Dr. Allen Back. Oct. 6, 2014

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1 Dr. Allen Back Oct. 6, 2014

2 Distribution Min=48 Max=100 Q1=74 Median=83 Q3=91.5 mean = std. dev. = 12.5

3 Distribution Distribution: (The letters will not be directly used.) Scores Freq is approximately a B is approximately a B is approximately a B is approximately a C is approximately a C Total 62

4 Distribution If you think there s a significant mistake in the way your test was scored... Solutions are posted on moodle. Please check those first. Challenges to the scoring of any question on this exam go back to the person who graded it. Write a brief explanation of what you think the mistake was and give your appeal (and exam) to one of us by Wed. Oct. 15. If you d like to discuss your appeal with us first, that fine. We can offer you advice. Keep in mind that rubrics and graders may make different decisions regarding partial credit; ultimately a very important thing is that everyone s paper is graded the same way.

5 Need Not Have Solutions Problem: Maximize f (x, y) = x 2 + y 2 2xy x 2 + y 2 on the domain 1 x 2 + y 2 4.

6 Need Not Have Solutions Problem: Maximize f (x, y) = x 2 + y 2 2xy x 2 + y 2 on the domain 1 x 2 + y 2 4. This function is invariant under T : R 2 R 2 defined by T (x, y) = (y, x). (i.e. f T = f.)

7 Need Not Have Solutions Problem: Maximize f (x, y) = x 2 + y 2 2xy x 2 + y 2 on the domain 1 x 2 + y 2 4. In polar coordinates, this is 1 sin (2θ) so it is clear that the maxima are along the x and y axes.

8 Need Not Have Solutions Problem: Maximize f (x, y) = x 2 + y 2 2xy x 2 + y 2 on the domain 1 x 2 + y 2 4. There are no maxima where x = y. What is true, is that if (x 0, y 0 ) is a max, then so is T (x 0, y 0 ) = (y 0, x 0 ).

9 Given f : U R n R, to find the formula for the k th polynomial P k of a C k function about the point p 0 = (a 1, a 2,..., a n ) U, we consider the 1-variable function g(t) defined by g(t) = f (p 0 + th) where h = (h 1, h 2,..., h n ) R n. If we are interested in relating the polynomial P k to the behavior of f at the point p = (x 1, x 2,..., x n ) U we might choose h = (x 1 a 1, x 2 a 2,..., x n a n ) as long as the entire line segment between p 0 and p lies in U.

10 g(t) = f (p 0 + th)

11 (2 variable case, f a C 3 function) g(t) = f (x 0 + th, y 0 + tk)

12 (2 variable case, f a C 3 function) g(t) = f (x 0 + th, y 0 + tk) g (t) = hf x (x 0 + th, y 0 + tk) +kf y (x 0 + th, y 0 + tk)

13 (2 variable case, f a C 3 function) g(t) = f (x 0 + th, y 0 + tk) g (t) = hf x (x 0 + th, y 0 + tk) +kf y (x 0 + th, y 0 + tk) g (t) = h(hf xx (x 0 + th, y 0 + tk) +kf xy (x 0 + th, y 0 + tk)) + k(hf yx (x 0 + th, y 0 + tk) +kf yy (x 0 + th, y 0 + tk))

14 g(t) = (2 variable case, f a C 3 function) f (x 0 + th, y 0 + tk) g (t) = hf x (x 0 + th, y 0 + tk) +kf y (x 0 + th, y 0 + tk) g (t) = h(hf xx (x 0 + th, y 0 + tk) +kf xy (x 0 + th, y 0 + tk)) + k(hf yx (x 0 + th, y 0 + tk) +kf yy (x 0 + th, y 0 + tk)) = [ h 2 f xx (x 0 + th, y 0 + tk) +2hkf xy (x 0 + th, y 0 + tk) +k 2 f yy (x 0 + th, y 0 + tk) ]

