Vector Calculus CHAPTER 9.5 ~ 9.9

Ch9.5~9.9_2 Contents  9.5 Directional Derivatives 9.5 Directional Derivatives  9.6 Tangent Planes and Normal Lines 9.6 Tangent Planes and Normal Lines  9.7 Divergence and Curl 9.7 Divergence and Curl  9.8 Lines Integrals 9.8 Lines Integrals  9.9 Independence of Path 9.9 Independence of Path

Ch9.5~9.9_3 9.5 Directional Derivative  Introduction See Fig 9.26.

Ch9.5~9.9_4 The Gradient of a Function  Define the vector differential operator as then (1) (2) are the gradients of the functions.

Ch9.5~9.9_5 Example 1 Compute Solution

Ch9.5~9.9_6 Example 2 If F(x, y, z) = xy 2 + 3x 2 – z 3, find the gradient at (2, –1, 4). Solution

Ch9.5~9.9_7 The directional derivative of z = f(x, y) in the direction of a unit vector u = cos  i + sin  j is (4) provided the limit exists. DEFINITION 9.5 Directional Derivatives

Ch9.5~9.9_8 Fig 9.27

Ch9.5~9.9_9 Proof Let x, y and  be fixed, then g(t) = f(x + t cos , y + t sin  ) is a function of one variable. If z = f(x, y) is a differentiable function of x and y, and u = cos  i + sin  j, then (5) THEOREM 9.6 Computing a Directional Derivative

Ch9.5~9.9_10 First Second by chain rule

Ch9.5~9.9_11 Here the subscripts 1 and 2 refer to partial derivatives of f(x + t cos , y + t sin  ) w.s.t. (x + t cos  ) and (y + t sin  ). When t = 0, x + t cos  and y + t sin  are simply x and y, then (7) becomes (8) Comparing (4), (6), (8), we have

Ch9.5~9.9_12 Example 3 Find the directional derivative of f(x, y) = 2x 2 y 3 + 6xy at (1, 1) in the direction of a unit vector whose angle with the positive x-axis is  /6. Solution

Ch9.5~9.9_13 Example 3 (2) Now,  =  /6, u = cos  i + sin  j becomes Then

Ch9.5~9.9_14 Example 4 Consider the plane perpendicular to xy-plane and passes through P(2, 1), Q(3, 2). What is the slope of the tangent line to the curve of intersection of this plane and f(x, y) = 4x 2 + y 2 at (2, 1, 17) in the direction of Q. Solution We want D u f(2, 1) in the direction given by, and form a unit vector

Ch9.5~9.9_15 Example 4 (2) Now then the requested slope is

Ch9.5~9.9_16 Functions of Three Variables  where , ,  are the direction angles of the vector u measured relative to the positive x, y, z axis. But as before, we can show that (9)

Ch9.5~9.9_17  Since u is a unit vector, from (10) in Sec 7.3 that In addition, (9) shows

Ch9.5~9.9_18 Example 5 Find the directional derivative of F(x, y, z) = xy 2 – 4x 2 y + z 2 at (1, –1, 2) in the direction 6i + 2j + 3k. Solution Since we have

Ch9.5~9.9_19 Example 5 (2) Since ‖ 6i + 2j + 3k ‖ = 7, then u = (6/7)i + (2/7)j + (3/7)k is a unit vector. It follows from (9) that

Ch9.5~9.9_20 Maximum Value of the Direction Derivative  From the fact that where  is the angle between and u. Because then

Ch9.5~9.9_21  In other words, The maximum value of the direction derivative is and it occurs when u has the same direction as (when cos  = 1), (10) and The minimum value of the direction derivative is and it occurs when u has opposite direction as (when cos  = −1) (11)

Ch9.5~9.9_22 Example 6  In Example 5, the maximum value of the directional derivative at (1, −1, 2) is and the minimum value is.

Ch9.5~9.9_23 Gradient points in Direction of Most Rapid Increase of f  Put another way, (10) and (11) state The gradient vector points in the direction in which f increase most rapidly, whereas points in the direction of the most rapid decrease of f.

