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 fx. 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 fp = 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 fp = Q. Place a small object at P. It is magnified by DfP by the first lens and that image is magnified by DgQ = DgF P by the second lens. The result is that g f is magnifies the original object by DfP DgQ. 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, GQ = R DF P : R n P R m Q, DGQ : 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 DGQ DF P. Special Case I Let f : R R, x fx = y, fa = b g : R R, y gy = z, gb = c, 2

3 so Dfa : R a R b Dgb : R b R c. The chain rule says that the derivative of g f at x = a is the composition Dgb Dfa, but Dfa, Dgb are 1 1 matrices, so their composition is just multiplication. We have the classic chain rule Dg fa = Dgb Dfa, or or Special Case II gfa = g faf a, dgfx 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 xt t f : R R 2 2, t ft = = yt 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 person in the car as it moves. What is d T f? It is the rate of change of the temperature perceived by the person in the car as it moves. The derivative of f at time t is x Dft = t 2t y =. t Write p = ft 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, y = 2x y x + 6y. The chain rule says that the derivative of T ft is the composition of linear maps given by the matrix multiplication 3t 2 3

4 DT ft = DT p Dft = T x T y = T dx x + T y x t y t dy = 2x y x 6y 2t. 3t 2 We interpret this physically. rate of change of temperature wrt t = rate due to change in x direction+ rate due to change in y direction = rate of change of T wrt x rate of change of x wrt t+ rate of change of T wrt x rate of change of y wrt t. More generally, let x 1 t f : R R n, t., t p R n, x n t T : R n R, x T x, p T. The derivative of T f at t is DT p Dft = T,, T x 1t x 1 x. n x nt = T x 1 dx T x n dx n. 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 d T ft = T Df. 4

5 Directional Derivative 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 ft = T Df =< T p, u >= T p u cosθ where θ is the angle between u and T p. Observation: Assume that u = 1, so that D u T p = T p u = T p cosθ where θ is the angle between T p and u As the direction of u varies D u T p has max absolute value at θ =, π. And D u T p is when u is orthogonal to T p. Definition Let T : R 3 R. If c R, the surface T x, y, z = c is called a level surface. Similarly we can define level curves. 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, y, z, then the equation of this plane is < T p, x x, y y, z z >= T x x x + T y y y + T z z z =. A similar approach allows us to find the equation of the tangent line to a curve fx, y = at a point x, y. The gradient vector of a function F x, y, z at a point P is orthogonal to the level surface of F passing through that point. More Gradients 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 5

Figure 3: Angles and the Derivative at t = 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.

6 Figure 3: Angles and the Derivative at t = 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. Let P x, y be air pressure. The curves P x, y = c, c R are isobars. The vector P at point q gives the velocity of the wind at point q. Let ax, y give the altitude above sea level of position x, y. On a geological survey map the curves ax, y are called contours. If we assume that the surface of the earth is very smooth and not porous, then a drop of water at point q will flow down hill with speed and direction given by aq. Let Ex, y, z be the force of an electrical field at point x, y, z. The equipotential surfaces are orthogonal to this vector field. Preservation of Angles 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 map 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? 6

Figure 4: Angles and the Derivative 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.

7 Figure 4: Angles and the Derivative 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 = fl 1 is tangent to fc 1 = D 1 and similarly M 2 is tangent to D 2. Thus we need to see how the linear map DfP 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,, 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,. At x, y =, F does NOT preserve angles. 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. 7

