3/14/2003, CHAIN RULE WRAPUP Math 21aHomework for Monday 3/17/2003:Section 11.6: 20, 26, 30, 42, 44DEFINITIONS.∇f(x, y) = (fx, fy), ∇f (x, y, z) = (fx, fy, fz), gradient ∇ is spelled ”Nabla”. (=name of an Egyptian harp, theHebrew word ”nevel”=harp seems to have the same aramaic origin).D~vf = ∇f · ~v directional derivativeFACTS:ddtf(~r(t)) = ∇f(~r(t)) · ~r0(t) chain rule.∇f(x0, y0, z0) is orthogonal to the level surface f (x, y, z) = c where c = f(x0, y0, z0).ddtf(~x + t~v) = D~vf by chain rule.x−x0fx(x0,y0,z0)=y−y0fy(x0,y0,z0)=z−z0fz(x0,y0,z0)normal line to f (x, y, z) = c at (x0, y0, z0).(x − x0)fx(x0, y0, z0) + (y − y0)fy(x0, y0, z0) + (z − z0)fz(x0, y0, z0) = 0 tangent plane at (x0, y0, z0)The directional derivative is maximal |∇f|2in the ~v = ∇f direction.f(x, y) increases if we walk on the xy-plane in the ∇f direction.Partial derivatives are special directional derivatives.If D~vf(~x) = 0 for all ~v, then ∇f(~x) =~0.ALGORITHMS.Implicit differentiation: f (x, y(x)) = 0, compute yxwithout knowing y(x): fx1 + fyy0(x) = 0 gives y0(x) =−fx/fy. Compute directional derivatives. Find direction where directional derivative is maximal or minimal.EXAMPLE PROBLEM. Show that the two surfaces x2+ y2= z2, x2+ y2+ z2= 1 intersect at a right angle atall points of intersection.SOLUTION. The angle of intersection is the angle between the normal vectors which are (2x, 2y, −2z) and(2x, 2y, 2z). The dot product is 4x2+ 4y2− 4z2= 0 (using the first equation).WHERE DOES THE CHAIN RULE MATTER (informal)• The chain rule is used in change of variable formu-las. For example, if f is a function in the plane andx = r cos(θ), y = r sin(θ) is a change of coordinatesinto polar coordinates, then calculating the gradientof f in the new coordinates requires the chain rule.• The chain rule will be used in the fundamental the-orem for line integrals:Rba∇f(~r(t)) · ~r0(t) dt =f(b) − f(a) which appears later in this course. Thechain rule said that inside the integral, we haved/dtf(~r(t)).• Gradients are orthogonal to level surfaces: as-sume we have a curve ~r(t) on a surface g(x, y, z) = c.Because we move on the surface where g is constant,we have d/dtg(~r(t)) = 0. The chain rule says∇g(~r(t)) · ~r0(t) = 0. In other words, the gradient ofg is orthogonal to the surface g = const. (We hadargued earlier using the linear approximation of g.)• The chain rule illustrates also the Lagrange multi-plier method which we will see later. If we want toextremize f(x, y) on the constraint g(x, y) and ~r(t) isa curve on g(x, y) = c, then at a critical point, wemust have d/dtf (~r(t)) = 0. With the chain rule onecan see that the velocity vector ~r0(t) is orthogonal to∇f(~r(t)). Because the velocity vector is orthogonalto ∇g The vectors ∇g and ∇f have to be parallel.• In fluid dynamics, you often see partial differentialequations starting with ut+ u∇u, where u is the ve-locity of the fluid. For example, the conservation ofmomentum of an ideal fluid in the presence of an ex-ternal force f is ut+ u · ∇u = (1/ρ)∇p + f, where pis the pressure and ρ is the density. What does thismean? The term ut+ u · ∇u is the change of velocityu in the coordinate frame of a particle in the fluid.(u(t, x, y, z) = (u1(t, x, y, z), u2(x, y, z), u3(x, y, z) is avector and ∇u means applying the gradient to eachcoordinate). For any quantity f like pressure, vortic-ity, density etc. , Df/Dt = ft+ u · ∇f means thetime derivative of f moving with the fluid.• In chaos theory, one wants to understand what hap-pens after iterating a map f. If f(n)(x) = ff(n−1)(x),then (f(n))0(x) = f0(f(n−1)x)f0(f(n−2)(x))...f0(x) isa product of matrices. Chaos means that (fn)0(x)grows exponentially for a large set of x. A mea-sure of chaos is the Lyapunov exponent λ =n−1log |(fn)0(x)| in the limit n → ∞. If this num-ber is positive, one has sensitive dependence on initialconditions. The map x 7→ 4x(1 − x) on the interval[0, 1] for example has this property. Small chaos oforder λ = 10−15is present in the solar system. Largechaos λ = 1015is present in a small volume of argongas at room temperature.3/14/2003, ICE Chain rule Math21aLet f (x, y, z) = z + x + sin(z) and ~r(t) = (cos(t), sin(t), t).Findddtf(r(t)) at the point t = 0 by differentiating the function t 7→ f(~r(t)).Findddtf(~r(t)) at the point t = 0 using the chain rule.The equation f(x, y, z) = 0 defines z = z(x, y). Find ∂z(x, y)/∂x at the point (0,
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