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Ch 15

Ch 15. 偏微分 Partial Derivatives. 學習內容. 多變數函數. 15.1 Functions of Several Variables 15.3 Partial Derivatives 15.4 Tangent Planes and Linear Approx. 15.5 Chain Rule 15.6 Directional Derivatives and Gradient 15.7 Maximum and Minimum Values. 多變數函數偏微分. 切平面與線性估計. 鎖鏈法則. 方向導函數與梯度向量. 極大值與極小值.

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Ch 15

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  1. Ch 15 偏微分 Partial Derivatives

  2. 學習內容 多變數函數 • 15.1 Functions of Several Variables • 15.3 Partial Derivatives • 15.4 Tangent Planes and Linear Approx. • 15.5 Chain Rule • 15.6 Directional Derivatives and Gradient • 15.7 Maximum and Minimum Values 多變數函數偏微分 切平面與線性估計 鎖鏈法則 方向導函數與梯度向量 極大值與極小值

  3. 15.5 Directional Derivatives and Gradient Vectors 方向導函數與梯度向量

  4. 單位方向向量 向量長度 v

  5. 旅行者的行囊 • RenoLas Vegas • SFO Los Angeles • 何者溫度變化比較大?

  6. 方向導函數 曲面上在P點的切線斜率 與u同方向時 h h h h h 切線 方向向量

  7. 方向導函數的符號 與 u同方向時，曲面z = f (x, y)上在P點P(x0 , y0)的切線斜率

  8. 方向導函數的計算方式 x y

  9. 方向導函數的計算方式 u= < a, b> u = < cos θ, sin θ> b a

  10. Example 2 • Find the directional derivative Du f(x, y) iff(x, y) = x3 – 3xy + 4y2. u is the unit vector given by angle θ = π/6 • What is Du f(1, 2)?

  11. f(x, y) = x3‧１– 3xy + 4y2 ‧１ θ θ θ = π/6

  12. What is Du f(1, 2)? y x x x y

  13. Q 1 Find the directional derivative of f(x, y) = x sin(xy) at (2,0) in the direction θ=π/3

  14. 方向導函數的計算方式 ?

  15. 梯度向量

  16. Example 3 • If f(x, y) = sin x + exy, • ▽ f(x, y) = ? • ▽f (0, 1) = ?

  17. f(x, y) = sin x‧1+ ex y 0 1 0 1 0 1 0

  18. Q 2 Evaluate the gradient of f(x, y) = y lnx at P(1, -3) (a) ＜-3, 0＞ (b) ＜-1/3, 0＞ (c) ＜0, -3＞ (d) ＜-1, 0＞

  19. Example 4 • Find the directional derivative of the function f(x, y) = x2y3 – 4yat the point (2, –1) in the direction of the vector v = 2 i + 5 j.

  20. f (x, y) = x2 y3 – 4y．1 v = 2 i + 5 j

  21. (-1) 2 -1 2 2 (-1)

  22. Example 5 • If f(x, y, z) = x sin yz, find: • The gradient of f • The directional derivative of f at (1, 3, 0) in the direction of v = i + 2 j – k.

  23. f(x, y, z) = x sin yz 0 1 3 × 0 3 × 0 1 3 ×0 ×3 1 3 × 0

  24. v = i + 2 j – k

  25. Q 3 Find the directional derivative of f (x, y) = x4-x2 y3 at P(2,1) in the direction v=4i+3j (a) ＜4/5, 3/5＞ (b) 148/76 (c) 76/5 (d) 16/5

  26. 曲線上同一點不同方向的切線斜率 遞增的切線 水平的切線 遞減的切線 Θ是梯度向量與單位向量u之夾角 u

  27. 最大與最小的切線斜率 斜率最大的切線 斜率是0的切線 斜率最小的切線 u u

  28. Example 6 • If f(x, y) = xey, find the rate of change of f at the point P(2, 0) in the direction from P to Q(½, 2). • In what direction does f have the maximum rate of change? What is this maximum rate of change?

  29. the rate of change of f(x, y) = xe y at P(2, 0) Q(½, 2) 2 0 0 0 2

  30. In direction, f have the maximum rate of change Fig. 15.6.8, p. 952

  31. Q 4 Find the maximum rate of change of f (x, y) = sin (xy) at (1,0) (a) (y cos(xy), x cos(xy)) (b) -1 (c) (0, 1) (d) 1

  32. 曲面上P(x0, y0, z0)點的切平面 t0 x-x0 y-y0 z-z0

  33. 曲面上P(x0, y0, z0)點的切平面之法線 法線=＜x-x0 , y-y0 ,z-z0 ＞ 法線//

  34. 切平面之特例 • z = f(x, y)  F(x, y, z) = f(x, y) – z = 0

  35. Example 8 • Find the equations of the tangent plane and normal line at the point (–2, 1, –3) to the ellipsoid

  36. at the point (–2, 1, –3) ▽ f (-2, 1, -3) = ＜-1, 2, -2/3＞ 法線方程式 切平面方程式

  37. Q 5 Find the tangent plane to the surface y = x2-z2 at (4, 7, 3) (a) 8( x - 4 ) －(y - 7 ) －6(z - 3 ) = 0 (b) ( x - 4 )+(y - 7 )+(z - 3 ) = 0 (c) 8( x - 4 )+(y - 7 ) －6(z - 3 ) = 0 (d) 16( x - 4 )+7(y - 7 )+9(z - 3 ) = 0

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