points in the same direction as v if s > 0, and in the opposite direction if s <

126 If s = 0 or v = 0, then sv = 0 Vector addition and subtraction can be realized by the triangle rule (or the parallelogram rule) and the difference rule. 180

127 Vectors can be conveniently represented using rectangular coordinates. If a vector v is positioned in a rectangular coordinate plane with its initial point at the origin (0, 0) and its terminal point at (v 1, v 2 ), then v 1 and v 2 are called the standard components of v. We write v = v 1, v 2 181

128 182

129 The following vector properties are important: a 1, b 1 = a 2, b 2 if and only if a 1 = a 2 k a, b = ka, kb a, b + c, d = a + c, b + d a, b c, d = a c, b d and b 1 = b 2 for constant k au + bv is called a linear combination of the vectors u and v and scalar a and b. 183

130 If u = u 1, u 2 and v = v 1, v 2 then au + bv = a u 1, u 2 + b v 1, v 2 = au 1 + bv 1, au 2 + bv 2 Theorem 16. [Properties of Vector Operations] For any vector u, v, and w in the plane and scalars s and t: Commutativity of vector addition u + v = v + u 184

131 Associativity of vector addition (u + v) + w = u + (v + w) Associativity of scalar multiplication (st)u = s(tu) Identity for addition u + 0 = u Inverse property for addition u + ( u) = 0 185

132 Distributive laws (s + t)u = su + tu s(u + v) = su + sv Some examples: 1. For the vectors u = 2, 3 and v = 1, 7, find (a) u + v (b) 3 4 u 186

133 (c) 3u 1 2 v Solution: (a) u + v = 2, 3 + 1, 7 = 2 + ( 1), = 1, 4 (b) 3 4 u = 3 4 2, 3 = 3 2, 9 4 (c) 3u 1 2 v = 3 2, , 7 = 13 2, Show that the line segment joining the midpoints of two sides of 187

134 a triangle is parallel to the third side and has half its length. Solution: Referring to the Figure. Consider ABC, and let P and Q be the midpoints of sides AC and BC respectively. 188

135 We have AP = 1 2 AC and BQ = 1 2BC. Also: AB = AP + P Q + QB = 1 2AC + P Q BQ = 1 2 (AB + BC) + P Q 1 2 BC = 1 2 AB + P Q 1 2 AB = P Q 189

136 The following facts about vectors are important: Given u = u 1, u 2, its length is u = u u2 2 The triangle inequality for any vectors u and v is u + v u + v These two properties are illustrated by the Figure: 190

137 A unit vector is a vector with length 1 A direction vector for a given nonzero vector v is a unit vector 191

138 u that points in the same direction as v: u = v v The unit vectors i = 1, 0 and j = 0, 1 point in the direction of the positive x and y axes, respectively. They are the standard basis vectors. Any vector v = v 1, v 2 in the plane can be represented as: v = v 1, v 2 = v 1 i + v 2 j 192

139 Extending to 3-dimensional space, we have v = v 1, v 2, v 3 = v 1 i + v 2 j + v 3 k Some examples: 1. Find a directional vector for v = 2, 3. Solution: Noting that v = ( 3) 2 = 13, we have: u = v v = 2,

140 = , 3 = 2, Note that we can also write v = 2i 3j, then u = 2i 3j 13 = 2 13 i 3 13 j 2. If u = 3i + 2j, v = 2i + 5j and w = i 4j, find the vector 2u + 5v w. 194

141 Solution: 2u + 5v w = 2(3i + 2j) + 5( 2i + 5j) (i 4j) = 5i + 33j 3. Two forces, F 1 and F 2 act on the same body. F 1 has a magnitude of 3 newtons and acts in the direction of i, whereas F 2 has 195

142 a magnitude of 2 newtons and acts in the direction of the unit vector u = 3 5 i 4 5 j Find the magnitude and direction of the additional force F 3 that must be applied to keep the body at rest. Solution: 196

143 From the given information, we have F 1 = 3( i) and F 2 = 2( 3 5 i 4 5 j) = 6 5 i 8 5 j Let F 3 = ai+bj then, for equilibrium, we need F 1 +F 2 +F 3 = 0. Thus: ( 3i) + ( 6 5 i 8 j) + (ai + bj) = 0i + 0j 5 Simplifying, we have ( a)i + ( b)j = 0i + 0j 197