15 g(t) = (2 variable case, f a C 3 function) f (x 0 + th, y 0 + tk) g (t) = hf x (x 0 + th, y 0 + tk) +kf y (x 0 + th, y 0 + tk) g (t) = h(hf xx (x 0 + th, y 0 + tk) +kf xy (x 0 + th, y 0 + tk)) + k(hf yx (x 0 + th, y 0 + tk) +kf yy (x 0 + th, y 0 + tk)) = [ h 2 f xx (x 0 + th, y 0 + tk) +2hkf xy (x 0 + th, y 0 + tk) +k 2 f yy (x 0 + th, y 0 + tk) ] g(0) = f (x 0, y 0 )

16 (2 variable case, f a C 3 function) g (t) = hf x (x 0 + th, y 0 + tk) +kf y (x 0 + th, y 0 + tk) g (t) = h(hf xx (x 0 + th, y 0 + tk) +kf xy (x 0 + th, y 0 + tk)) + k(hf yx (x 0 + th, y 0 + tk) +kf yy (x 0 + th, y 0 + tk)) = [ h 2 f xx (x 0 + th, y 0 + tk) +2hkf xy (x 0 + th, y 0 + tk) +k 2 f yy (x 0 + th, y 0 + tk) ] g(0) = f (x 0, y 0 ) g (0) = hf x (x 0, y 0 ) +kf y (x 0, y 0 )

17 (2 variable case, f a C 3 function) g (t) = h(hf xx (x 0 + th, y 0 + tk) +kf xy (x 0 + th, y 0 + tk)) + k(hf yx (x 0 + th, y 0 + tk) +kf yy (x 0 + th, y 0 + tk)) = [ h 2 f xx (x 0 + th, y 0 + tk) +2hkf xy (x 0 + th, y 0 + tk) +k 2 f yy (x 0 + th, y 0 + tk) ] g(0) = f (x 0, y 0 ) g (0) = hf x (x 0, y 0 ) +kf y (x 0, y 0 ) g (0) = [ h 2 f xx (x 0, y 0 ) +2hkf xy (x 0, y 0 ) +k 2 f yy (x 0, y 0 ) ]

18 So the 1-variable (about t = 0) with Remainder at t = 1 g(t) = P 1 + R 1 g(1) = g(0) + g (0)(1 0) g (c)(1 0) 2 for some c (0, 1) becomes the multivariable

19 becomes the multivariable f (x 0 + th, y 0 + tk) = P 1 + R 1 f (x 0 + h, y 0 + k) = f (x 0, y 0 ) + hf x (x 0, y 0 ) + kf y (x 0, y 0 ) (i.e. f (x, y)) = f (x 0, y 0 ) + hf x (x 0, y 0 ) + kf y (x 0, y 0 ) + 1 2[ h 2 f xx (x, y ) + 2hkf xy (x, y ) + k 2 f yy (x, y ) ] for some point (x, y ) on the (interior of the) line segment between (x 0, y 0 ) and (x, y).

20 Similarly g (0) = h 3 f xxx (x 0, y 0 ) + 3h 2 kf xxy (x 0, y 0 ) +3hk 2 f xyy (x 0, y 0 ) + k 3 f yyy (x 0, y 0 ) (with g (t) similar) and 1-variable g(1) = P 2 + R 2 become for a C 3 function f (x, y) = f (x 0, y 0 ) + hf x (x 0, y 0 ) + kf y (x 0, y 0 ) + 1 2[ h 2 f xx (x 0, y 0 ) + 2hkf xy (x 0, y 0 ) + k 2 f yy (x 0, y 0 ) ] + 1 3![ h 3 f xxx (x, y ) + 3h 2 kf xxy (x, y ) +3hk 2 f xyy (x, y )+ k 3 f yyy (x, y ) ]

21 The coefficients 1 1, and in the above derivative expressions are in fact the binomial coefficients from Pascal s triangle which describes the expansion of (x + y) m.

22 Problem: Find the first and second polynomials of about (x, y) = ( π 2, π). f (x, y) = cos (x + 3y) + sin x

23 Problem: Find the first and second polynomials of f (x, y) = cos (x + 3y) + sin x about (x, y) = ( π 2, π). Estimate the error in approximating f by P 1 for x π 2 <.1 and y π <.1.

24 Problem: Find the first and second polynomials of f (x, y) = cos (x + 3y) + sin x about (x, y) = ( π 2, π). Estimate the error in approximating f by P 1 for x π 2 <.01 and y π <.01.

