Coordinate Systems and Examples of the Chain Rule

Transcription

1 Coordinate Sstems and Eamples of the Chain Rule Ale Nita Abstract One of the reasons the chain rule is so important is that we often want to change coordinates in order to make difficult problems easier b eploiting internal smmetries or other nice properties that are hidden in the Cartesian coordinate sstem. We will see how this works for eample when tring to solve first order linear partial differential equations or when working with differential equations on circles, spheres and clinders. It is one of the achievements of vector calculus that it has a develped language for describing all of these different aspects in a unified wa, in particular using coordinate changes and differentiating functions under these coordinate changes to provide simplified epressions. The ke idea is the following: if ϕ : R n R n is a change-of-coordinates function and f : R n R is a differentiable function, then instead of working with Df(, which ma be difficult in practice, we ma work with D(f ϕ( = Df(ϕ(Dϕ(. 1 Different Coordinate Sstems in R 2 and R 3 Eample 1.1 Let s start with something simple, like rotating our coordinates through an angle of. In a previous eample, in the section on differentiabilit, we showed that a rotation through is a linear function R : R 2 R 2 which operates on vectors (,, written here as column vectors, b left multipling them b the rotation matri R, ( ( u = R = v ( cos sin sin cos ( = ( cos sin sin + cos (1.1 v u = (, = (u, v In terms of the component functions of R, we have u = R 1 (, = cos sin v = R 2 (, = sin + cos We have thus introduced new rotated coordinates (u, v in R 2. 1

2 Eample 1.2 There are other coordinate sstems in R 2 besides rotated Cartesian coordinates, but which are still linear in a sense. Consider for eample two vectors u and v in R 2 which are not scalar multiples of each other (i.e. the are not collinear. Then I claim that an vector in R 2 can be described as a scalar multiple of u plus a scalar multiple of v (hence the describe a coordinate sstem, telling ou how to find b going some wa in the direction of u then some wa in the direction of v, much like in regular Cartesian coordinates we go some wa in the direction of i and some wa in the direction of j. Let s see how this works. Let u = (a, b and v = (c, d and suppose the are not scalar multiples of each other (i.e. (c, d k(a, b = (ka, kb for an k. Then consider an = (,. To sa that = k 1 u + k 2 v is to sa (, = k 1 (a, b + k 2 (c, d = ( k 1 a + k 2 c, k 1 b + k 2 d which, if we want to discover what k 1 and k 2 are, means we have to solve a sstem of two equations in two unknowns, = ak 1 + bk 2 = bk 1 + dk 2 (Recall that here a, b, c, d and, are all known, onl k 1 and k 2 are unkown! OK, well, we know how to solve such a sstem, namel add c/a times the first equation to the second equation, { = ak1 + bk 2 = ck 1 + dk 2 = and then solve the second equation for k 2, k 2 = a c a = ak 1 + bk 2 c a + = 0k 1 + a a c = ad bc ad bc ( bc a + d k 2 Finall, substitute k 2 back into either equation and solve the result for k 1. For eample, substitute it into the first equation: ( a c = ak 1 + bk 2 = ak 1 + b = k 1 = b( a c ad bc ad bc a d b = ad bc Thus, = k 1 u + k 2 v = ( d b u + ad bc ( a c v (1.2 ad bc If we want to define a function ϕ : R 2 R 2 changing ij-coordinates to uv-coordinates, it is the function sending (, to (k 1, k 2, namel ϕ : R 2 R 2 (k 1, k 2 = ϕ(, = ( d b a c, ad bc ad bc (1.3 We can epress this in matri notation (and this eplains the linearit of the uv-coordinates: ( ( ( ( k1 1 d b = ϕ = (1.4 k 2 ad bc c a 2