Ch9.5~9.9_24 Example 7  Each year in L.A. there is a bicycle race up to the top of a hill by a road known to be the steepest in the city. To understand why a bicyclist with a modicum of sanity will zigzag up the road, let us suppose the graph of shown in Fig 9.28(a) is a mathematical model of the hill. The gradient of f is

Ch9.5~9.9_25 Example 7 (2) where r = – xi – yj is a vector pointing to the center of circular base. Thus the steepest ascent up the hill is a straight road whose projection in the xy-plane is a radius of the circular base. Since a bicyclist will zigzag a direction u other than to reduce this component. See Fig 9.28.

Ch9.5~9.9_26 Fig 9.28

Ch9.5~9.9_27 Example 8 The temperature in a rectangular box is approximated by If a mosquito is located at ( ½, 1, 1), in which the direction should it fly up to cool off as rapidly as possible?

Ch9.5~9.9_28 Example 8 (2) Solution The gradient of T is Therefore, To cool off most rapidly, it should fly in the direction −¼k, that is, it should dive for the floor of the box, where the temperature is T(x, y, 0) = 0

Ch9.5~9.9_ Tangent Plane and Normal Lines  Geometric Interpretation of the Gradient : Functions of Two Variables Suppose f(x, y) = c is the level curve of z = f(x, y) passes through P(x 0, y 0 ), that is, f(x 0, y 0 ) = c. If x = g(t), y = h(t) such that x 0 = g(t 0 ), y 0 = h(t 0 ), then the derivative of f w.s.t. t is (1) When we introduce

Ch9.5~9.9_30  then (1) becomes When at t = t 0, we have (2) Thus, if, is orthogonal to at P(x 0, y 0 ). See Fig 9.30.

Ch9.5~9.9_31 Fig 9.30

Ch9.5~9.9_32 Example 1 Find the level curves of f(x, y) = −x 2 + y 2 passing through (2, 3). Graph the gradient at the point. Solution Since f(2, 3) = 5, we have −x 2 + y 2 = 5. Now See Fig 9.31.

Ch9.5~9.9_33 Fig 9.31

Ch9.5~9.9_34 Geometric Interpretation of the Gradient : Functions of Three Variables  Similar concepts to two variables, the derivative of F(f(t), g(t), h(t)) = c implies (3) In particular, at t = t 0, (3) is (4) See Fig 9.32.

Ch9.5~9.9_35 Fig 9.32

Ch9.5~9.9_36 Example 2 Find the level surfaces of F(x, y, z) = x 2 + y 2 + z 2 passing through (1, 1, 1). Graph the gradient at the point. Solution Since F(1, 1, 1) = 3 ， then x 2 + y 2 + z 2 = 3 See Fig 9.33.

Ch9.5~9.9_37 Fig 9.33

Ch9.5~9.9_38 Let P(x 0, y 0, z 0 ) is a point on the graph of F(x, y, z) = c, where  F is not 0. The tangent plane at P is a plane through P and is perpendicular to  F evaluated at P. DEFINITION 9.6 Tangent Plane

Ch9.5~9.9_39 That is,. See Fig Let P(x 0, y 0, z 0 ) is a point on the graph of F(x, y, z) = c, where  F is not 0. Then an equation of this tangent plane at P is F x (x 0, y 0, z 0 )(x – x 0 ) + F y (x 0, y 0, z 0 )(y – y 0 ) + F z (x 0, y 0, z 0 )(z – z 0 ) = 0 (5) THEOREM 2.1 Criterion for an Extra Differential

Ch9.5~9.9_40 Fig 9.34

Ch9.5~9.9_41 Example 3 Find the equation of the tangent plane to x 2 – 4y 2 + z 2 = 16 at (2, 1, 4). Solution F(2, 1, 4) = 16, the did graph passes (2, 1, 4). Now F x (x, y, z) = 2x, F y (x, y, z) = – 8y, F z (x, y, z)= 2z, then From (5) we have the equation: 4(x – 2) – 8(y – 1) + 8(z – 4) = 0 or x – 2y + 2z = 8.

