1 / 110

Directional Derivatives and Tangent Planes in Multivariable Calculus

This article explores directional derivatives, tangent planes, and normal lines in multivariable calculus. It covers the concept of the gradient, computation of directional derivatives, and the geometric interpretation of gradients in two and three variables. The article also includes examples and equations for finding tangent planes and level curves/surfaces.

garlock
Download Presentation

Directional Derivatives and Tangent Planes in Multivariable Calculus

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related