144 and thus a = 0 and b = 0 a = 9 5 b = 8 5 The required force is, F 3 = 9 5 i + 8 5j. This is a force of magnitude F 3 = ( 9 5 )2 + ( 8 5 )2 = newtons which acts in the direction of the unit vector v = F 3 F 3 = ( 9 5 i j) = i j 198

145 The Dot Product The dot product of vectors v = a 1 i + a 2 j + a 3 k and w = b 1 i + b 2 j + b 3 k is the scalar denoted by v wand defined by v w = a 1 b 1 + a 2 b 2 + a 3 b 3 Theorem 17. [Properties of the dot products] If u, v and w are vectors in R 2 or R 3 and c is a scalar, then Magnitude of a vector 199

146 v v = v 2 Zero product 0 v = 0 Commutativity v w = w v Multiple of a dot product c(v w) = (cv) w = v (cw) 200

147 Distributivity u (v + w) = u v + u w An example: Find the dot product of v = 3i + 2j + k and w = 4i j + 2k. Solution: v w = 3(4) + 2( 1) + 1(2) =

148 Angle between Vectors The angle between two vectors plays an important role in certain applications. If θ is the angle between the nonzero vectors v and w, then cos θ = v w v w Geometrically, we can also use v w = v w cos θ 202

149 Two vectors are perpendicular or orthogonal, if the angle between them is θ = π/2 Nonzero vectors v and w are orthogonal if and only if v w = 0 Projections Let v and w be two vectors in R 3 as shown in the Figure. 203

150 The vector projection of v onto w, or proj w is the vector u on the line determined by w. Note that u = tw for some scalar t and 204

151 that v tw is orthogonal to w. Thus, (v tw) w = 0 v w = t(w w) t = v w w w and the vector projection is ( v w ) u = w w w Since w w = w 2, we have ( ) v w proj w v = w 2 w 205

152 = The scalar ( ) v w w w w ( ) is called the scalar projection of v onto w, or v w w the component of v in the direction of w, or comp w v. Note that w w is the direction vector for w, comp w v = v w w = v cos θ, since (v w = v w cos θ) 206

153 . An example: Find the vector and scalar projections of v = 2i + 3j + 5k onto w = 2i 2j k. Solution: The vector projection of v onto w is ( v w ) proj w v = w w w 207

154 = ( ) 2(2) + (3)( 2) + 5( 1) ( 2) 2 + ( 1) 2 (2i 2j k) = 7 9 (2i 2j k) = 14 9 i j k comp w v = v w w = Work as a Dot Product ( 2) + 5 ( 1) 22 + ( 2) 2 + ( 1) 2 =

155 The work W done by a constant force F on an object moving along the line from a point P to a point Q is given by W = F PQ where PQ is the displacement vector of the object s motion. 209

156 An example: A boat sails north aided by a wind blowing in a direction of N30 E with magnitude 50 N. How much work is performed by the wind as the boat moves 10 m? Solution: The wind force is F = 50 N, acting in the direction θ = 30, as shown in the Figure. The displacement vector is PQ = 10j. The 210

157 force is F = 50 cos 60 i + 50 sin 60 j The work done is then W = F PQ = 10(25 3) = N m 211

158 212

159 The Cross Product If v = a 1 i+a 2 j+a 3 k and w = b 1 i+b 2 j+b 3 k, the cross product of the vector v with the vector w, written as v w, is the vector v w = (a 2 b 3 a 3 b 2 )i + (a 3 b 1 a 1 b 3 )j + (a 1 b 2 a 2 b 1 )k A convenient notation for the above is the following determinant 213

160 form: i j k v w = a 1 a 2 a 3 b 1 b 2 b 3 Expanding the determinant about the first row: i j k a v w = a 1 a 2 a 3 = 2 a 3 b 2 b 3 i b 1 b 2 b 3 214

161 EE /05 a 1 a 3 b 1 b 3 j + a 1 a 2 b 1 b 2 k = (a 2 b 3 a 3 b 2 )i (a 1 b 3 a 3 b 1 )j +(a 1 b 2 a 2 b 1 )k = (a 2 b 3 a 3 b 2 )i + (a 3 b 1 a 1 b 3 )j +(a 1 b 2 a 2 b 1 )k 215