3 Eample 1.3 Let s tr this out with some specific vectors. Let u = (1, 1 and v = (1, 2 and let = (3, 5. v k 2 k 1 u Let s find k 1 and k 2 such that = k 1 u + k 2 v, and check that the above formula works. Well, and indeed d b k 1 = ad bc = = 1 a c k 2 = ad bc = = 2 k 1 u + k 2 v = 1(1, 1 + 2(1, 2 = (1, 1 + (2, 4 = (1 + 2, = (3, 5 = And we can do this for an in R 2. There are of course other coordinate sstems, and the most common are polar, clindrical and spherical. Let us discuss these in turn. Eample 1.4 Polar coordinates are used in R 2, and specif an point other than the origin, given in Cartesian coordinates b = (,, b giving the length r of and the angle which it makes with the -ais, r = = ( and = arctan (1.5 Notice that the domain of arctan is naturall ( π 2, π 2, so this set of coordinates onl works on the right half-plane {(, > 0}. We can adjoin the -ais and the left half of the plane, but there we have to define differentl. We leave the tedious details to the reader. We then get the composite function ϕ : R 2 R 2 ( 2 ( (r, = ϕ(, = + 2, arctan (1.6 The thing to note is that we can go back and forth between Cartesian and polar coordinates. The reverse direction, from (r, to (, is ψ = ϕ 1 : R 2 R 2 (, = ψ(r, = ( r cos, r sin (1.7 3

4 We usuall see this in the form, = r cos and = r sin (1.8 r = (, = (r cos, r sin Eample 1.5 Clindrical coordinates are like polar coordinates, but we include a third dimension as well, which we call z (we ma as well identif it with the z-coordinate in Cartesian coordinates, though we can just as well identif it with an line through the origin. Thus, our change-of-coordinates functions changing Cartesian into clindrical coordinates is ϕ : R 3 R 3 ( 2 ( (r,, z = ϕ(,, z = + 2, arctan, z (1.9 and its inverse is the map ψ = ϕ 1 : R 3 R 3 (,, z = ψ(r,, z = ( r cos, r sin, z (1.10 z 4

5 Eample 1.6 Spherical coordinates are used when working with a sstem having inherent spherical smmetr, for eample the gravitational or the electric field surrounding a point particle. z z ϕ ϕ Figure 1.1: Spherical Coordinates The basic idea behind spherical coordinates is that a point = (,, z can be entirel determined not onl b the coordinates, and z in the coordinate directions i, j and k, but b it s length ρ = and two angles, one between and k (the z-ais, ( k ϕ = arccos k ( z = arccos ρ and one between the projection of onto the, -plane and the -ais, i.e. between (,, 0 and i = (1, 0, 0, ( ( (,, 0 (1, 0, 0 = arccos = arccos (,, 0 (1, 0, This could also be described as the arctangent of /, if we draw a right triangle with legs and and hpotenuse ( = arctan and there is a similar epression involving arcsine. This gives us a function from R 3 to R 3, converting Cartesian coordinates into spherical coordinates: C : R 3 R 3 ( ( z ( (ρ,, ϕ = C(,, z =, arccos, arctan (1.11 5

6 In components this is ρ = ( z = arccos ( ( ( ϕ = arctan = arccos = arcsin (1.12 The reverse transformation, S = C 1, converting spherical to Cartesian coordinates, is which, in components is S = C 1 : R 3 R 3 (,, z = S(ρ,, ϕ = ( (1.13 ρ sin ϕ cos, ρ sin ϕ sin, ρ cos ϕ = ρ sin ϕ cos = ρ sin ϕ sin z = ρ cos ϕ (1.14 Refer to Figure 1.6 for the derivation to follow: First, note that in that figure, the projection of = (,, z onto the -plane is (,, 0, so is eas to compute using the triangle = r (,, 0 = (r cos, r sin, 0 Indeed, is easil seen to be the arctangent of /, or the arccosine of / 2 + 2, or the arcsine of / In fact, in the -plane we have the standard polar coordinates. Hence we also have = r cos and = r sin. Now let s look at the 3D picture. The vertical triangle has horizontal leg r = 2 + 2, 6

7 vertical leg z, and hpotenuse ρ = r 2 + z 2 = z 2 =, z ϕ ρ ϕ z r From this picture we can see that r ρ = sin ϕ, so that r = ρ sin ϕ. We can substitute this epression into our earlier epressions for and in polar coordinates: = r cos = ρ sin ϕ cos = r sin = ρ sin ϕ sin Finall, z ρ = cos ϕ, so we also have z = ρ cos ϕ 2 New Coordinates and the Chain Rule What are coordinates good for? The answer is the are basicall the most important thing in vector calculus, because the are precisel what simplifies complicated problems. The onl difficult in switching to more convenient coordinates is that ou have to then translate the whole problem ou re looking at into these new coordinates, and there ma well be need to use the chain rule. The reason is most interesting problems in phsics and engineering are equations involving partial derivatives, that is partial differential equations. These equations normall have phsical interpretations and are derived from observations and eperimentation. Once handed to us, however, we must treat the problems mathematicall, and our purpose here is to elucidate some of the basic methods of this mathematical treatment. Let us look at some eamples. Eample 2.1 Suppose ou are asked to solve the following first order linear partial differential equation, 3u + 4u = 0 or, in other notation, 3 u + 4 u = 0 A solution of this equation is a function u : R 2 R, z = u(,, whose partial derivatives satisf this equation at all points (, in the plane. How are we to solve such a problem? Well, the first thing to notice is that what we have here is in fact a matri (or dot product on the left hand side, namel 3u + 4u = ( ( 3 u u = Du v 4 7