25 Problem: Find the first and second polynomials of about (x, y) = ( π 2, π). f (x, y) = cos (x + 3y) + sin x Find an ɛ so that the error in approximating f by P 1 for x π < ɛ and y π < ɛ is at most Or at most 10 6.

26 Now the 1-variable function g(t) = f (p 0 + th) where p 0 R n and h = (h 1, h 2,..., h n ) satisfies g (t) = Σ n i=1h i (D i g)(p 0 + th i ) where D i denotes the i th partial derivative.

27 And or g (t) = Σ n i=1h i ( Σ n j=1 h j D j (D i g)(p 0 + th i ) ) g (t) = Σ n i=1σ n j=1h i h j (D ij g)(p 0 + th i ).

28 And g (t) = Σ n i=1h i ( Σ n j=1 h j D j (D i g)(p 0 + th i ) ) or g (t) = Σ n i=1σ n j=1h i h j (D ij g)(p 0 + th i ). Noting that (h 1 + h n ) 2 = Σ n i=1h i Σ n j=1h j = Σ n i=1σ n j=1h i h j we can see how the binomial (actually multinomial) coefficients in (h 1 + h n ) m enter when we collect terms in the polynomials.

29 A good way to write the degree m part of the m th polynomial in terms of multi-index notation is ( ) 1 I! Σ I I! i 1!i 2!... i n! hi D I f where the n powers are arranged in a vector I = (i 1, i 2,..., i n ) and I = i 1 + i i n (= m for terms contributing to P m ) h I = h i 1 1 h i hn in D I f = D i 1 1 D i D in n f where for example D 5 3 f means partially differentiate f 5 times with respect to the third variable x 3.

30 Graph of f (x, y) = 1 x+y 2 for.5 x, y 1.5

31 P 2 about (1, 1) for f (x, y) = 1 x+y 2 for.5 x, y 1.5

32 Both superimposed for.5 x, y 1.5

33 Error in using P 2 for f for.5 x, y 1.5

34 Error in using P 2 for f for.75 x, y 1.25

35 Error in using P 2 for f for 0 x, y 3

36 Let x = r cos θ, y = r sin θ. Given f (x, y), express 2 f r 2 in terms of partials of f with respect to x and y.

37 Really there are two functions here; namely f (x, y) and g(r, θ) = f (r cos θ, r sin θ). But many applied fields routinely use the same letter f for both functions.

38 The key step is realizing that the task (via the chain rule) of relating f f f to and is just like the task of doing the r x y analagous thing with expressions like ( ) f. r x (The right hand side operator notation means partially differentiate the function f of (x, y) with respect to r.) x

39 f r = = ( ) f = r x = x = r cos θ y = r sin θ f x + f y x r y r f f cos θ + x y sin θ ( ( )) ( ( f x f + x x r y x ( 2 ) ( f 2 f x 2 cos θ + y x )) y r ) sin θ

40 A similar computation with 2 f would show that the laplacian θ2 2 f x f y 2 is given in polar coordinates by 2 f r f r r + 1 ( 2 ) f r 2 θ 2.

41 Problem 33 on page 146 from your homework asks you to show using the chain rule that if z = f ( y ) for a 1-variable x differentiable function f, then x z x + y z y = 0.

42 Problem 33 on page 146 from your homework asks you to show using the chain rule that if z = f ( y ) for a 1-variable x differentiable function f, then x z x + y z y = 0. You don t need it to do this homework problem, but a natural question is Does x z x + y z y = 0 imply z = f (y x ) for some 1-variable differentiable function f?

43 The answer is yes, and we can easily show it using what we learned in section 2.6! (i.e. essentially chain rule ideas applied to curves.)

44 The beautiful idea for the equation a(x, y) z x + b(x, y) z y = c(x, y, z) is to consider curves (x(t), y(t) called characteristics) satisfying dx dt = dy dt = a(x(t), y(t)) b(x(t), y(t)).

