The Chain Rule. d dx x(t) dx. dt (t)

Transcription

1 The Chain Rule The Problem You alreay routinely use the one imensional chain rule t f xt = f x xt x t t in oing computations like t sint2 = cost 2 2t In this example, fx = sinx an xt = t 2. We now generalize the chain rule to functions of more than one variable. For concreteness, we consier the case in which all functions are functions of two variables. That is, we fin the partial erivatives an of a function gs,t that is efine as a composition gs,t = f xs,t,ys,t The Solution By efinition, gs+h,t gs,t f xs+h,t,ys+h,t f xs,t,ys,t s,t = lim = lim h 0 h h 0 h The first step in evaluating this limit is to use the Taylor expansion to tell us how the arguments xs+h,t an ys+h,t behave for very small h. In first year calculus, you learne a whole series of successively better approximations for a function F near a point s: Fs+h Fs Fs+h Fs+F sh Fs+h Fs+F sh+ 1 2 F sh 2 Fs+h Fs+F sh+ 1 2 F sh ! F3 sh 3. You also learne that the error introuce by truncating the expansion at the linear term can be written as 1 2 F s h 2 for some s between s an s+h. So the formula Fs+h = Fs+F sh+ 1 2 F s h 2 is exact for some unknown s between s an s + h. This is a generalization of the Mean Value Theorem. We shall use the notation O h 2 to enote any function of h that is boune by a constant times h 2 for small h. In this notation Fs+h = Fs+F sh+o h 2 c Joel Felman All rights reserve. 1

2 Applying this with Fs = xs,t an with Fs = ys,t, for any fixe t, gives Think of these as xs+h,t = xs,t+ s,th+oh2 ys+h,t = ys,t+ s,th+oh2 xs+h,t = X + x with X = xs,t an x = s,th+oh2 ys+h,t = Y + y with Y = ys,t an y = s,th+oh2 1 We have seen that fx + x,y + y fx,y+ X,Y x+ X,Y y We shall later see that, much as for single variable calculus, Subbing in 1, fx + x,y + y = fx,y+ X,Y x+ X,Y y +O x2 + y 2 gs+h,t gs,t = f xs+h,t,ys+h,t f xs,t,ys,t = fx + x,y + y fx,y = = X,Y x+ X,Y y +O x2 + y 2 xs,t,ys,t s,th+ Diviing by h an taking the limit, we get the upper formula in xs,t,ys,t s,th+oh2 s,t = s,t = xs,t,ys,t xs,t,ys,t s,t+ xs,t,ys,t s,t+ s,t xs,t,ys,t s,t The lower formula is erive similarly. The Memory Ai To help remember these formulae, it is useful to preten that our variables are physical quantities with f,g having units of grams, x,y having units of meters an s,t having units of secons. Note that a The function g appears once in the numerator on the left. The function f, which is gotten from g by a change of variables, appears once in the numerator of each term on the right. b The variable in the enominator on the left appears once in the enominator of each term on the right. c f is a functionof two inepenentvariables, x an y. Thereare two terms on the right, one for each inepenent variable of f. Each term contains the partial erivative of f with respect to c Joel Felman All rights reserve. 2

3 a ifferent inepenent variable. That inepenent variable appears once in the enominator an once in the numerator, so that its units cancel out. Thus all terms on the right han sie have the same units as that on the left han sie. Namely, grams per secon. The left han sie is a function of s an t. Hence the right han sie must also be a function of s an t. As f is a function of x an y this is achieve by evaluating f at x = xs,t an y = ys,t. I recommen strongly that you use the following proceure, without leaving out any steps, the first couple of ozen times that you use the chain rule. Step 1 List explicitly all the functions involve an what each is a function of. Ensure that all ifferent functions have ifferent names. Invent new names for some of the functions if necessary. In the example on the previous page, the list woul be fx,y xs,t ys,t gs,t = f xs,t,ys,t While the functions f an g are closely relate, they are not the same. One is a function of x an y while the other is a function of s an t. Step 2 Write own the template = Note that the template satisfies a an b above. Step 3 Fill in the blanks with everything that makes sense. In particular, since f is a function of x an y, it may only be ifferentiate with respect to x an y. = + Note that x an y are functions of s so that the erivatives an make sense. Also note that the units work out right. See c above. Step 4 Put in the functional epenence explicitly. Fortunately, there is only one functional epenence that makes sense. See above. s,t = xs,t,ys,t s,t+ xs,t,ys,t s,t Example 1: Fin t f xt,yt, for fx,y = x 2 y 2, xt = cost an yt = sint. Define gt = f xt,yt. The appropriate chain rule for this example is For the given functions g t t = xt,yt x t t+ xt,yt y t t fx,y = x 2 y 2 x,y = 2x x,y = 2y xt,yt = 2xt = 2cost xt,yt = 2yt = 2sint c Joel Felman All rights reserve. 3

