Dr. Allen Back. Sep. 8, 2014

Transcription

1 in R 3 Dr. Allen Back Sep. 8, 2014

2 in R 3

3 in R 3 Def: For f (x, y), the partial derivative with respect to x at p 0 = (x 0, y 0 ) is f x = lim f (x 0 + h, y 0 ) f (x 0, y 0 ) h 0 h or f x = lim f (p 0 + he 1 ) f (p 0 ) h 0 h [ ] 1 where e 1 = is the first standard basis vector of R 0 2.[]

4 A Basic Application: f f x x + f y y. in R 3

5 Example: If the sides of a rectangle increase by 2% and 3%, approximately how much will the area increase? in R 3

6 A Basic Application: f f x x + f y y. in R 3 In the above, the terms more precisely are: f = f (x, y) f (x 0, y 0 ) x = x x 0 y = y y 0 f x = f x (x 0, y 0 ) f y = f y (x 0, y 0 )

7 in R 3 For f : U R n R m, we write f (p) = (f 1 (p), f 2 (p),... f m (p)) where each f i : U R n R is an ordinary real valued function called a component function of f.

8 in R 3 For example the polar coordinate formulas x = y = r cos θ r sin θ may be expressed in terms of the function f : R 2 R 2 defined by f (r, θ) = (r cos θ, r sin θ). (Specifically (x, y) = f (r, θ).) Here the first component function is and the second is f 1 (r, θ) = r cos θ f 2 (r, θ) = r sin θ.

9 Def: A linear transformation T is DF(p) if lim x p f (x) f (p) T (x p) x p = 0 in R 3

10 in R 3 Def: A linear transformation T is DF(p) if Compare to lim x p f (x) f (p) T (x p) x p f x which can be rewritten as = 0 f (p 0 + he 1 ) f (p 0 ) == lim h 0 h f (p 0 + he 1 ) f (p 0 ) hf x (p 0 ) lim = 0 h 0 h

11 Thm: If f : U R n R m is diff., then the entries of Df(p) must be the partial of the component functions of f. in R 3

12 in R 3 Thm: If f : U R n R m is diff., then the entries of Df(p) must be the partial of the component functions of f. So for f : R n R m, the matrix T = Df (p 0 ) will be m n. One row for each component function. The row i col j entry will be f i x j.

13 in R 3 Def: A linear transformation T is DF(p) if lim x p f (x) f (p) T (x p) x p = 0 Thm: If f : U R n R m is diff., then the entries of Df(p) must be the partial of the component functions of f. Problem: f (r, θ) = (r cos θ, r sin θ). Find the derivative.

14 in R 3 Def: A linear transformation T is DF(p) if lim x p f (x) f (p) T (x p) x p = 0 Thm: If f : U R n R m is diff., then the entries of Df(p) must be the partial of the component functions of f. Why?

15 First remember from linear algebra, that for an m n matrix T, the i th column of T is where e i is mapped by Y ; i.e. Te i. in R 3

16 If the limit exists, restrict x = p + te i to lie on a line through p in the direction of the i th standard basis vector e i. in R 3

17 in R 3 If the limit exists, restrict x = p + te i to lie on a line through p in the direction of the i th standard basis vector e i. Then f (p + te i ) f (p) lim t 0 + = Te i t showing that the i th column of T consists of the i th partial of the component functions of f.

18 Thm: If f is differentiable at p 0 then it is continuous there. in R 3

19 Thm: If the partial of f : U R n R exist and are continuous in a neighborhood of p 0 U, then f is differentiable at p 0. in R 3

20 Thm: A function f : U R n R m is differentiable at p 0 if and only each of its component functions f i : U R n R is. (Recall f = (f 1,..., f m ).) in R 3

21 in R 3 The chain rule in multivariable calculus is in some ways very simple. But it can lead to extremely intricate sorts of relationships (try thermodynamics in physical chemistry... ) as well as counter-intuitive looking formulas like z y x = x. z y (The above in a context where f (x, y, z) = C.)

22 First let s try the conceptually simple point of view, using the fact that of functions are linear transformations. (Matrices.) in R 3

23 in R 3 Think about differentiable functions and g : U R n R m f : V R m R p where the image of f (f (U)) is a subset of the domain V of g. The chain rule is about the derivative of the composition f g.

