Figure 1: Functions Viewed Optically. Math 425 Lecture 4. The Chain Rule

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Figure 1: Functions Viewed Optically Math 425 Lecture 4 The Chain Rule The Chain Rule in the One Variable Case We explain the chain rule in the one variable case.

1 Figure 1: Functions Viewed Optically Math 425 Lecture 4 The Chain Rule The Chain Rule in the One Variable Case We explain the chain rule in the one variable case. To do this we view a function f as a (impossible) lens. A light ray leaves the domain copy of the reals at x and is bent by a lens so that it hits the range copy of the reals at f(x). From this point of view the derivative of F at P is the magnification of the lens at P. See the figure Functions Viewed Optically. 1

Figure 2: Chain Rule Optically We compose two functions, f and g that is, in terms of lenses, we place one lens after another. Assume that f(p ) = Q. Place a small object at P.

2 Figure 2: Chain Rule Optically We compose two functions, f and g that is, in terms of lenses, we place one lens after another. Assume that f(p ) = Q. Place a small object at P. It is magnified by Df(P ) by the first lens and that image is magnified by Dg(Q) = Dg(F (P ) by the second lens. The result is that the original object is magnified by the composition g f by Df(P ) Dg(Q). Note that the product of two 1 1 matrices is just multiplication of their entries. That is what is happening here. In terms of best linear approximations the chair rule says that the best linear approximation of the composition of two functions is the composition of the best linear approximations to the individual functions. The Chain Rule in General We set up our notation: so that F : R n R m, F (P ) = Q, G : R m R p, G(Q) = R DF (P ) : R n P R m Q, DG(Q) : R m Q R p R. The chain rule says that the best linear approximation to or, if you wish, the derivative of G F, at P is DG(Q) DF (P ). Special Case I Let f : R R, x f(x) = y, f(a) = b g : R R, y g(y) = z, g(b) = c, 2

3 so Df(a) : R a R b Dg(b) : R b R c. The chain rule says that the derivative of g f at x = a is the composition Dg(b) Df(a), but Df(a), Dg(b) are 1 1 matrices, so their composition is just multiplication. We have the classic chain rule D(g f)(a) = Dg(b) Df(a), or or Special Case II (g(f(a)) = g (f(a))f (a), d(g(f(x)) dx = dg df dy dx. Let f : R R n, T : R n R. We wish to understand what the derivative of T f is in terms of the derivative of f and g. We look at a physical situation where this arises. Let ( ) ( ) x(t) t f : R R 2 2, t f(t) = = y(t) t 3, ( ) x T : R 2 R, 2x 2 3xy. y We physically interpret f as giving the position of a car along a road in terms of time t, and T as giving the temperature at each point. Then what is T f? it is the temperature perceived by the prson in the car as it moves. What is d (T f)? It is the rate of change dt of the temperature perceived by the person in the car as it moves. It is the rate of change of temperature perceived by a person in the car. The derivative of f at time t 0 is ( ) ( ) x Df(t 0 ) = (t 0 ) 2t0 y =. (t 0 ) Write p = f(t 0 ) and q = T (p). The derivative of T at postion p is a linear map DT (p) : R 2 p R q. It is DT (p) = ( ) T T x y. If T (x, y) = x 2 xy + 3y 2, then DT (x 0, y 0 ) = ( 2x y x + 6y ). The chain rule says that the derivative of T f(t 0 ) is the composition of linear maps given by the matrix multiplication 3t 2 0 3

