3/10/2003, CHAIN RULE Math 21a, O. KnillHomework: Section 11.5: 6, 18, 24, 28, 30REMEMBER? If f and g are functions of one variable t, then d/dtf (g(t)) = f0(g(t))g0(t). For example,d/dt sin(log(t)) = cos(log(t))/t.THE CHAIN RULE. If ~r(t) is curve in space and f is a function of three variables, we get a function of onevariables t 7→ f(~r(t)). The chain rule looks like the 1D chain rule, but where derivative f0is replaced with thegradient ∇f = (fx, fy, fz). The derivative isd/dtf(~r(t)) = ∇f(~r(t)) · ~r0(t)WRITING IT OUT. Writing the dot product out givesddtf(~r(t)) = fx(x(t), y(t), z(t))x0(t) + fy(x(t), y(t), z(t))y0(t) + fz(x(t), y(t), z(t))z0(t) .In the case of two variablesddtf(x(t), y(t)) = fx(x(t), y(t))dx(t)dt+ fy(x(t), y(t))dy(t)dt(see case 1 in the book).EXAMPLE. Let z = sin(x + 2y), where x and y are functions of t: x = et, y = cos(t). What isdzdt?Here, z = f(x, y) = sin(x + 2y), zx= cos(x + 2y), and zy= 2 cos(x + 2y) anddxdt= et,dydt= − sin(t) anddzdt= cos(x + 2y)ex− 2 cos(x + 2y) sin(t).EXAMPLE. If f is the temperature distribution in a room and~r(t) is the path of a spider then f(~r(t)) is the temperature, thespider experiences at time t. The rate of change depends on thevelocity ~r0(t) of the spider as well as the temperature gradient ∇fand the angle between gradient and velocity. For example, if thespider moves perpendicular to the gradient, its velocity is tangentto a level curve and the rate of change is zero.EXAMPLE. Assume the spider moves along a circle ~r(t) = (cos(t), sin(t)) on a table with temperaturedistribution T (x, y) = x2− y3. Find the rate of change of the temperature, the spider experiences in time.SOLUTION. ∇T (x, y) = (2x, −3y2), ~r0(t) = (− sin(t), cos(t)) d/dtT (~r(t)) = ∇T(~r(t)) · ~r0(t) =(2 cos(t), −3 sin(t)2) · (− sin(t), cos(t)) = −2 cos(t) sin(t) − 3 sin2(t) cos(t).APPLICATION: IMPLICIT DIFFERENTIATION.1)2D From f(x, y) = 0 one can express y as a function of x. From d/df(x, y(x)) = ∇f ·(1, y0(x)) = fx+fyy0= 0we obtainy0= −fx/fy.2)3D If z = g(x, y) is given by f(x, y, z) = 0 then fx+ fzgx= 0 and fy+ fzgy= 0 so that gx= −fx/fzandgy= −fy/fz.EXAMPLE. f(x, y) = x4+ x sin(xy) = 0 defines y = g(x). If f (x, g(x)) = 0, then gx(x) = −fx/fy=−(4x3+ sin(xy) + xy cos(xy))/(x2cos(xy)).APPLICATION: If f(x, y, z) = 0, then x = x(y, z), y = y(x, z) and z = z(x, y). From yx= −fx/fy, xz=−fz/fx, zy= −fy/fzwe get the relation yxxzzy= −1 . This formula appears in thermodynamics.DIETERICI EQUATION. In thermodynamics the temperature T , the pressure p and the volume Vof a gase are related. One refinement of the ideal gas law pV = RT is the Dieterici equationf(p, V, T ) = p(V − b)ea/RV T− RT = 0. The constant b depends on the volume of the molecules and adepends on the interaction of the molecules. (A different variation of the ideal gas law is van der Waals law(ICE)). Problem: compute VT.If V = V (T, p), the chain rule says fVVT+ fT= 0, so thatVT= −fT/fV= −(−ap(V − b)ea/RV T/(RV T2) −R)/(pV ea/RV T− p(V − b)ea/RV T/(RV2T )). (This could be simplified to (R + a/T V )/(RT/(V − b) − a/V2)).MORE GENERAL CHAIN RULE. (case 2 in the book)If~f is a vector-valued function, we can apply the chain rule for each of the components. The derivative~f0of~fis then a vector too: d/dtfi(r(t)) = ∇fi(r(t)) · r0(t).DERIVATIVE. If f : Rn→ Rmis a map, its derivative f0is the m × n matrix [f0]ij=∂∂xjfi.EXAMPLES OF DERIVATIVES.f : R → R3curvef0velocity vector.f : R3→ Rscalar functionf0gradient vector.f : R2→ R3surfacef0tangent matrixf : R2→ R2coordinate changef0Jacobean matrix.f : R3→ R3gradient fieldf0Hessian matrix.THE GENERAL CHAIN RULE.If f : Rn→ Rmand g : Rk→ Rn, we can compose f ◦ g, which is a map from Rkto Rm. The chain ruleexpresses the derivative of f ◦ g(x) = f(g(x)) in terms of the derivatives of f and g.∂∂xjf(g(x))i=Pk∂∂xkfi(g(x))∂∂xjgk(x)or short(f ◦ g)0(x) = f0(g(x)) · g0(x)Here f0(g(x)) and g0(x) are matrices and · is the matrix multiplication. The chain rule in higher dimensionslooks like the chain rule in one dimension, only that the objects are matrices and the multiplication is matrixmultiplication.EXAMPLE. GRADIENT IN POLAR COORDINATES. In polar coordinates, the gradient is defined as ∇f =(fr, fθ/r). Using the chain rule, we can relate this to the usual gradient: d/drf(x(r, θ), y(r, θ)) = fx(x, y) cos(θ)+fy(x, y) sin(θ) and d/(rdθ)f(x(r, θ), y(r, θ)) = −fx(x, y) sin(θ) + fy(x, y) cos(θ) means that the length of ∇f isthe same in both coordinate systems.PROOFS OF THE CHAIN RULE.1. Proof) By linear approximation f by a function L it is enough to check the chain rule forlinear functions f(x, y) = ax + by − c and where ~r(t) = (x0+ tu, y0+ tv) is a line. Thenddtf(~r(t)) =ddt(a(x0+ tu) + b(y0+ tv)) = au + bv and this is the dot product of ∇f = (a, b) with~r0(t) = (u, v).2. Proof) Plugging in the definitions of the derivatives using limits.WHERE DO WE NEED THE CHAIN RULE (informal).While the chain rule is useful in calculations using the composition of functions, the iteration of maps or indoing change of variables, it is also useful for understanding some theoretical aspects. Examples:• In the proof of the fact that gradients are orthogonal to level surfaces.• It appears in change of variable formulas.• It will be used in the fundamental theorem for line integrals coming up later in the course.• The chain rule illustrates also the Lagrange multiplier method which we will see later.• In fluid dynamics, PDE’s often involve terms ut+ u∇u which give the change of the velocity in theframe of a fluid particle.• In chaos theory, one wants to understand what happens after iterating a map
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