8 Change of Variables and Taking the Derivative The potential of a gravitational field away from mass and or an electrical field away from charge satisfies Laplace s equation 2 u x u y 2 =. In addition steady state temperature satisfies this equation. We take the first step in expressing this equation in polar coordinates. We use the chain rule. Consider the composition of functions By the chair rule we have or, more concretely, R, φ P x = R cosφ, y = R sinφ u R. u x, u y Du P = Du DP, x R y R x φ y φ = u R, u φ. This gives u R = u x x R + u y y R and u φ = u x x φ + u y y φ, or, u R = cosφ u x + sinφ u u and = R sinφ u y φ x + R cosφ u y, or, more suggestively, R = cosφ x + sinφ y It is an exercise to show that Vector Fields 2 u x + 2 u 2 y = 1 2 R and R φ = R sinφ x + R cosφ y. R + 1R 2 R 2 φ. 2 A vector field is a rule that associates to each point in R n a vector in the tangent plane to R n. Vector fields arise as force fields where one associates the force vector at a point to that point. Vector fields also arise in fluid flows. One associates to each point the velocity of the flow at that point. Mathematically a vector field is given by a map F : R n R n. More generally we can have vector fields that vary with time. In that case we have a map F : R R n R n. The most direct way of studying a vector field F is to study 8

9 Figure 5: A Vector Field with Integral curve or Pathline in the case F is force field the path a particle takes when it is subject to the given force field, in the case F is a fluid filed, the path a particle flows along in the fluid. Suchn a path is called a pathline. This is not the route we take in this course. We limit ourselves to a couple of examples. We can view a vector field as giving the velocity of a particles at each point in space. Here is an example in R 2. dx dy = y x To find the path we must solve the system of differential equations given above. For each initial point a, b there is the unique solution. The curve that this particle sweeps out is called an integral curve or a pathline. xt yt =. a cost + b sint. b cost a sint The above example is a steady flow, that is, the vector field is independent of time. We look an example of a vector field that is Not steady. Let x cost f = y sint We can solve the differential equations = dx dy cost sint 9

10 by hand. We get xt sint + a = yt cost + b The integral curve or pathline with initial point, is the circle of radius one with center at, 1. Another way of getting a picture of what the vector field is doing is to consider it as a fluid flow and release a dye at a single point for a period of time and see what curve results. This is called a streakline. An example of this is the trial of smoke made by a smoke. We compute the streakline for the above example. Fix a time t when stop and photography the streakline. For an arbitrary time s we wish to see the position at time t of the dye that was released at position, at time t = s. The requirement that the particle was at, at time t = s means that xt = sint sins, yt = cost + coss. The streakline is gotten by looking at the position of the dye at time t = t for all of the different values of s.. Thus it is the curve parametrized by s xs sint sins = ys cost + coss This is a circle of radius 1 with center at sint, cost. The Derivative and Fluid Flow We consider a fluid flow F on R 2 and give an interpretation of the or rather a part of the derivative of F in terms of the flow. We look at how a small piece of the fluid flows and study how its area changes. The same considerations work in R 3. There we would look at the volume rather than the area. p Given an initial point P = the flow F, determines how this point flows across the plane. This is done by the differential equations x x = F, = y y dx dy p 1 p p 1. Check that this is indeed solution to the given differential equation. We give some notation i. F : R 2 R 2 is the vector field. ii. t is a small time interval. iii. P is an arbitrary point. iv. h is a small distance. 1

11 Figure 6: A Blob Flows We are concerned with how the area of the square changes per unit time. The vertices of our initial square are h h {P, P + }, P +, P + }. h h Let ut be the image of the vector after time t of the vector with tail at P and head 1 at P + h. Similarly let vt be the image of the vector after time t of the vector with tail at P and head at P + h }. Our goal is to calculate the area of the parallelogram 1 spanned by ut, vt at time t and then find how fast it is changing as t. To do this we calculate ut, vt and take the determinant of the matrix with columns ut, vt. h To find out ut we see where P and P + travel to in time t. The point P moves to P + tf P. On the other hand P + h moves to P + h + F P + h t. But Hence P + h F P + moves to h F P + DF P P + h. h + F P + DF P h t. 11

12 Thus subtracting the two red lines we get h h ut + tdf P Similarly Write vt + tdf P h h a b DF P =. c d The area of our blob at time t is approximately h a b detut vt = det + t = a + d. h c d This is the divergence of the function F. If we write F = divf = F 1 x + F 2 y. F1 F 2, then 12

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)