162 Theorem 18. [Properties of the cross product] If u, w, a, b, and c are vectors in R 3 and s and t are scalars, then Scalar distributivity (sv) (tw) = st(v w) Distributivity for cross product over addition u (v + w) = (u v) + (u w) (u + v) w = (u w) + (v w) 216

163 Anticommutativity v w = (w v) Product of a multiple v sv = 0 Zero product v 0 = 0 v = 0 217

164 Lagrange s identity v w 2 = v 2 w 2 (v w) 2 cab-bac formula a (b c) = (c a)b (b a)c Note in particular w v = (v w) i i = 0 i j = k i k = j j i = k 7j j = 0 j k = i k i = j k j = i k k = 0 218

165 Theorem 19. [Orthogonality property of cross product] If v and w are nonzero vectors in R 3 that are not multiples of one another, then v w is orthogonal to both v and w 219

166 The Right-Hand Rule To determine the direction of the vector (v w), place the little finger of the right hand along v and curl the fingers toward w to cover the smaller angle between v and w, then the thumb points in the direction of v w. 220

167 Theorem 20. [Magnitude of the cross product] If v and w are nonzero vectors in R 3 with θ the angle between v 221

168 and w (0 θ π), then v w = v w sin θ Applications of cross product: Area and Torque Referring to the Figure, let v be defined by AB and w by AC. Then the parallelogram with adjacent side AB and AC has AREA = AB AC 222

169 The triangle ABC has AREA = 1 AB AC 2 223

170 Referring to the Figure,consider a force F acting on a rigid body at a point given by the position vector r.( For instance, tightening a bolt using a wrench, producing a turning effect.) 224

171 The torque τ (relative to the origin), is defined as τ = r F and measures the tendency of the body to rotate about the origin. The direction of the torque vector indicates the axis of rotation. The magnitude of the torque vector is τ = r F = r F sin θ where θ is the angle between the position and force vectors. It is 225

172 interesting to note that only the component of F perpendicular to r causes rotation! Some examples: 1. Find the area of the triangle with vertices P ( 2, 4, 5), Q(0, 7, 4), and R( 1, 5, 0). Solution: 226

173 The area of the triangle P QR has area A = 1 PQ PR 2 First find: PQ = (0 + 2)i + (7 4)j + ( 4 5)k = 2i + 3j 9k PR = i + j 5k 227

174 and compute the cross product: i j k PQ PR = = ( )i ( )j + (2 3)k = 6i + j k 228

175 Thus the area is A = 1 PQ PR 2 = 1 ( 6) ( 1) 2 = The figure shows a half open door that is 3 ft wide. A force of 30 lb is applied in a horizontal direction at the edge of the door. 229

176 Find the torque of the force about the hinges on the door. Solution: Using the coordinate system assigned, F = 30i. The door is 230

177 half open, it makes an angle of π 4 as shown. The position vector PQ is represented by PQ = 3(cos π 4 i + sin π 4 j) = i j The torque can thus be found: T = PQ F 231

178 = i j k = 45 2k 232

179 Lines and Planes in R 3 Parametric Equations - Parametric Representation for a Curve in R 2 Let f and g be continuous functions of t on an interval I; then the equations x = f(t) and y = g(t) are called parametric equations with parameter t. As t varies 233

180 over the parametric set I, the points (x, y) = (f(t), g(t)) trace out a parametric curve. Examples: 1. Sketch the path of the curve x = t 2 9, y = 1 3t for 3 t 2. Solution: see the Figure. 234

181 235

182 2. Describe the path x = sin πt, y = cos 2πt, for 0 t 0.5. Solution: Using a double angle identity, we have cos 2πt = 1 2 sin 2 πt so that y = 1 2x 2 236

183 This is an equation for a parabola. Because t is restricted to an interval, the parametric representation involves only part of the right side of the parabola. 237

184 238

185 Parametric Form of a Line in R 3 If L is a line that contains the point (x 0, y 0, z 0 ) and is parallel to the vector v = Ai + Bj + Ck, then L has parametric form x = x 0 + ta y = y 0 + tb z = z 0 + tc The quantities A, B, C are called the direction numbers, or in notation [A, B, C]. The vector v is called a direction vector of the line L.(See Figure) 239

