PARTIAL DERIVATIVES

# PARTIAL DERIVATIVES

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## PARTIAL DERIVATIVES

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1. 14 PARTIAL DERIVATIVES

2. PARTIAL DERIVATIVES • As we saw in Chapter 4, one of the main uses of ordinary derivatives is in finding maximum and minimum values.

3. PARTIAL DERIVATIVES 14.7 Maximum and Minimum Values • In this section, we will learn how to: • Use partial derivatives to locate • maxima and minima of functions of two variables.

4. MAXIMUM & MINIMUM VALUES • Look at the hills and valleys in the graph of f shown here.

5. ABSOLUTE MAXIMUM • There are two points (a, b) where f has a local maximum—that is, where f(a, b) is larger than nearby values of f(x, y). • The larger of these two values is the absolute maximum.

6. ABSOLUTE MINIMUM • Likewise, f has two local minima—where f(a, b) is smaller than nearby values. • The smaller of these two values is the absolute minimum.

7. LOCAL MAX. & LOCAL MAX. VAL. Definition 1 • A function of two variables has a local maximumat (a, b) if f(x, y) ≤ f(a, b) when (x, y) is near (a, b). • This means that f(x, y) ≤ f(a, b) for all points (x, y) in some disk with center (a, b). • The number f(a, b) is called a local maximum value.

8. LOCAL MIN. & LOCAL MIN. VALUE Definition 1 • If f(x, y) ≥ f(a, b) when (x, y) is near (a, b), then f has a local minimumat (a, b). • f(a, b) is a local minimum value.

9. ABSOLUTE MAXIMUM & MINIMUM • If the inequalities in Definition 1 hold for allpoints (x, y) in the domain of f, then f has an absolute maximum (or absolute minimum) at (a, b).

10. LOCAL MAXIMUM & MINIMUM Theorem 2 • If f has a local maximum or minimum at (a, b) and the first-order partial derivatives of f exist there, then fx(a, b) = 0 and fy(a, b) = 0

11. LOCAL MAXIMUM & MINIMUM Proof • Let g(x) = f(x, b). • If f has a local maximum (or minimum) at (a, b), then g has a local maximum (or minimum) at a. • So, g’(a) = 0 by Fermat’s Theorem.

12. LOCAL MAXIMUM & MINIMUM Proof • However, g’(a) = fx(a, b) • See Equation 1 in Section 14.3 • So, fx(a, b) = 0.

13. LOCAL MAXIMUM & MINIMUM Proof • Similarly, by applying Fermat’s Theorem to the function G(y) = f(a, y), we obtain: fy(a, b) = 0

14. LOCAL MAXIMUM & MINIMUM • If we put fx(a, b) = 0 and fy(a, b) = 0 in the equation of a tangent plane (Equation 2 in Section 14.4), we get: z = z0

15. THEOREM 2—GEOMETRIC INTERPRETATION • Thus, the geometric interpretation of Theorem 2 is: • If the graph of f has a tangent plane at a local maximum or minimum, then the tangent plane must be horizontal.

16. CRITICAL POINT • A point (a, b) is called a critical point(or stationary point) of f if either: • fx(a, b) = 0 and fy(a, b) = 0 • One of these partial derivatives does not exist.

17. CRITICAL POINTS • Theorem 2 says that, if f has a local maximum or minimum at (a, b), then (a, b) is a critical point of f.

18. CRITICAL POINTS • However, as in single-variable calculus, not all critical points give rise to maxima or minima. • At a critical point, a function could have a local maximum or a local minimum or neither.

19. LOCAL MINIMUM Example 1 • Let f(x, y) = x2 + y2 – 2x – 6y + 14 • Then, fx(x, y) = 2x – 2fy(x, y) = 2y – 6 • These partial derivatives are equal to 0 when x = 1 and y = 3. • So, the only critical point is (1, 3).

20. LOCAL MINIMUM Example 1 • By completing the square, we find:f(x, y) = 4 + (x – 1)2 + (y – 3)2 • Since (x – 1)2≥ 0 and (y – 3)2 ≥ 0, we have f(x, y) ≥ 4 for all values of x and y. • So, f(1, 3) = 4 is a local minimum. • In fact, it is the absolute minimum of f.

21. LOCAL MINIMUM Example 1 • This can be confirmed geometrically from the graph of f, which is the elliptic paraboloid with vertex (1, 3, 4).

22. EXTREME VALUES Example 2 • Find the extreme values of f(x, y) = y2 – x2 • Since fx = –2x and fy = –2y, the only critical point is (0, 0).

23. EXTREME VALUES Example 2 • Notice that, for points on the x-axis, we have y = 0. • So, f(x, y) = –x2 < 0 (if x≠ 0). • For points on the y-axis, we have x = 0. • So, f(x, y) = y2 > 0 (if y≠ 0).

24. EXTREME VALUES Example 2 • Thus, every disk with center (0, 0) contains points where f takes positive values as well as points where f takes negative values. • So, f(0, 0) = 0 can’t be an extreme value for f. • Hence, f has no extreme value.