4 D(T f)(t 0 ) = DT (p) Df(t 0 ) = ( ( ) ) T T x (t x y 0 ) y (t 0 ) = ( 2x y x 6y ) ( ) 2t0. 3t 2 0 More generally, let x 1 (t) f : R R n, t., t 0 p R n, x n (t) T : R n R, x T (x), p T 0. The derivative of T f at t 0 is DT (p) Df(t 0 ) = ( T,, T x 1(t 0 ) ) x 1 x. n x n(t 0 ) = T x 1 dx 1 dt + + T x n dx n dt. Definition/Notation: Let T : R n R. The derivative of T at p R n is called the gradient of T and is denoted DF (p) = T (p) = ( T x 1 (p), T x 2 (p),, T x n (p)). Let f : R R n, t (x 1 (t),, x n (t). The chain rules says Directional Derivative d (T f)(t) = T Df. dt Fix a vector u R n and let f : R R n be the function t p + tu. This represents a particle passing through p with constant velocity u. Definition 1. The directional derivative of T in the direction u at the point p is D u (T )(p) = d (T f)(t) = T Df =< T (p), u >= T (p) u cos(θ) dt where θ is the angle between u and T (p). 4

5 Observation: Assume that u = 1, so that D u (T )(p) = T (p) cos(θ). As the direction of u varies D u (T )(p) is has max absolute value at θ = 0, π. And D u (T )(p) is 0 when u is orthogonal to T (p). Application: For this application assume that T : R R 3. If c R, the surface T (x, y, z) = c is called a level surface. Example: The surface defined by x 2 + 4y 2 = z is a level surface for the function F (x, y, z) = x 2 + 4y 2 z 2. We continue with our application. Assume that the level surface S given by T (x, y, z) = c passes through the point p so that T (p) = c. We define a line L through p given in the form {p+tu t R} to be tangent to S provided D u (T )(p) is zero. Equivalently we say that a line L linearly parametrized by t is tangent to S at p if the restriction of the function T to L has zero derivative with respect to t. The union of all the tangent lines to S through p is the tangent plane to S at p. We see that this is the plane through p orthogonal to T (p). If the coordinates of p are (x 0, y 0, z 0 ), then the equation of this plane is < T (p), x x 0, y y 0, z z 0 >= 0. A similar approach allows us to find the equation of the tangent line to a curve f(x, y) = 0 at a point (x 0, y 0 ). More on the Gradient Let T : R 2 R given the temperature at each point (x, y) in the plane. Heat flows from high temperature to low temperature. Let u be a vector of norm 1. The derivative of the function T (p + tu) T (p) at t = 0 is just the directional derivative of T at p in the direction u. It is given by < T (p), u >= D u (T )(p). Physically the heat flow is D u (T )(p). The rule associating to each point the heat flow at that point is given by the vector T. An Example Let C 1, C 2 be two curves in R 2 meeting at a point p. We define the angle between C 1, C 2 at p to be the angle between their tangent lines L 1, L 2 at p. Let F : R 2 R 2 mapping p q, C 1 D 1, C 2 D 2. Then D 1, D 2 meet at q. What is the relation between the angle between C 1 and C 2 at p and the angle between D 1, D 2 at q? Answer in General Case Since the derivative is linear it maps lines to lines. In particular it maps L 1, L 2 to lines M 1, M 2. Since L 1, L 2 is tangent to C 1, C 2 we have M 1 = f(l 1 ) is 5

6 Figure 3: Angles and the Derivative tangent to f(c 1 ) = D 1 and similarly M 2 is tangent to D 2. Thus we need to see how the linear map Df(P ) treats angles. Subexample: Let F : R 2 R 2 ( ) ( ) x x F : 2 y 2. y 2xy The derivative of F at a general point is ( ) 2x 2y. 2y 2x If (x, y) (0, 0), this matrix is the composition of a dilation and a rotation. Hence it preserves angles. This implies that F also preserves angles provided (x, y) (0, 0). Note that F maps the x-axis to the x-axis. It maps the positive y-axis to the negative x-axis. At the origin the map does not preserve angles. 6

7 Figure 4: Angles and the Derivative 7

Figure 1: Functions Viewed Optically. Math 425 Lecture 4. The Chain Rule

Figure 1: Functions Viewed Optically Math 425 Lecture 4 The Chain Rule The Chain Rule in the One Variable Case We explain the chain rule in the one variable case. To do this we view a function f as a impossible