186 An Example: Find parametric equations for the line that contains the point (3, 1, 4) and is parallel to the vector v = i + j 2k. Find where this line passes through the coordinate planes. 240

187 Solution: The direction numbers are [ 1, 1, 2] and x 0 = 3, y 0 = 1, z 0 = 4, so the line has the parametric form x = 3 t y = 1 + t z = 4 2t This line intersects the x y-plane when z = 0: 0 = 4 2t implies t = 2 Thus, x = 3 2 = 1 and y = 3. This is the point (1, 3, 0). 241

188 Similarly, the line intersects the xz-plane at (4, 0, 6) and the yzplane at (0, 4, 2). 242

189 In the special case where none of the direction numbers A, B or C is 0, we can obtain the following symmetric equations for a line: Symmetric Form of a Line in R 3 : If L is a line that contains the point (x 0, y 0, z 0 ) and is parallel to the vector v = Ai + Bj + Ck, (A, B, C nonzero numbers), then 243

190 the point (x, y, z) is on L if and only if its coordinates satisfy x x 0 A = y y 0 B = z z 0 C An Example: Find symmetric equations for the line L through the points P ( 1, 3, 7) and Q(4, 2, 1) 244

191 Solution: The required line passes through P and is parallel to the vector PQ = (4 + 1)i + (2 3)j + ( 1 7)k = 5i j 8k Thus, the direction numbers are [5, 1, 8]. Choosing P as (x 0, y 0, z 0 ), we obtain x = y 3 1 = z

192 Equation of a Plane in R 3 Planes in space can be characterized by vector methods, Any plane is completely determined by any one of its points and its orientation, i.e. the direction it faces, The direction of a plane is specify by a vector N that is orthogonal to every vector in the plane the normal to the plane. 246

193 Example: Find an equation for the plane that contains the point Q(3, 7, 2) and is normal to the vector N = 2i + j 3k. Solution: The normal vector V is orthogonal to every vector in the plane. If point P (x, y, z) is in the plane, then N must be orthogonal to the vector QP = (x 3)i + (y + 7)j + (z 2)z 247

194 Hence N QP = 2(x 3) + (1)(y + 7) + ( 3)(z 2) = 0 2x 6 + y + 7 3z + 6 = 0 2x + y 3z + 7 = 0 Therefore, 2x + y 3z + 7 = 0 is the equation of the plane. Generalizing the approach illustrated in the above example, we take note of the following: 248

195 The plane that contains the point (x 0, y 0, z 0 ) and having a normal vector N = Ai + Bj + Ck must have the equation of the plane as A(x x 0 ) + B(y y 0 ) + C(z z 0 ) = 0 This is called the point-normal form of the equation of plane; By rearranging the terms, we can have the standard form of the 249

196 equation of plane as Ax + By + Cz + D = 0 The numbers [A, B, C] are the attitude numbers of the plane. 250

197 The attitude numbers of a plane are the same as direction numbers of a normal vector! Some Examples: 1. Find the a normal vector to the plane 5x + 7y 3z = 0 Solution: 251

198 A normal to the plane 5x + 7y 3z = 0 is N = 5i + 7j 3k 2. Find the standard form equation of a plane containing P ( 1, 2, 1), Q(0, 3, 2), and R(1, 1, 4). Solution: 252

199 The normal N to the required plane is orthogonal to the vectors PQ and PR, i.e. N = PR PQ. PQ = (0 + 1)i + ( 3 2)j + (2 1)k = i 5j + k PR = (1 + 1)i + (1 2)j + ( 4 1)k = 2i j 5k 253

200 N = PQ PR = i j k = 26i + 7j + 9k Using the point P, we can then find the equation of the plane as: 26(x + 1) + 7(y 2) + 9(z 1) = 0 254

201 Simplifying, we have 26x + 7y + 9z + 3 = 0 3. Find the equation of the line that passes through the point Q(2, 1, 3) and is orthogonal to the plane 3x 7y + 5z + 55 = 0. Where does the line intersect the plane? Solution: 255