25. MAXIMUM & MINIMUM VALUES • Example 2 illustrates the fact that a function need not have a maximum or minimum value at a critical point.

26. MAXIMUM & MINIMUM VALUES • The figure shows how this is possible. • The graph of f is the hyperbolic paraboloid z = y2 – x2. • It hasa horizontal tangent plane (z = 0) at the origin.

27. MAXIMUM & MINIMUM VALUES • You can see that f(0, 0) = 0 is: • A maximum in the direction of the x-axis. • A minimum in the direction of the y-axis.

28. SADDLE POINT • Near the origin, the graph has the shape of a saddle. • So, (0, 0) is called a saddle pointof f.

29. EXTREME VALUE AT CRITICAL POINT • We need to be able to determine whether or not a function has an extreme value at a critical point. • The following test is analogous to the Second Derivative Test for functions of one variable.

30. SECOND DERIVATIVES TEST Theorem 3 • Suppose that: • The second partial derivatives of f are continuous on a disk with center (a, b). • fx(a, b) = 0 and fy(a, b) = 0 [that is, (a, b) is a critical point of f].

31. SECOND DERIVATIVES TEST Theorem 3 • Let D = D(a, b) = fxx(a, b) fyy(a, b) – [fxy(a, b)]2 • If D > 0 and fxx(a, b) > 0, f(a, b) is a local minimum. • If D > 0 and fxx(a, b) < 0, f(a, b) is a local maximum. • If D < 0, f(a, b) is not a local maximum or minimum.

32. SECOND DERIVATIVES TEST Note 1 • In case c, • The point (a, b) is called a saddle point of f . • The graph of f crosses its tangent plane at (a, b).

33. SECOND DERIVATIVES TEST Note 2 • If D = 0, the test gives no information: • f could have a local maximum or local minimum at (a, b), or (a, b) could be a saddle point of f.

34. SECOND DERIVATIVES TEST Note 3 • To remember the formula for D, it’s helpful to write it as a determinant:

35. SECOND DERIVATIVES TEST Example 3 • Find the local maximum and minimum values and saddle points of f(x, y) = x4 + y4 – 4xy + 1

36. SECOND DERIVATIVES TEST Example 3 • We first locate the critical points:fx = 4x3 – 4yfy = 4y3 – 4x

37. SECOND DERIVATIVES TEST Example 3 • Setting these partial derivatives equal to 0, we obtain: x3 – y = 0y3 – x = 0 • To solve these equations, we substitute y = x3 from the first equation into the second one.

38. SECOND DERIVATIVES TEST Example 3 • This gives:

39. SECOND DERIVATIVES TEST Example 3 • So, there are three real roots: x = 0, 1, –1 • The three critical points are: (0, 0), (1, 1), (–1, –1)

40. SECOND DERIVATIVES TEST Example 3 • Next, we calculate the second partial derivatives and D(x, y):fxx = 12x2fxy = –4 fyy = 12y2D(x, y) = fxx fyy – (fxy)2 = 144x2y2 – 16

41. SECOND DERIVATIVES TEST Example 3 • As D(0, 0) = –16 < 0, it follows from case c of the Second Derivatives Test that the origin is a saddle point. • That is, f has no local maximum or minimum at (0, 0).

42. SECOND DERIVATIVES TEST Example 3 • As D(1, 1) = 128 > 0 and fxx(1, 1) = 12 > 0, we see from case a of the test that f(1, 1) = –1 is a local minimum. • Similarly, we have D(–1, –1) = 128 > 0 and fxx(–1, –1) = 12 > 0. • So f(–1, –1) = –1 is also a local minimum.

43. SECOND DERIVATIVES TEST Example 3 • The graph of f is shown here.

44. CONTOUR MAP • A contour map of the function in Example 3 is shown here.

45. CONTOUR MAP • The level curves near (1, 1) and (–1, –1) are oval in shape. • They indicate that: • As we move away from (1, 1) or (–1, –1) in any direction, the values of f are increasing.

46. CONTOUR MAP • The level curves near (0, 0) resemble hyperbolas. • They reveal that: • As we move away from the origin (where the value of fis 1), the values of fdecrease in some directions but increase in other directions.

47. CONTOUR MAP • Thus, the map suggests the presence of the minima and saddle point that we found in Example 3.

48. MAXIMUM & MINIMUM VALUES Example 4 • Find and classify the critical points of the function • f(x, y) = 10x2y – 5x2 – 4y2 – x4 – 2y4 • Also, find the highest point on the graph of f.

49. MAXIMUM & MINIMUM VALUES Example 4 • The first-order partial derivatives are:fx = 20xy – 10x – 4x3fy = 10x2 – 8y – 8y3

50. MAXIMUM & MINIMUM VALUES E. g. 4—Eqns. 4 & 5 • So, to find the critical points, we need to solve the equations 2x(10y – 5 – 2x2) = 0 5x2 – 4y – 4y3 = 0
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