Ch9.5~9.9_42 Surfaces Given by z = f(x, y)  When the equation is given by z = f(x, y), then we can set F = z – f(x, y) or F = f(x, y) – z.

Ch9.5~9.9_43 Example 4 Find the equation of the tangent plane to z = ½x 2 + ½ y at (1, –1, 5). Solution Let F(x, y, z) = ½x 2 + ½ y 2 – z + 4. This graph did pass (1, –1, 5), since F(1, –1, 5) = 0. Now F x = x, F y = y, F z = –1, then From (5), the desired equation is (x + 1) – (y – 1) – (z – 5) = 0 or –x + y + z = 7

Ch9.5~9.9_44 Fig 9.35

Ch9.5~9.9_45 Normal Line  Let P(x 0, y 0, z 0 ) is on the graph of F(x, y, z) = c, where  F  0. The line containing P that is parallel to  F(x 0, y 0, z 0 ) is called the normal line to the surface at P.

Ch9.5~9.9_46 Example 5 Find parametric equations for the normal line to the surface in Example 4 at (1, –1, 5). Solution A direction vector for the normal line at (1, –1, 5) is  F(1, –1, 5) = i – j – k then the desired equations are x = 1 + t, y = –1– t, z = 5 – t

Ch9.5~9.9_ Divergence and Curl  Vector Functions F(x, y) = P(x, y)i+ Q(x, y)j F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k

Ch9.5~9.9_48 Fig 9.37 (a) ~ (b)

Ch9.5~9.9_49 Fig 9.37 (c) ~ (d)

Ch9.5~9.9_50 Example 1 Graph F(x, y) = – yi + xj Solution Since let For and k = 2, we have (i) x 2 + y 2 = 1 ： at (1, 0), (0, 1), (–1, 0), (0, –1), the corresponding vectors j, –i,– j, i have the same length 1.

Ch9.5~9.9_51 Example 1 (2) (ii) x 2 + y 2 = 2 ： at (1, 1), (–1, 1), (–1, –1), (1, –1), the corresponding vectors – i + j, – i – j, i – j, i + j have the same length. (iii) x 2 + y 2 = 4 ： at (2, 0), (0, 2), (–2, 0), (0, –2), the corresponding vectors 2j, –2i, –2j, 2i have the same length 2. See Fig 9.38.

Ch9.5~9.9_52 Example 1 (3)

Ch9.5~9.9_53  In practice, we usually use this form: (1) The curl of a vector field F = Pi + Qj + Rk is the vector field DEFINITION 9.7 Curl

Ch9.5~9.9_54  Observe that we can also use this form: (4) The divergence of a vector field F = Pi + Qj + Rk is the scalar function DEFINITION 9.8 Divergence

Ch9.5~9.9_55 Example 2 If F = (x 2 y 3 – z 4 )i + 4x 5 y 2 zj – y 4 z 6 k, find curl F and div F 。 Solution

Ch9.5~9.9_56 Example 2 (2)

Ch9.5~9.9_57 Please prove  If f is a scalar function with continuous second partial derivatives, then (5) If F is a vector field with continuous second partial derivatives, then (6)

Ch9.5~9.9_58 Physical Interpretations  See Fig The curl F is a measure of tendency of the fluid to turn the device about its vertical axis w. If curl F = 0, then the flow of the fluid is said to be irrotational. Also see Fig 9.42.

Ch9.5~9.9_59 Fig 9.41

Ch9.5~9.9_60 Fig 9.42

Ch9.5~9.9_61  From definition 9.8 we saw that div F near a point is the flux per unit volume. (i) If div F(P) > 0: source for F. (ii) If div F(P) < 0: sink for F. (iii) If div F(P) = 0: no sources or sinks near P. Besides, If   F = 0: incompressible or solenoidal. See Fig 9.43.

