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MA101-Lecturenotes(2019-20)-Module 13 (1).pdf

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MA101-Lecturenotes(2019-20)-Module 13 (1).pdf

1. 1. Scalar Fields and Vector fields
2. 2. Definition • A scalar field is an assignment of a scalar to each point in region in the space. E.g. the temperature at a point on the earth is a scalar field. • A vector field is an assignment of a vector to each point in a region in the space. e.g. the velocity field of a moving fluid is a vector field as it associates a velocity vector to each point in the fluid.
3. 3. Definition • A scalar field is a map from D to ℜ, where D is a subset of ℜn. • A vector field is a map from D to ℜn, where D is a subset of ℜn. is a subset of ℜn. • For n=2: vector field in plane, • for n=3: vector field in space • Example: Gradient field
4. 4. Line integral • Line integral in a scalar field • Line integral in a vector field
5. 5. LINE INTEGRAL IN A SCALAR FIELD MOTIVATION A rescue team follows a path in a danger area where for each position the degree of radiation is defined. Compute the total amount of radiation gathered by the rescue team along the path. path. RESCUE BASE
6. 6. Piecewise Smooth Curves
7. 7. Piecewise Smooth Curves A classic property of gravitational fields is that, subject to certain physical constraints, the work done by gravity on an object moving between two points in the field is independent of the path taken by the object. One of the constraints is that the path must be a piecewise smooth curve. Recall that a plane curve C given by smooth curve. Recall that a plane curve C given by r(t) = x(t)i + y(t)j, a ≤ t ≤ b is smooth if are continuous on [a, b] and not simultaneously 0 on (a, b).
8. 8. Piecewise Smooth Curves Similarly, a space curve C given by r(t) = x(t)i + y(t)j + z(t)k, a ≤ t ≤ b is smooth if is smooth if are continuous on [a, b] and not simultaneously 0 on (a, b). A curve C is piecewise smooth if the interval [a, b] can be partitioned into a finite number of subintervals, on each of which C is smooth.
9. 9. Example 1 – Finding a Piecewise Smooth Parametrization Find a piecewise smooth parametrization of the graph of C shown in Figure.
10. 10. Because C consists of three line segments C1, C2, and C3, you can construct a smooth parametrization for each segment and piece them together by making the last t-value in Ci correspond to the first t- value in Ci + 1, as follows. Example 1 – Solution value in Ci + 1, as follows.
11. 11. So, C is given by Example 1 – Solution Because C1, C2, and C3 are smooth, it follows that C is piecewise smooth.
12. 12. Parametrization of a curve induces an orientation to the curve. For instance, in Example 1, the curve is Piecewise Smooth Curves For instance, in Example 1, the curve is oriented such that the positive direction is from (0, 0, 0), following the curve to (1, 2, 1).
13. 13. Line Integrals You will study a new type of integral called a line integral for which you integrate over a piecewise smooth curve C. To introduce the concept of a line integral, consider the mass of a wire of finite length, given by a curve C in space. The density (mass per unit length) of the wire at the point (x, y, z) is given by f(x, y, z).
14. 14. Line Integrals Partition the curve C by the points P0, P1, …, Pn producing n subarcs, as shown in Figure. producing n subarcs, as shown in Figure.
15. 15. Line Integrals The length of the ith subarc is given by ∆si. Next, choose a point (xi, yi, zi) in each subarc. If the length of each subarc is small, the total mass of the wire can be approximated by the sum of the wire can be approximated by the sum If you let ||∆|| denote the length of the longest subarc and let ||∆|| approach 0, it seems reasonable that the limit of this sum approaches the mass of the wire.
16. 16. Line Integrals
17. 17. Line Integrals To evaluate a line integral over a plane curve C given by r(t) = x(t)i + y(t)j, use the fact that A similar formula holds for a space curve.