8 where v = (3, 4 is thought of as a column vector here. But we also know that this product is precisel the directional derivative of u in the direction of v, D v u = Du v = 3u + 4u v Thus, our original equation boils down to solving the directional derivative equation D v u = 0 Well, what does it sa about u that it s directional derivative in the direction of v = (3, 4 is 0? It means u is not changing in that direction, at an point in R 2! Pick a line parallel to v, that is having slope 4/3, = 4 + b or 4 3 = 3b 3 where b is arbitrar. Then, on this line, u will be constant. The onl change in u will occur if we move from one line to another, that is in a direction perpendicular to v (for we note, of course, that the equation of the line can be epressed as a dot product, the vector (4, 3 with the vector (,, and the vector (4, 3 is orthogonal to v!. Thus, an differentiable function f : R R taking η = 4 3 as input will work for us, so a general solution is u = f g, whereg : R 2 R, g(, = 4 3, and f : R R is arbitrar If we chose coordinates as in Eample 1.2 in such a wa that one of the coordinate vectors were v, then the fact that D v u = 0 would mean that it s partial derivative with respect to that new coordinate would be zero, alwas! Consequentl, we would onl have to worr about the other partial of u, in the other coordinate direction. Let s call the other coordinate w, which we ma as well pick to be orthogonal to v. Thus, if sa ξ and η are our new coordinates, meaning that ξv + ηw = (, for all (, in R 2, then our original partial differential equation would simplif to 3u + 4u = 0 = u η = 0 Let us verif that this is indeed a solution: Let u(, = (f g(, = f(4 3. Then b the chain rule we have 3u + 4u = 3 f( f(4 3 = 3f ( f (4 3 ( 3 = (12 12f (4 3 = 0 For eample, we could choose f(t = e t, or f(t = sin t, or f(t = t 3 + 5, or anthing we want as long as it s differentiable on R! In these cases, u(, = e 4 3, or u(, = sin(4 3, or u(, = ( are all solutions. There is another wa to view this problem which highlights the change of coordinates. This still involves the observation that u is constant along the lines 4 3 = c. The idea, then, is to pick coordinates, one parallel to v, the other perpendicular to v, for then the partial in one direction will be identicall zero and the result will be a simpler partial differential equation. Let us eplain. 8

9 Define new coordinates ξ = η = 4 3 Note that ξ is in the direction of v, and η is perpendicular to it. If we solve this sstem for and this will be clearer: and therefore (, = = 3 25 ξ η = 4 25 ξ 3 25 η ( 3 25 ξ η, 4 25 ξ 3 25 η = ξ 1 1 (3, 4 + η (4, so (, is epressed as a multiple of v plus a multiple of w, the vector orthogonal to v. What happens when we consider u(ξ, η, that is the epression of u in these new coordinates? Well, let s find out b computing some partials: and consequentl, u = u ξ ξ + u η η = 3u ξ + 4u η u = u ξ ξ + u η η = 4u ξ 3u η 0 = 3u + 4u = 3(3u ξ + 4u η + 4(4u ξ 3u η = 25u ξ + (12 12u η = 25u ξ Therefore, dividing b 25 gives u ξ = 0 a much simpler partial differential equation than 3u + 4u = 0! Indeed, we can solve this b integrating with respect to ξ. When we do, since u is a function of ξ and η, we will get an arbitrar differentiable function of η, u(ξ, η = f(η Converting back to (, coordinates now we see that u(, = f(4 3 Eercise 2.2 Show that an homogeneous linear partial differential equation with constant coefficients (the technical term for these things, au + bu = 0 has a general solution given b u(, = f(b a for an twice differentiable function f : R R. 9