Ch9.5~9.9_62 Fig 9.43

Ch9.5~9.9_ Line Integrals  Terminology In Fig 9.46, we show four new terminologies.  Fig 9.46.

Ch9.5~9.9_64 Line Integral in the Plane  If C is a smooth curve defined by x = f(t), y = g(t), a  t  b. Since dx = f’(t) dt, dy = g’(t) dt, and (which is called the differential of arc length), then we have (1) (2) (3)

Ch9.5~9.9_65 Example 1 Evaluate (a) (b) (c) on the ¼ circle C defined by Fig 9.47:

Ch9.5~9.9_66 Example 1 (2) Solution (a)

Ch9.5~9.9_67 Example 1 (3) (b)

Ch9.5~9.9_68 Example 1 (4) (c)

Ch9.5~9.9_69 Method of Evaluation  If the curve C is defined by y = f(x), a  x  b, then we have dy = f ’(x) dx and Thus (4) (5) (6)  Note: If C is composed of two smooth curves C 1 and C 2, then

Ch9.5~9.9_70  Notation: In many applications, we have we usually write as or(7) A line integral along a close curve:

Ch9.5~9.9_71 Example 2 Evaluate where C: Solution See Fig Using dy = 3x 2 dx,

Ch9.5~9.9_72 Fig 9.48

Ch9.5~9.9_73 Example 3 Evaluate, where C: Solution

Ch9.5~9.9_74 Example 4 Evaluate, where C is shown in Fig 9.49(a). Solution Since C is piecewise smooth, we express the integral as See Fig 9.49(b).

Ch9.5~9.9_75 Fig 9.49

Ch9.5~9.9_76 Example 4 (2) (i) On C 1, we use x as a parameter. Since y = 0, dy = 0, (ii) On C 2, we use y as a parameter. Since x = 2, dx = 0,

Ch9.5~9.9_77 (iii) On C 3, we use x as a parameter. Since y = x 2, dy = 2x dx, Hence, Example 4 (3)

Ch9.5~9.9_78  Note: See Fig 9.50, where −C denote the curve having opposite orientation, then Equivalently, (8) For example, in (a) of Example 1,

Ch9.5~9.9_79 Fig 9.50

Ch9.5~9.9_80 Lines Integrals in Space

Ch9.5~9.9_81 Method to Evaluate Line Integral in Space  If C is defined by then we have Similar method can be used for

Ch9.5~9.9_82 and We usually use the following form

Ch9.5~9.9_83 Example 5 Evaluate, where C is Solution Since we have we get

Ch9.5~9.9_84 Another Notation  Let r(t) = f(t)i + g(t)j, then dr(t)/dt = f’(t)i + g’(t)j = (dx/dt)i + (dy/dt)j Now if F(x, y) = P(x, y)i + Q(x, y)j thus (10) When on a space (11) where F(x, y, z) = Pi + Qj + Rk dr = dx i + dy j + dz k,

Ch9.5~9.9_85 Work  If A and B are the points (f(a), g(a)) and (f(b), g(b)). Suppose C is divided into n subarcs of lengths ∆s k. On each subarc F(x * k, y * k ) is a constant force. See Fig 5.91(a).  If, as shown in Figure 9.51(b), the length of the vector is an approximation to the length of the kth subarc, then the approximate work done by F over the subarc is

Ch9.5~9.9_86  The work done by F along C is as the line integral or(12) Since we let dr = Tds, where T = dr / ds is a unit tangent to C. (13) The work done by a force F along a curve C is due entirely to the tangential component of F.

Ch9.5~9.9_87 Fig 9.51

Ch9.5~9.9_88 Example 6 Find the work done by (a) F(x, y) = xi + yj (b) F = (¾ i + ½ j) along the curve C traced by r(t) = cos ti + sin tj, from t = 0 to t = . Solution (a) dr(t) = (− sin ti + cos tj) dt, then

Ch9.5~9.9_89 Fig 9.52

Ch9.5~9.9_90 Example 6 (2) (b) See Fig 9.53.