202 By inspection, N = 3i 7j + 5k is a normal vector. The required line must be parallel to this normal. Thus the line contains the point Q(2, 1, 3) and has the direction numbers [3, 7, 5], so its parametric form is x = 2 + 3t, y = 1 7t, z = 3 + 5t Substituting into equation of the plane: 3(2 + 3t) 7( 1 7t) + 5(3 + 5t) =

203 t = 1 The point of intersection is found by putting t = 1 into equation of the line: x = 2 + 3( 1) = 1 y = 1 7( 1) = 6 z = 3 + 5( 1) = 2 The point of intersection is ( 1, 6, 2) 257

204 4. Find the equation of a line passing through ( 1, 2, 3) that is parallel to the line of intersection of the planes 3x 2y + z = 4 and x + 2y + 3z = 5. Solution: By inspection, normals to the planes are N 1 = 3i 2j + k and N 2 = i + 2j + 3k. The desired line is perpendicular to both of these normals (can you vusualise this?) so a vector parallel to the 258

205 line is found by computing the cross product: N 1 N 2 = i j k = ( 6 2)i (9 1)j + (6 + 2)k = 8(i + j k) Thus the required line passes through ( 1, 2, 3) and is parallel to 259

206 the vector 1, 1, 1 so it has the parametric form x = 1 + t, y = 2 + t, z = 3 t 260

207 Vector Methods For Measuring Distances in R 3 Theorem 21. [Distance from a point to a plane in R 3 ] The distance from a point P (x 0, y 0, z 0 ) to the plane Ax + By + Cz + D = 0 is given by d = QP N N = Ax 0 + By 0 + Cz 0 + D A2 + B 2 + C 2 where Q is any point in the given plane and N is a normal to the given plane. 261

208 Theorem 22. [Distance from a point to a line] 262

209 The distance from a point P to the line L is given by the formula d = v QP v where v is a vector parallel to L and Q is any point on L. Examples: 263

210 1. Find an equation for the sphere with center C( 3, 1, 5) that is tangent to the the plane 6x 2y + 3z = 9. Solution: The radius r of the sphere is the distance from the center C to 264

211 the given plane: r = 6( 3) + ( 2)(1) + 3(5) 9 = 62 + ( 2) = 2 The equation of the sphere is (x + 3) 2 + (y 1) 2 + (z 5) 2 =

212 2. Find the distance from the point P (3, 8, 1) to the line x 3 3 = y = z Solution: We need a point Q on the line. By inspection, Q(3, 7, 2) is on the line and that QP = 0, 1, 3. A vector parallel to L is 266

213 v = 3, 1, 5, so that i j k v QP = = 2i 9j 3k Thus d = v QP v = 22 + ( 9) 2 + ( 3) ( 1)

214 Introduction 1. Denition f(t) is a given function with variable t, for all t 0, Laplace Transform (L:T:) of f(t) is L[f(t)] = F (s) = Z 1 0 f(t)e ;st dt (0.1) F (s) is a function of s where s is generally a complex variable s = + jw. Original function f(t) is the "inverse" of F (s) dened by f(t) = L ;1 [F (s)] Laplace Transforms slide-3

215 Examples Example 1: Let f(t) = 1, for t 0. Find F (s). Solution: L[f(t)] = = 1 s Unit Step Function: Z 1 1e ;st dt 0 = ; e;st s t Laplace Transforms slide-4

216 Examples The unit step function u(t) is dened as follows Hence u(t) = 1 t 0 = 0 t < 0 L[u(t)] = 1 s Suppose given g(t) only in the interval t 0, we can rewrite it as g(t)u(t). g(t) 1 u(t) t X t = g(t)u(t) t Laplace Transforms slide-5

217 Examples Example 2: Find the L:T: of f(t) = e at t 0 where a is a constant. Solution : Hence, L[f(t)] = = Z 1 e at e ;st dt Z0 1 e ;(s;a)t dt 0 = ; e;(s;a)t (s ; a) 1 = (s ; a) > 0 (s ; a) F (s) = (s ; a) for s > a Laplace Transforms slide-6

218 Examples Note that from the denition of f(t) = e at t 0, f(t) can be written as f(t) = e at u(t) and L[e at u(t)] = 1 (s ; a) for s > a Laplace Transforms slide-7