24 Here s a picture: in R 3

25 in R 3 For g : U R n R m and f : V R m R p, let s use p to denote a point of R n q to denote a point of R m r to denote a point of R p. So more colloquially, we might write q = r = g(p) f (q) and so of course f g gives the relationship r = f (g(p)). (The latter is (f g)(p).)

26 in R 3 Fix a point p 0 with g(p 0 ) = q 0 and f (q 0 ) = r 0. Let the of g and f at the relevant points be T = Dg(p 0 ) S = Df (q 0 ). How are the changes in p, q, and r related?

27 in R 3 Fix a point p 0 with g(p 0 ) = q 0 and f (q 0 ) = r 0. Let the of g and f at the relevant points be T = Dg(p 0 ) S = Df (q 0 ). How are the changes in p, q, and r related? By the linear approximation properties of the derivative, q T p r S q And so plugging the first approximate equality into the second gives the approximation r S(T p) = (ST ) p.

28 r (ST ) p. in R 3 What is this saying?

29 in R 3 r (ST ) p. What is this saying? For g : U R n R m and f : V R m R p, T = Df (p 0 ) is an m n matrix S = Dg(q 0 ) is an p m matrix So the product ST is a p n matrix representing the derivative at p 0 of g f.

30 So the chain rule theorem says that if f is differentiable at p 0 with f (p 0 ) = q 0 and g is differentiable at q 0, then g f is also differentiable at p 0 with derivative the matrix product (Dg(q 0 )) (Df (p 0 )). in R 3

31 in R 3 Problem: Suppose we have the polar coordinate map g(r, θ) = (r cos θ, r sin θ) and (r, θ) = f (u, v) is given by f (u, v) = (uv, v). Find the derivative of g f.

32 f : U R 2 R and g : U R R 2. Derivatives/ of f g and g f? in R 3

33 Tree diagrams for chain rule applications. in R 3

34 Cases like f (x, u(x, y), v(y)). in R 3

35 The tangent plane to the graph of z = f (x, y) at (x, y) = (x 0, y 0 ) is defined to be the plane given by z z 0 = f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ). in R 3 (The approximation z f x x + f y y is replaced by an exact equality on the tangent plane.)

36 plane to z = x 2 y 2 at ( 1, 0, 1). in R 3 Note the tangent plane needn t meet the surface in just one point.

37 in R 3 in R 3

38 in R 3 in R 3 If the normal n =< a, b, c >, r =< x, y, z > is a general point and P 0 = (x 0, y 0, z 0 ), then n ( r P 0 ) = 0 becomes a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0.

39 in R 3 in R 3 Find the equation of the plane through the three points P 0 = (1, 0, 1), P 1 = ( 1, 1, 2) and P 2 = (1, 2, 3).

40 in R 3 in R 3 Find the equation of the plane through the three points P 0 = (1, 0, 1), P 1 = ( 1, 1, 2) and P 2 = (1, 2, 3). Solution: First find the normal n = P 0 P 1 P 0 P 2

41 in R 3 Find the equation of the plane through the three points P 0 = (1, 0, 1), P 1 = ( 1, 1, 2) and P 2 = (1, 2, 3). Solution: First find the normal n = P 0 P 1 P 0 P 2 in R 3 The cross product: which is î ĵ ˆk

42 in R 3 Find the equation of the plane through the three points P 0 = (1, 0, 1), P 1 = ( 1, 1, 2) and P 2 = (1, 2, 3). Solution: First find the normal n = P 0 P 1 P 0 P 2 in R 3 The cross product: î ĵ ˆk which is î ĵ ˆk =< 0, 4, 4 >.

43 in R 3 in R 3 Find the equation of the plane through the three points P 0 = (1, 0, 1), P 1 = ( 1, 1, 2) and P 2 = (1, 2, 3). î ĵ ˆk =< 0, 4, 4 >. So our plane is < 0, 4, 4 > ( r < 1, 0, 1 >) = 0

44 in R 3 Find the equation of the plane through the three points P 0 = (1, 0, 1), P 1 = ( 1, 1, 2) and P 2 = (1, 2, 3). So our plane is < 0, 4, 4 > ( r < 1, 0, 1 >) = 0 in R 3 or 0(x 1) + 4(y 0) 4(z 1) = 0 or 4y 4z + 4 = 0.

