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**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**(6) Second by chain rule**Here the subscripts 1 and 2 refer to partial derivatives of**f(x + t cos , y + t sin ) w.r.t. (x + t cos )and (y + t sin ), respectively. When t = 0, x + t cos and y + t sin are simply x and y, respectively, 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**Functions of Three Variables**• w=F(x,y,z)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 it in another way, (10) and (11) state that: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.**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 to 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.r.t. t is (1)When we introduce**then (1) becomes When at t = t0, we have (2)Thus, if**, is orthogonal to at 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 to the concepts of 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) be 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) be a point on the graph of F(x, y, z) = c, where F is not 0. Then an equation of the 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 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, and F=0 represents the equation.**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 = -3**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 Functions of two or three variables • 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)kare also called vector fields. • The concept of velocity field and force field plays an important role in mechanics, electricity and magnetism.**Example 1: Two-dimensional VF**Graph the vector field 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.**Example 1 (2)**(ii) x2 + y2 = 2：at (1, 1), (–1, 1), (–1, –1), (1, –1), the corresponding vectors – i + j, – i – j, i – j, i + jhave the same length . (iii) x2 + y2 = 4：at (2, 0), (0, 2), (–2, 0), (0, –2), the corresponding vectors 2j, –2i, –2j, 2ihave the same length 2. See Fig 9.38.**DEFINITION 9.7**• In practice, we usually use this form: (1) Curl The curl of a vector field F = Pi + Qj + Rk is the vector field**DEFINITION 9.8**• Observe that we can also use this form: (4) Divergence The divergence of a vector field F = Pi + Qj + Rk is the scalar function**Example 2**IfF = (x2y3 – z4)i + 4x5y2zj – y4z6 k, find curl Fand div F。 Solution