219 Examples Example 3: Find the L:T: of f(t) = sinh(wt). First note that sinh(wt) = 1 2 [ewt ; e ;wt ] Hence L[sinh(wt)] = 1 2 = 1 2 Z Z 1 1 e!t e ;st dt ; s ; ; 1 w s + w (s + w) ; (s ; w) 0 e ;!t e ;st dt = 1 2 = s 2 ; w 2 w s 2 ; w 2 Laplace Transforms slide-8

220 Examples Example 4: Find the L:T: of f(t) = cosh(wt). First note that cosh(wt) = 1 2 [ewt + e ;wt ] Hence L[cosh(wt)] = 1 2 = 1 2 Z Z 1 1 e!t e ;st dt s ; + 1 w s + w (s + w) + (s ; w) 0 e ;!t e ;st dt = 1 2 = s 2 ; w 2 s s 2 ; w 2 Laplace Transforms slide-9

221 Linearity of Laplace Transforms 1(a) Linearity of Laplace Transforms Theorem 0.1. The Laplace transformation is a linear operation. That is for any functions f(t) and g(t) whose L:T: exist and for any constants a and b L[af(t) + bg(t)] = al[f(t)] + bl[g(t)] Proof. Z 1 LHS = [af(t) + bg(t)]e ;st dt 0Z Z 1 1 = a f(t)e ;st dt + b 0 = al[f(t)] + bl[g(t)] 0 g(t)e ;st dt Laplace Transforms slide-10

222 Examples Example 5: L[sin(wt)] = 1 2j = L " Z 1 0 e j!t e ;st dt ; 1 2j e jwt ; e ;jwt 2j # Z 1 0 e ;j!t e ;st dt = 1 2j L[ejwt ; e ;jwt ] = 1 2j 1 s ; ; 1 jw s + jw (s + jw) ; (s ; jw) = 1 2j = w s 2 + w 2 s 2 + w 2 Laplace Transforms slide-11

223 Examples Example 6: L[cos(wt)] = L " e jwt + e ;jwt 2 # = 1 2 L[ejwt + e ;jwt ] = s ; + 1 jw s + jw (s + jw) + (s ; jw) = 1 2 = s s 2 + w 2 s 2 + w 2 Laplace Transforms slide-12

224 Examples Example 7: L[t] = Z 1 0 te ;st dt Integrate by parts Revision: Integration by parts: Dierentiation of a product of two functions u(t) and v(t) - from here on simply called u and v. d dt (uv) = udv + dt vdu dt Integrate both sides w.r.t. t and rearrange: Z u dv dt dt = uv ; Z v du dt dt Laplace Transforms slide-13

225 Examples Let L[t] = Z 1 0 te ;st dt = Z 1 0 u dv dt dt ie: Let u = t and dv dt = e;st. L[t] = = " uv ; te ;st ;s = s = 1 s 2 " Z 1 v du dt dt # 0 1 Z 1 0 e ;st ;s ; 0 # 1 0 e ;st ;s! 1dt Laplace Transforms slide-14

226 Examples Example 8: L[t 2 ] = Z 1 t 2 e ;st dt Z0 1 = u dv 0 dt dt = t2 e ;st Z 1 2te ;st dt ;s Z 0 s 0 1 te ;st dt = 2 s 0 = 2 1 ss 2 = 2! s 3 It can be shown by induction that L[t n ] = n! s n+1 Laplace Transforms slide-15

227 Important Laplace Transforms f(t) F (s) = L[f(t)] = R 1 0 f(t)e ;st dt 1 u(t) 2 e at 1 3 sin(wt) 4 cos(wt) 5 sinh(wt) 6 cosh(wt) 1 s s;a w s 2 +w 2 s s 2 +w 2 w s 2 ;w 2 s s 2 ;w 2 7 t n n! s n+1 Laplace Transforms slide-16

228 Inverse Laplace Transform 1(b) Inverse Laplace Transform The procedure to obtain the inverse Laplace transform of F (s) is given in the following steps: 1. First use partial fractions to rewrite the given F (s) as a sum of simple terms for which the inverse Laplace Transform is straightforward. 2. Then take the Inverse L.T. as the sum of the Inverse L.T. of the simple terms. Laplace Transforms slide-17