45 in R 3

46 in R 3 Find the equation of the line through the points P 0 = (1, 1, 0) and P 1 = (2, 2, 2).

47 Find the equation of the line through the points P 0 = (1, 1, 0) and P 1 = (2, 2, 2). in R 3

48 Find the equation of the line through the points P 0 = (1, 1, 0) and P 1 = (2, 2, 2). in R 3 Solution: P 0 P 1 = (2, 2, 2) (1, 1, 0) =< 1, 1, 2 >. So our line is r =< 1, 1, 0 > +t < 1, 1, 2 >=< 1 + t, 1 + t, 2t >. where t is any real number.

49 in R 3 Solution: P 0 P 1 = (2, 2, 2) (1, 1, 0) =< 1, 1, 2 >. So our line is r =< 1, 1, 0 > +t < 1, 1, 2 >=< 1 + t, 1 + t, 2t >. where t is any real number. This is called the vector form of the equation of a line.

50 in R 3 Solution: P 0 P 1 = (2, 2, 2) (1, 1, 0) =< 1, 1, 2 >. So our line is r =< 1, 1, 0 > +t < 1, 1, 2 >=< 1 + t, 1 + t, 2t >. where t is any real number. This is called the vector form of the equation of a line. Thinking our general position vector r =< x, y, z >, we can express this as the parametric form: x = y = z = 1 + t 1 + t 2t

51 Thinking our general position vector r =< x, y, z >, we can express this as the parametric form: x = y = z = 1 + t 1 + t 2t in R 3 Solving for t shows t = x 1 = y 1 = z 2 which realizes this line as the intersection of the planes x = y and z = 2(y 1) but there are many other pairs of planes containing this line.

52 The cross product of vectors in R 3 is another vector. in R 3

53 in R 3 The cross product of vectors in R 3 is another vector. It is good because: it is geometrically meaningful it is straightforward to calculate it is useful (e.g. torque, angular momentum)

54 in R 3 The cross product of vectors in R 3 is another vector. u = v w is geometrically determined by the properties: u is perpendicular to both v and w. u is the v w sin θ, the area of the parallelogram spanned by v and w. Choice from the remaining two possibilities is now made based on the right hand rule.

55 u = v w is geometrically determined by the properties: u is perpendicular to both v and w. u is the v w sin θ, the area of the parallelogram spanned by v and w. Choice from the remaining two possibilities is now made based on the right hand rule. in R 3

56 in R 3 u = v w is geometrically determined by the properties: u is perpendicular to both v and w. u is the v w sin θ, the area of the parallelogram spanned by v and w. Choice from the remaining two possibilities is now made based on the right hand rule.

57 u = v w is geometrically determined by the properties: u is perpendicular to both v and w. u is the v w sin θ, the area of the parallelogram spanned by v and w. Choice from the remaining two possibilities is now made based on the right hand rule. in R 3

58 in R 3 The algebraic definition of the cross product is based on the determinant î ĵ ˆk v w = v 1 v 2 v 3 w 1 w 2 w 3 which means v w = î v 2 v 3 w 2 w 3 ĵ v 1 v 3 w 1 w 3 + ˆk v 1 v 2 w 1 w 2. where a b c d = ad bc and î =< 1, 0, 0 >, ĵ =< 0, 1, 0 >, and ˆk =< 0, 0, 1 >,

59 v w = w v in R 3

60 Since î ĵ = ˆk and cyclic (so ĵ ˆk = î and ˆk î = ĵ) it is sometimes easiest to use that algebra or comparison with the picture below to determine cross products or use the right hand rule. in R 3

61 in R 3 For example < 1, 1, 0 > < 0, 0, 1 >= (î + ĵ) ˆk = ĵ + î =< 1, 1, 0 > is easier than writing out the 3 3 determinant.

62 in R 3 s can be used to find the area of a parallelogram or triangle spanned by two vectors in R 3. find the volume of a parallelopiped using the scalar triple product u ( v w) = ( u v) w.