18. 18. Line Integrals Note that if f(x, y, z) = 1, the line integral gives the arc length of the curve C. That is,
19. 19. Evaluate where C is the line segment shown in Figure. Example 2 – Evaluating a Line Integral
20. 20. Begin by writing a parametric form of the equation of the line segment: x = t, y = 2t, and z = t, 0 ≤ t ≤ 1. Example 2 – Solution Therefore, x'(t) = 1, y'(t) = 2, and z'(t) = 1, which implies that
21. 21. So, the line integral takes the following form. Example 2 – Solution
22. 22. Line Integrals For parametrizations given by r(t) = x(t)i + y(t)j + z(t)k, it is helpful to remember the form of ds as
23. 23. • Just as for an ordinary single integral, we can interpret the line integral of a positive function as an area. • In fact, if f(x, y) ≥ 0, represents the area of one side of the “fence” or “curtain” shown here, ( ) , C f x y ds ∫ shown here, whose: – Base is C. – Height above the point (x, y) is f(x, y).
24. 24. • Now, let C be a piecewise-smooth curve. – That is, C is a union of a finite number of smooth curves C1, C2, …, Cn, where the initial point of Ci+1 is the terminal point of Ci.
25. 25. • Then, we define the integral of f along C as the sum of the integrals of f along each of the smooth pieces of C: ( ) , f x y ds ∫ ( ) ( ) ( ) ( ) 1 2 , , , ... , n C C C C f x y ds f x y ds f x y ds f x y ds = + + + ∫ ∫ ∫ ∫
26. 26. LINE INTEGRAL IN A VECTOR FIELD MOTIVATION A ship sails from an island to another one along a fixed route. Knowing all the sea currents, how much fuel will be needed ?
27. 27. One of the most important physical applications of line integrals is that of finding the work done on an object moving in a force field. For example, Figure shows an inverse square force field similar to the gravitational field of the sun. sun.
28. 28. Line Integrals of Vector Fields To see how a line integral can be used to find work done in a force field F, consider an object moving along a path C in the field, as shown in Figure. Figure15.13
29. 29. Line Integrals of Vector Fields To determine the work done by the force, you need consider only that part of the force that is acting in the same direction as that in which the object is moving. This means that at each point on C, you can consider the projection F T of the force vector F onto the unit tangent projection F T of the force vector F onto the unit tangent vector T. On a small subarc of length ∆si, the increment of work is ∆Wi = (force)(distance) ≈ [F(xi, yi, zi) T(xi, yi, zi)] ∆si where (xi, yi, zi) is a point in the ith subarc.
30. 30. Line Integrals of Vector Fields Consequently, the total work done is given by the following integral. This line integral appears in other contexts and is the basis of the following definition of the line integral of a vector field. Note in the definition that
31. 31. Line Integrals of Vector Fields
32. 32. Find the work done by the force field on a particle as it moves along the helix given by Example – Work Done by a Force from the point (1, 0, 0) to (–1, 0, 3π), as shown in Figure.
33. 33. Because r(t) = x(t)i + y(t)j + z(t)k = cos ti + sin tj + tk it follows that x(t) = cos t, y(t) = sin t, and z(t) = t. Example – Solution So, the force field can be written as
34. 34. To find the work done by the force field in moving a particle along the curve C, use the fact that r'(t) = –sin ti + cos tj + k and write the following. Example – Solution
35. 35. Line Integrals of Vector Fields For line integrals of vector functions, the orientation of the curve C is important. If the orientation of the curve is reversed, the If the orientation of the curve is reversed, the unit tangent vector T(t) is changed to –T(t), and you obtain
36. 36. Line Integrals in Differential Form Line Integrals in Differential Form
37. 37. Line Integrals in Differential Form A second commonly used form of line integrals is derived from the vector field notation used in the preceding section. If F is a vector field of the form F(x, y) = Mi + Nj, and C is given by r(t) = x(t)i + y(t)j, then F • dr is often written as given by r(t) = x(t)i + y(t)j, then F • dr is often written as M dx + N dy.
38. 38. Line Integrals in Differential Form This differential form can be extended to three variables. The parentheses are often omitted, as follows.