229 Examples Example 9: Given F (s) = 1 (s ; a)(s ; b) Find f(t) = L ;1 [F (s)]. Solution - Use Partial Fractions: 1 (s ; a)(s ; b) = A s ; a + B s ; b Multiplying both sides by (s ; a)(s ; b), we have Let s = a, 1 = A(s ; b) + B(s ; a) 1 = A(a ; b) or A = 1 (a ; b) Laplace Transforms slide-18

230 Examples or Let s = b, 1 = B(b ; a) B = 1 (b ; a) Hence 1 (s ; a)(s ; b) = 1 1 (a ; b) (s ; a) (b ; a) (s ; b) 1 = (a ; b) [ 1 (s ; a) ; 1 (s ; b) ] Laplace Transforms slide-19

231 Examples Therefore L ;1 1 (s ; a)(s ; b) = L ;1 1 (a ; b) 1 (s ; a) ; 1 (s ; b) = 1 (a ; b) 1 L ;1 (s ; a) 1 ; L ;1 (s ; b) = 1 (a ; b) [eat ; e bt ] Laplace Transforms slide-20

232 Properties of Laplace Transforms 2. Properties of Laplace Transforms 2(a) Dierentiation Theorem 0.2. L df dt = L[f 0 (t)] = sl[f(t)] ; f(0) Proof. Consider the case where f 0 (t) is piecewise continuous for all t 0. Then L[f 0 (t)] = Z 1 e ;st 0 f (t)dt = 0 Z 1 v du dt dt = uv ; 0 = e ;st f(t) 1 0 ; Z 1 = ;f(0) + s Z 1 0 Z 0 1 e ;st f(t)dt 0 Laplace Transforms u dv dt dt f(t)(;s)e ;st dt slide-21

233 Properties of Laplace Transforms Hence L[f 0 (t)] = sl[f(t)] ; f(0) Remark 0.1. Dierentiation of a function f(t) is equivalent to multiplication of F (s) by s. Laplace Transforms slide-22

234 Properties of Laplace Transforms Similarly, L[f 00 (t)] = L[ d2 f(t) dt 2 ] Extending this, = sl[f 0 (t)] ; f 0 (0) = s[sl[f(t)] ; f(0)] ; f 0 (0) = s 2 L[f(t)] ; sf(0) ; f 0 (0) = s 2 F (s) ; sf(0) ; f 0 (0) L[f 000 (t)] = s 3 F (s) ; s 2 f(0) ; s 1 f 0 (0) ; f 00 (0) By induction, it can be shown that L[f n (t)] = s n F (s) ; s n;1 f(0) ; s n;2 f 1 (0) ; : : : ;sf n;2 (0) ; f n;1 (0) Laplace Transforms slide-23

235 Examples Example 10: f(t) = t 2 0 f (t) = 00 2t f (t) = 2 f(0) = 0 0 f (0) = 0 Using the above formula, we have 00 L[f (t)] = s 2 L[f(t)] ; sf(0) ; 0 f (0) But L[f 00 (t)] = L[2] = 2 s Hence or 2 s = s2 L[f(t)] ; sf(0) ; f 0 (0) = s 2 L[f(t)] L[f(t)] = 2 s 3 Laplace Transforms slide-24

236 Properties of Laplace Transforms 2(b) Integration Theorem 0.3. Suppose f(t) is piecewise continuous Then (Z t ) L 0 f()d = 1 L[f(t)] (s > 0 s > ) s or L ;1 1 s F (s) = Z t 0 f()d Laplace Transforms slide-25

237 Properties of Laplace Transforms Remark 0.2. Suppose G(s) = 1 s F (s) First nd Then f(t) = L ;1 [F (s)] g(t) = Z t 0 f()d Laplace Transforms slide-26

238 Examples. Example 11: Find the inverse transform of 1 G(s) = s(s 2 + w 2 ) Solution: L ;1 1 s 2 + w 2 = L ;1 1 w = 1 w sin(wt) w s 2 + w 2 Therefore 1 L ;1 1 ss 2 + w 2 = Z t 0 = 1 w 1 w sin(w)d ; cos(w) w t = 1 w2(1 ; cos(wt)) 0 Laplace Transforms slide-27