Ch9.5~9.9_91 Fig 9.53

Ch9.5~9.9_92  Note: Let dr = T ds, where T = dr / ds, then  Circulation of F around C Circulation

Ch9.5~9.9_93 Fig 9.54

Ch9.5~9.9_ Independence of Path  Differential For two variables: For three variables:

Ch9.5~9.9_95 Path Independence  A line integral whose value is the same for every curve or path connecting A and B.

Ch9.5~9.9_96 Example 1  has the same value on each path between (0, 0) and (1, 1) shown in Fig Recall that

Ch9.5~9.9_97 Fig 9.65

Ch9.5~9.9_98 Suppose there exists a  (x, y) such that d  = Pdx + Qdy, that is, Pdx + Qdy is an exact differential. Then depends on only the endpoints A and B, and THEOREM 9.8 Fundamental Theorem for Line Integrals

Ch9.5~9.9_99 THEOREM 9.8 Proof  Let C be a smooth curve: The endpoints are (f(a), g(a)) and (f(b), g(b)), then

Ch9.5~9.9_100 Two facts  (i) This is also valid for piecewise smooth curves. (ii) The converse of this theorem is also true. is independent of path iff P dx + Q dy is an exact differential.(1)  Notation for a line integral independent of path:

Ch9.5~9.9_101 Example 2  Since d(xy) = y dx + x dy, y dx + x dy is an exact differential. Hence is independent of path. Especially, if the endpoints are (0, 0) and (1, 1), we have

Ch9.5~9.9_102 Simply Connected Region in the Plane  Refer to Fig Besides, a simply connected region is open if it contains no boundary points.

Ch9.5~9.9_103 Fig 9.66

Ch9.5~9.9_104 Let P and Q have continuous first partial derivatives in an open simply connected region. Then is independent of the path C if and only if for all (x, y) in the region. THEOREM 9.9 Test for Path Independence

Ch9.5~9.9_105 Example 3 Show that is not independent of path C. Solution We have P = x 2 – 2y 3 and Q = x + 5y, then and Since, we complete the proof.

Ch9.5~9.9_106 Example 4 Show that is independent of any path between (−1, 0) and (3, 4). Evaluate. Solution We have P = y 2 – 6xy + 6 and Q = 2xy – 3x 2, then and This is an exact differential.

Ch9.5~9.9_107 Example 4 (2) Suppose there exists a  such that Integrating the first, we have then we have, g(y) = C.

Ch9.5~9.9_108 Example 4 (3) Since d(y 2 x – 3x2y + 6x + C) = d(y2x – 3x2y + 6x), we simply use then

Ch9.5~9.9_109 Another Approach  We know y = x + 1 is one of the paths connecting (−1, 0) and (3, 4). Then

Ch9.5~9.9_110 Conservative Vector Fields  If a vector field is independent of path, we have where F = Pi + Qj is a vector field and  In other words, F is the gradient of . Since F = , F is said to be a gradient field and the  is called the potential function of F. Besides, we all call this kind of vector field to be conservative.

Ch9.5~9.9_111 Example 5 Show that F = (y 2 + 5)i + (2xy – 8)j is a gradient field. Find a potential function for F. Solution Since then we have

Ch9.5~9.9_112 Let P, Q, and R have continuous first partial derivatives in an open simply connected region of space. Then is independent of the path C if and only if THEOREM 9.10 Test for Path Independence

Ch9.5~9.9_113 Example 6 Show that is independent of path between (1, 1, 1) and (2, 1, 4). Solution Since it is independent of path.

Ch9.5~9.9_114 Example 6 (2) Suppose there exists a function , such that Integrating the first w.s.t. x, then It is the fact thus

Ch9.5~9.9_115 Example 6 (3) Now and we have and h(z) = – z + C. Disregarding C, we get (2)

Ch9.5~9.9_116 Example 6 (4) Finally,

Ch9.5~9.9_117  From example 6, we know that F is a conservative vector field, and can be written as F = . Remember in Sec 9.7, we have  = 0, thus F is a conservative vector field iff   F = 0