39. 39. Example – Evaluating a Line Integral in Differential Form Let C be the circle of radius 3 given by r(t) = 3 cos ti + 3 sin tj, 0 ≤ t ≤ 2π as shown in Figure. Evaluate the line integral
40. 40. Example – Solution Because x = 3 cos t and y = 3 sin t, you have dx = –3 sin t dt and dy = 3 cos t dt. So, the line integral is
41. 41. Example – Solution
42. 42. Suppose instead of being a force field, suppose that F represents the velocity field of a fluid flowing through a region in space. Under these circumstances, the integral of F .T along a curve in the region gives the fluid’s flow along the curve.
43. 43. EXAMPLE: Finding Circulation Around a Circle
44. 44. Flux Across a Plane Curve To find the rate at which a fluid is entering or leaving a region enclosed by a smooth curve C in the xy-plane, we calculate the line integral over C of F.n, the scalar component of the fluid’s velocity field in the direction of the curve’s outward-pointing normal vector.
45. 45. Notice the difference between flux and circulation: Flux is the integral of the normal component of F; circulation is the integral of the tangential component of F. the tangential component of F.
46. 46. How to evaluate Flux of F across C we choose a smooth parameterization that traces the curve C exactly once as t increases from a to b. We can find the outward unit normal vector n by crossing the curve’s vector n by crossing the curve’s unit tangent vector T with the vector k. • If the motion is clockwise, k×T points outward; • if the motion is counterclockwise, T×k points outward We choose: n = T × k
47. 47. Now, Here the circle on the integral shows that the integration around the closed curve C is to be in the counterclockwise direction.
48. 48. EXAMPLE: Finding Flux Across a Circle Note that the flux of F across the circle is positive, implies the net flow across the curve is outward. A net inward flow would have given a negative flux.
49. 49. Path Independence Under differentiability conditions, a field F is conservative iff it is the gradient field of a scalar function ƒ; i.e., iff for some ƒ. The function ƒ then has a special name.
50. 50. once we have found a potential function ƒ for a field F, we can evaluate all the work integrals in the domain of F over any path between A and B by
51. 51. Connectivity and Simple Connectivity • All curves are piecewise smooth, that is, made up of finitely many smooth pieces connected end to end. • The components of F have continuous first • The components of F have continuous first partial derivatives implies that when this continuity requirement guarantees that the mixed second derivatives of the potential function ƒ are equal.
52. 52. Simple curve: A curve that doesn’t intersect itself anywhere between its endpoints. r(a) = r(b) for a simple closed curve But r(a) ≠ r(b) when a t1 t2 b
53. 53. Simply-connected region: A simply-connected region in the plane is a connected region D such that every simple closed curve in D encloses only points that are in D. Intuitively speaking, a simply-connected region contains no hole and can’t consist of two separate pieces. of two separate pieces. An open connected region means that every point can be connected to every other point by a smooth curve that lies in the region. Note that connectivity and simple connectivity are not the same, and neither implies the other. Think of connected regions as being in “one piece” and simply connected regions as not having any “holes that catch loops.”
54. 54. INDEPENDENCE OF PATH Suppose C1 and C2 are two piecewise-smooth curves (which are called paths) that have the same initial point A and terminal point B. We have Note: F is conservative on D is equivalent to saying that the integral of F around every closed path in D is zero. In other words, the line integral of a conservative vector field depends only on the initial point and terminal point of a curve.
55. 55. EXAMPLE: Finding Work Done by a Conservative Field
56. 56. Proof that Part 1 Part 2 Proof that Part 1 Part 2 If we have two paths from A to B, one of them can be reversed to make a loop.
57. 57. Finding Potentials for Conservative Fields
58. 58. EXAMPLE: Finding a Potential Function
59. 59. Integrating first equation w.r.t. ‘x’ Differentiating it w.r.t. ‘y’ and equating with the Computing ‘g’ as a function of ‘y’ gives Computing ‘g’ as a function of ‘y’ gives Further differentiating ‘f’ w.r.t. to ‘z’ and equating it with Integrating, we have
60. 60. EXAMPLE: Showing That a Differential Form Is Exact