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CHAPTER 9.5 ~ 9.9. Vector Calculus. Contents. 9.5 Directional Derivatives 9.6 Tangent Planes and Normal Lines 9.7 Divergence and Curl 9.8 Lines Integrals 9.9 Independence of Path. 9.5 Directional Derivative. Introduction See Fig 9.26. The Gradient of a Function.

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## Vector Calculus

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**CHAPTER 9.5 ~ 9.9**Vector Calculus**Contents**• 9.5 Directional Derivatives • 9.6 Tangent Planes and Normal Lines • 9.7 Divergence and Curl • 9.8 Lines Integrals • 9.9 Independence of Path**9.5 Directional Derivative**• IntroductionSee Fig 9.26.**The Gradient of a Function**• Define the vector differential operator asthen (1) (2)are the gradients of the functions.**Example 1**Compute Solution**Example 2**If F(x, y, z) = xy2 + 3x2 – z3, find the gradient at (2, –1, 4). Solution**DEFINITION 9.5**Directional Derivatives The directional derivative of z = f(x, y) in the direction of a unit vector u = cos i + sin jis (4)provided the limit exists.**THEOREM 9.6**ProofLet 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 xand y, and u = cos i + sin j, then (5) Computing a Directional Derivative**First**Second by chain rule**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**Example 3**Find the directional derivative of f(x, y) = 2x2y3 + 6xyat (1, 1) in the direction of a unit vector whose angle with the positive x-axis is /6. Solution**Example 3 (2)**Now, = /6, u = cos i + sin j becomesThen**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) = 4x2 + y2 at (2, 1, 17) in the direction of Q. SolutionWe want Duf(2, 1)in the direction given by ,and form a unit vector**Example 4 (2)**Nowthen the requested slope is**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)**Since u is a unit vector, from (10) in Sec 7.3 thatIn**addition, (9) shows**Example 5**Find the directional derivative of F(x,y, z) = xy2 – 4x2y + z2at (1, –1, 2) in the direction 6i + 2j + 3k. SolutionSincewe have**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**Maximum Value of the Direction Derivative**• From the fact thatwhere is the angle between and u. Becausethen**In other words, The maximum value of the direction**derivative is and it occurs when uhas the same direction as (when cos = 1), (10)andThe minimum value of the direction derivative is and it occurs when uhas opposite direction as (when cos = −1)(11)**Example 6**• In Example 5, the maximum value of the directional derivative at (1, −1, 2) is and the minimum value is .**Gradient points in Direction of Most Rapid Increase of f**• Put another way, (10) and (11) stateThe gradient vector points in the direction in which f increase most rapidly, whereas points in the direction of the most rapid decrease of f.**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 ofshown in Fig 9.28(a) is a mathematical model of the hill. The gradient of f is**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. Sincea bicyclist will zigzag a direction u other than to reduce this component. See Fig 9.28.**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?**Example 8 (2)**SolutionThe 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**9.6 Tangent Plane and Normal Lines**• Geometric Interpretation of the Gradient : Functions of Two VariablesSuppose f(x, y) = c is the level curve of z = f(x, y) passes through P(x0,y0),that is, f(x0, y0) = c.If x = g(t), y = h(t) such that x0 = g(t0), y0= h(t0),then the derivative of f w.s.t. t is (1)When we introduce**then (1) becomes When at t = t0, we have (2)Thus, if**, is orthogonal toat P(x0, y0).See Fig 9.30.**Example 1**Find the level curves of f(x, y) = −x2+ y2 passing through (2, 3). Graph the gradient at the point. Solution Since f(2, 3) = 5, we have −x2 + y2= 5.NowSee Fig 9.31.**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 = t0, (3) is (4)See Fig 9.32.**Example 2**Find the level surfaces of F(x, y, z) = x2 + y2 + z2 passing through (1, 1, 1). Graph the gradient at the point. SolutionSince F(1, 1, 1) = 3，then x2 + y2 + z2 = 3See Fig 9.33.**DEFINITION 9.6**Tangent Plane Let P(x0, y0, z0) 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.**That is, . See Fig 9.34.**THEOREM 2.1 Let P(x0, y0, z0) 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 isFx(x0, y0, z0)(x – x0) + Fy(x0, y0, z0)(y – y0) + Fz(x0, y0, z0)(z – z0) = 0 (5) Criterion for an Extra Differential**Example 3**Find the equation of the tangent plane to x2 – 4y2 + z2 = 16 at (2, 1, 4). SolutionF(2, 1, 4) = 16, the did graph passes (2, 1, 4). Now Fx(x, y, z) = 2x, Fy(x, y, z) = – 8y, Fz(x, y, z)= 2z, thenFrom (5) we have the equation: 4(x – 2) – 8(y – 1) + 8(z – 4) = 0 or x – 2y + 2z = 8.**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.**Example 4**Find the equation of the tangent plane to z = ½x2 + ½ y2 + 4 at (1, –1, 5). SolutionLet F(x, y, z) = ½x2 + ½ y2 – z + 4. This graph did pass (1, –1, 5), since F(1, –1, 5)= 0. Now Fx = x, Fy = y, Fz = –1, then From (5), the desired equation is (x + 1) – (y – 1) – (z – 5) = 0or –x + y + z = 7**Normal Line**• Let P(x0, y0, z0) is on the graph of F(x, y, z) = c, where F 0. The line containing P that is parallel to F(x0, y0, z0) is called the normal line to the surface at P.**Example 5**Find parametric equations for the normal line to the surface in Example 4 at (1, –1, 5). SolutionA direction vector for the normal line at (1, –1, 5) is F(1, –1, 5) = i – j – kthen the desired equations arex = 1 + t, y = –1– t, z = 5 – t**9.7 Divergence and Curl**• Vector FunctionsF(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**Example 1**Graph F(x, y) = – yi + xj SolutionSinceletFor and k = 2, we have(i) x2 + y2 = 1：at (1, 0), (0, 1), (–1, 0), (0, –1), the corresponding vectors j, –i,– j , ihave the same length 1.

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