45 Here for x z x + y z y = 0 we have dx dt = dy dt = with solutions x(t) = x 0 e t and y(t) = y 0 e t. x y

46 Here for we have x z x + y z y = 0 dx dt = dy dt = with solutions x(t) = x 0 e t and y(t) = y 0 e t. The key observation is that for a solution z(x, y) d [z(x(t), y(t))] = z dx dt x dt + z = x x+ = 0! x y z dy y dt z y y

47 Here for x z x + y z y = 0 with solutions x(t) = x 0 e t and y(t) = y 0 e t. The key observation is that for a solution z(x, y) d [z(x(t), y(t))] = z dx dt x dt + z = x x+ = 0! So z(x, y) is constant along lines y x as we wanted to show. z(x, y) = z(1, y x ) = f (y x ) z dy y dt z y y = k for any k and thus

48 Problem: Maximize f (x, y) = x 4 + 4x 2 y 2 among points satisfying the constraint g(x, y) = x 2 + y 2 = 1.

49 Problem: Maximize f (x, y) = x 4 + 4x 2 y 2 among points satisfying the constraint g(x, y) = x 2 + y 2 = 1. Only points on the circle x 2 + y 2 = 1 count!

50 Problem: Maximize f (x, y) = x 4 + 4x 2 y 2 among points satisfying the constraint g(x, y) = x 2 + y 2 = 1. Parameterizing the circle as x = cos t and y = sin t is one approach, but even when you can nicely parameterize the constraint set LM is often useful.

51 Problem: Maximize f (x, y) = x 4 + 4x 2 y 2 among points satisfying the constraint g(x, y) = x 2 + y 2 = 1. The method of LM says to look for constrained local extrema, seek solutions of f = λ g for some unknown constant λ. (As well as places on the constraint set where g = 0.)

52 The need to check g = 0 too: Problem: Minimize f (x, y) = y on the constraint set (y x 2 )(y 4x 2 ) = 0.

53 The need to check g = 0 too: Problem: Minimize f (x, y) = y on the constraint set (y x 2 )(y 4x 2 ) = 0. Clearly (x, y) = (0, 0) is the answer, but f λ g there since g is 0 at (0, 0).

54 Maximize f (x, y) = ye x. On the constraint set x 2 + xy + 4y 2 = 1. The constraint set is an ellipse.

55 Maximize f (x, y) = ye x. On the constraint set x 2 + xy + 4y 2 = 1. With the level curve f (x, y) = 2.

56 Maximize f (x, y) = ye x. On the constraint set x 2 + xy + 4y 2 = 1. The level curve f (x, y) =.5. meets the ellipse obliquely.

57 Maximize f (x, y) = ye x. On the constraint set x 2 + xy + 4y 2 = 1. The level curve f (x, y) =.5. meets the ellipse obliquely. The essence of LM is that at a non-tangent intersection point, b/c D u f = f u, f changes to first order as you move along the constraint set. So not a local constr. min or max.

58 Maximize f (x, y) = ye x. On the constraint set x 2 + xy + 4y 2 = 1. The level curve f (x, y) = is tangent.

59 Maximize f (x, y) = ye x. On the constraint set x 2 + xy + 4y 2 = 1. The level curve f (x, y) = is tangent. This value was found by solving f = λ g numerically. ((x, y) = (.776,.427).)

60 Maximize f (x, y) = ye x. On the constraint set x 2 + xy + 4y 2 = 1. The level curve f (x, y) = is tangent. This value was found by solving f = λ g numerically. ((x, y) = (.776,.427).) It appears to be the minimum value of f on the constraint set.

61 Maximize f (x, y) = ye x. On the constraint set x 2 + xy + 4y 2 = 1. The level curve f (x, y) = is also tangent.

62 Maximize f (x, y) = ye x. On the constraint set x 2 + xy + 4y 2 = 1. The level curve f (x, y) = is also tangent. Also found numerically. ((x, y) = (.459,.381).)

63 Maximize f (x, y) = ye x. On the constraint set x 2 + xy + 4y 2 = 1. The level curve f (x, y) = is also tangent. Also found numerically. ((x, y) = (.459,.381).) It appears to be the maximum value of f on the constraint set.

3 Applications of partial differentiation

Advanced Calculus Chapter 3 Applications of partial differentiation 37 3 Applications of partial differentiation 3.1 Stationary points Higher derivatives Let U R 2 and f : U R. The partial derivatives