4 so that g t = 2cost sint+ 2sintcost = 4sintcost t Of course, in this example we can compute gt explicitly an then ifferentiate gt = f xt,yt = xt 2 yt 2 = cos 2 t sin 2 t g t = 2cost sint 2sintcost = 4sintcost Example 2: Fin fx+ct. Define ux,t = x+ct an wx,t = fx+ct = f ux,t. Then fx+ct = w f x,t = u ux,t u x,t = cf x+ct Example 3: Fin 2 2 fx+ct. Define Wx,t = w x,t = cf x+ct = F ux,t where Fu = cf u. Then 2 2 fx+ct = W x,t = F u ux,t u x,t = cf x+ctc = c 2 f x+ct Example 4: Suppose that FP,V,T = 0. Fin P. Beforewecansolvethisproblem,wefirsthavetoeciewhatitmeans. Thishappensregularly in applications. In fact, this particular problem comes from thermoynamics. The variables P, V, T are the pressure, volume an temperature, respectively, of some gas. These three variables are not inepenent. They are relate by an equation of state, here enote FP,V,T = 0. Given values forany two of P, V, T, thethircanbe founby solving FP,V,T = 0. We arebeingaske to fin P. This implicitly instructs us to treat P, in this problem, as the epenent variable. So a careful woring of this problem which you will never encounter in the real worl woul be the following. The function PV,T is efine by F PV,T,V,T = 0. Fin P, that is the V rate of change of pressure as the temperature is varie, while holing the volume fixe. Since we are not tol explicitly what F is, we cannot solve explicitly for PV,T. So, instea we ifferentiate both sies of F PV,T,V,T = 0 with respect to T, while holing V fixe. P P + V V + = 0 Recalling that V an T are the inepenent variables an that, in computing, V is to be treate as a constant, V = 0 = 1 c Joel Felman All rights reserve. 4

5 Now putting in the functional epenence P PV,T,V,T P V,T+ PV,T,V,T = 0 an solving P V,T = PV,T,V,T PV,T,V,T P Example 5: Suppose that fx,y = 0. Fin 2 y x 2. Once again, x an y are not inepenent variables. Given a value for either x or y, the other is etermine by solving fx,y = 0. Since we are aske to fin 2 y x, it is y that is to be viewe as 2 a function of x, rather than the other way aroun. So fx,y = 0 really means, in this problem, f x,yx = 0 for all x. Differentiating both sies of this equation with respect to x, f x,yx = 0 = x f x,yx = 0 for all x Note that x f x,yx is not the same as f x x,yx. The former is, by efinition, the rate of change with respect to x of gx = f x,yx. Precisely, On the other han g x = lim gx+ x gx f x+ x,yx+ x f x,yx = lim fx+ x,y fx,y f x x,y = lim f x+ x,yx f x,yx = f x x,yx = lim 3 The two right han sies 2 an 3 are not the same. In 2, as x varies the value of y that is substitute into the first f, namely yx+ x varies. In 3, the corresponing value of y is yx an is inepenent of x. As a concrete example, suppose that fx,y = x + y. Then, yx = x so that x f x,yx = x fx, x = x [x+ x] = x 0 = 0 But fx,y = x+y implies that f x x,y = 1 for all x an y so that Now back to f x,yx = 0 f x x,yx = f x x,y y= x = 1 y= x = 1 for all x = x f x,yx = 0 = f x x,yx x x +f y x,yx y x x = 0 = y x x = f x x,yx f y x,yx = 2 y [ x 2x = fx x,yx ] x f y x,yx = f y x,yx x [f x by the chain rule x,yx ] fx x,yx x [f y x,yx ] 2 4 f y x,yx c Joel Felman All rights reserve. 5 2

6 by the quotient rule. Now it suffices to substitute in x [f x x,yx ] an x [f y x,yx ]. For the former apply the chain rule to γx = ϕ x,yx with ϕx,y = f x x,y. Substituting this an x [f x x [f y x,yx ] = γ x x = ϕ x,yx x x + ϕ x,yx y x x x,yx y x x = f xx x,yx x x +f xy [ f x x,yx = f xx x,yx fxy x,yx ] f y x,yx x,yx ] = fyx x,yx x x +f yy x,yx y x x [ f x x,yx = f yx x,yx fyy x,yx into the right han sie of 4 gives the final answer. ] f y x,yx Example 6: Fin the graient of a function given in polar coorinates. Once again, figuring out what the question means is half the battle. The graient is a quantity that frequently appears in applications e.g. the temperature graient. By efinition, the graient of a function gx,y is the vector g x x,y,g y x,y. In this question we are tol that we are given some function fr,θ of the polar coorinates r an θ. We are suppose to convert this function to Cartesian coorinates. This means that we are to consier the function gx,y = f rx,y,θx,y with rx,y = x 2 +y 2 θx,y = arctan y x Then we are to compute the graient of gx,y an express the answer in terms of r an θ. By the chain rule = r r + θ θ Our main job is to compute r, θ, r So r = 1 2 an θ 2x x 2 +y 2 = r = 1 2y 2 = x 2 +y 2 θ = y/x2 1+y/x 2 θ = 1/x 1+y/x 2 or, with all the arguments omitte, ] = r r + θ θ an express them in terms of r an θ. x x = rcosθ 2 +y 2 r y x 2 +y 2 = rsinθ r = y x 2 +y = rsinθ 2 r 2 = x x 2 +y 2 = rcosθ r 2 g x x,y = r rx,y,θx,y cosθx,y 1 rx,y = cosθ = sinθ = sinθ r = cosθ r θ rx,y,θx,y sinθx,y g y x,y = r rx,y,θx,y sinθx,y [ gx g y + 1 rx,y = r θ rx,y,θx,y cosθx,y [ ] cosθ + sinθ 1 r θ [ ] sinθ cosθ For clarity, I ve written the vectors as columns, rather than rows. c Joel Felman All rights reserve. 6