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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. Introduction See Fig 9.26. The Gradient of a Function.

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  1. 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

  2. 9.5 Directional Derivative • IntroductionSee Fig 9.26.

  3. The Gradient of a Function • Define the vector differential operator asthen (1) (2)are the gradients of the functions.

  4. Example 1 Compute Solution

  5. Example 2 If F(x, y, z) = xy2 + 3x2 – z3, find the gradient at (2, –1, 4). Solution

  6. 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.

  7. Fig 9.27

  8. 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

  9. First (6) Second by chain rule

  10. 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

  11. 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

  12. Example 3 (2) Now, = /6, u = cos i + sin j becomesThen

  13. 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)

  14. Since u is a unit vector, from (10) in Sec 7.3 thatIn addition, (9) shows

  15. 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

  16. 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

  17. Maximum Value of the Direction Derivative • From the fact thatwhere  is the angle between and u. Becausethen

  18. 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)

  19. Example 6 • In Example 5, the maximum value of the directional derivative at (1, −1, 2) is and the minimum value is .

  20. 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.

  21. 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?

  22. 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

  23. 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

  24. then (1) becomes When at t = t0, we have (2)Thus, if , is orthogonal to at P(x0, y0).See Fig 9.30.

  25. Fig 9.30

  26. 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.

  27. Fig 9.31

  28. 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.

  29. Fig 9.32

  30. 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.

  31. Fig 9.33

  32. Fig 9.32

  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.

  34. 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

  35. Fig 9.34

  36. 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.

  37. 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.

  38. 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

  39. Fig 9.35

  40. 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.

  41. 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

  42. 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.

  43. Fig 9.37 (a) ~ (b)

  44. Fig 9.37 (c) ~ (d)

  45. 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.

  46. 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.

  47. Example 1 (3)

  48. 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

  49. 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

  50. Example 2 IfF = (x2y3 – z4)i + 4x5y2zj – y4z6 k, find curl Fand div F。 Solution

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