239 Properties of Laplace Transforms 2(c) Shifting on the s-axis Theorem 0.4. Suppose Then Similarly L[f(t)] = F (s) L[e ;at f(t)] = F (s + a) L ;1 [F (s + a)] = e ;at f(t) Proof. F (s + a) = = Z 1 e ;(s+a)t f(t)dt Z0 1 e ;st [e ;at f(t)]dt 0 = L[e ;at f(t)] Laplace Transforms slide-28

240 Properties of Laplace Transforms Remark 0.3. L.T. of e ;at f(t) can be found by rst nding the L.T. of f(t) (neglecting the factor e ;at ) and then changing s with s + a in F (s). Remark 0.4. Similarly and L[e at f(t)] = F (s ; a) L ;1 [F (s ; a)] = e at f(t) Laplace Transforms slide-29

241 Properties of Laplace Transforms Example 12: Given Then L[sin(wt)] = L[e ;at sin(wt)] = Example 13: Given Then L[cos(wt)] = L[e ;at cos(wt)] = w s 2 + w 2 w (s + a) 2 + w 2 s s 2 + w 2 (s + a) (s + a) 2 + w 2 Laplace Transforms slide-30

242 Properties of Laplace Transforms 2(d) Shifting on the t-axis Theorem 0.5. Suppose Then or L[f(t)] = F (s) L[f(t ; a)u(t ; a)] = e ;as F (s) L ;1 [e ;as F (s)] = f(t ; a)u(t ; a) Remark 0.5. f(t ; a)u(t ; a) describes the result of translating f(t) to the right by an amount a and setting the function to zero for all t < a. Laplace Transforms slide-31

243 Properties of Laplace Transforms f(t) 1 u(t-1) 1 t 1 t = f(t-1)u(t-1) 1 t Laplace Transforms slide-32

244 Properties of Laplace Transforms Example 14: Suppose f(t) = 0 t < 1 = t 2 ; 3t + 2 t 1 Then f(t) = (t 2 ; 3t + 2)u(t ; 1) = [(t ; 1) 2 ; (t ; 1)]u(t ; 1) Let g(t) = t 2 ; t. Therefore L[f(t)] = L[g(t ; 1)u(t ; 1)] = e ;s L[g(t)] = e ;s [ 2 ; 1 s 3 s 2] Laplace Transforms slide-33

245 Properties of Laplace Transforms Consider the following dierent functions :- 1. f(t ; t 0 ) 2. f(t ; t 0 )u(t) 3. f(t)u(t ; t 0 ) 4. f(t ; t 0 )u(t ; t 0 ) All the functions are dierent. Let f(t) = sin(wt). Laplace Transforms slide-34

246 Properties of Laplace Transforms 1 2 sin(ω(t-t )) 0 sin(ω (t-t ))u(t) 0 t 0 0 t t t 3 4 sin(ω t)u(t-t 0 ) sin( ω (t-t ))u(t-t 0 ) 0 t 0 t t 0 t Laplace Transforms slide-35

247 Properties of Laplace Transforms Example 15: Find the L.T. of the three functions (graphs 2, 3, and 4). Graph 2: L[sin(w(t ; t 0 ))u(t)] = L[sin(w(t ; t 0 ))] = L[sin(wt) cos(wt 0 ) ; cos(wt) sin(wt 0 )] w = cos(wt 0 ) s 2 + ; w 2 sin(wt 0) = cos(wt 0)w ; sin(wt 0 )s s 2 + w 2 s s 2 + w 2 Laplace Transforms slide-36

248 Properties of Laplace Transforms Graph 3: L[sin(wt)u(t ; t 0 )] = Z 1 t0 sin(wt)e ;st dt = 1 2j Z 1 [e (;s+jw)t ; e (;s;jw)t ]dt t0 = 1 2j [e(;s+jw)t 0 s ; jw ; e (;s;jw)t0 s + ] jw = e ;t 0 s [ cos(wt 0)w + sin(wt 0 )s ] (s 2 + w 2 ) Graph 4: L[sin(w(t ; t 0 ))u(t ; t 0 )] = e ;t 0 s L[sin(wt)] = e ;t 0 s w (s 2 + w 2 ) Note that the Laplace Transforms are all dierent. Laplace Transforms slide-37