10 Eercise 2.3 Show that an inhomogeneous linear partial differential equation with constant coefficients (the technical term for these things, au + bu = c (where c is an nonzero constant has a general solution given b u(, = f(b ae c/(a2 + 2 for an twice differentiable function f : R R. [Hint: An ordinar differential equation of the form a + b = 0 for a function of (which we hope to discover, for that would be solving the differential equation, and constants a and b can be solved b noticing that we can rewrite it as / = b/a, then noting that / = (ln via the chain rule, so that in fact we have (ln = b/a. Integrating with respect to gives ln = (b/a + c for some arbitrar constant of integration c. Taking the eponential of both sides gives = e (b/a+c = Ce (b/a, where C = e c. Thus, we have solved our differential equation, up to a constant C.] Eample 2.4 Consider the Laplace operator on R 2, = This operator takes twice differentiable functions f : R 2 R to at least continuous functions from R 2 to R, f = 2 f f 2 : R2 R This is the epression of in standard Cartesian coordinates, but what if we wanted to eploit some internal smmetr of a problem. Suppose that we were tring to solve a second order differential equation, called the wave equation (which obviousl models wave propagation through some medium, 2 u t 2 = u (Here u : R 3 R is given also as a function of time, u(t,,, and the z-value of u at a point (, in R 2 at time t represents the height of the wave. Suppose further that we know the wave has some circular smmetr, that is the wave propagates outward smmetricall about some point. Then it would obviousl be more convenient to treat this problem, that is to look for solutions to u tt = u, if we convert to polar coordinates. M claim is that the Laplace operator in polar coordinates takes the epression = 1 r r + 2 r r Let s prove this b starting with the right-hand side and then showing it s equal to the left-hand side. Well, this means using the polar coordinates of Eample 1.4, = r cos and = r sin and the chain rule on an twice-differentiable function f : R 2 R, since what we want to compute is the partials of f(r,, which means the partials of the composition f ψ : R 2 R where ψ = ϕ 1 : R 2 R 2 (, = ψ(r, = ( r cos, r sin 10

11 is the change-of-coordinate function from polar to Cartesian. Well, the chain rule in components is f r (r, = f (ψ(r, r (r, + f (ψ(r, r (r, f (r, = f (ψ(r, (r, + f (ψ(r, (r, which we ll abbreviate, in the interest of space, to f r = f r + f r f = f + f thus avoiding mention of where we evaluate each (but as ou have noted in our WebWork and homework, where ou evaluate e.g. f is different from where ou evaluate r!. Then, since r = cos, r = sin, = r sin, = r cos we have f r = f cos + f sin f = f r sin + f r cos Let s now compute the second partials. This will involve using the chain rule again on f and f, for eample (f r = (f r + (f r = f r + f r = f cos + f sin. So, and f rr = (f r r ( = f cos + f sin r = ( f f r r sin = (f r + f r cos + (f r + f r sin f = (f ( = f r sin + f r cos = f cos 2 + 2f sin cos + f sin 2 = r ( f sin + r( f cos ( (f = r sin + f ( ( (f sin + r cos + f ( cos (f (f = r( + f sin + f cos + r( + f cos f sin (f = r( ( r sin + f (r cos sin + f cos (f +r( ( r sin + f (r cos cos f sin = f r 2 sin 2 2f r 2 sin cos + f r 2 cos 2 rf cos rf sin 11

12 Therefore, the punchline is 1 r f r + f rr + 1 r 2 f = 1 ( ( f cos + f sin + f cos 2 + 2f sin cos + f sin 2 r + 1 (f r 2 r 2 sin 2 2f r 2 sin cos + f r 2 cos 2 rf cos rf sin 1 = r f cos + 1 r f sin + f cos 2 + 2f sin cos + f sin 2 +f sin 2 2f sin cos + f cos 2 1 r f cos 1 r f sin ( = f sin 2 + cos 2 ( + f sin 2 + cos 2 = f + f This completes the proof. Eample 2.5 Show that the Laplace operator on R 3, = z 2 has the following epression in spherical coordinates: = 2 r r r 2 1 sin 2 or, in even more compact notation, = 1 r 2 ( r 2 + r r 1 r 2 sin 2 φ r ( sin + r + 1 cos r 2 sin 1 r 2 sin 2 You can check out the Plane Math page for a derivation of this, but I would recommend that onl as a wa to check our work. 2 φ 2 12