11/19/2002, REVIEW (second Midterm) Math 21a, O. KnillMIDTERM TOPICS.• Partial derivatives, gradient• Extrema of functions of two variables.• Extrema of functions with constraints.• Parameterized surfaces.• Integration of functions of two variables.• Chain rule• Linear approximation.• Estimation using linear approximation.• Tangent planes.• Integration in Polar coordinatesLEVELS OF UNDERSTANDING.1. I) KNOW. Know what the objects, definitions, theorems, names are. Know jargon and history. (Youlearn this mostly in class or by reading the book).2. II) DO. Know how to work with the objects. Persue algorithms. (You learn this mostly by doinghomework, doing in class exercices).3. III) UNDERSTAND. See different aspects, contexts. Why is it done like this? (You learn this mostlyfrom listening to lectures, discussions and by gaining experience).4. IV) APPLY. Extend the theory, apply to new situations, invent new objects. Ask: Why not...? (Youlearn this mostly by doing challenging problems, experiment with technology, writing papers).I) Definitions and objects.CONSTRAINED EXTREMUM of f constrained by G = c are obtained where ∇f = λ∇g, g = c.CHAIN RULE. If r(t) = (x(t), y(t)) is a curve and f(x, y) is a function, then d/dtf(r(t)) = ∇f(r(t)) · r0(t).CHAIN RULE. If g(x) and f(x, y) are functions, then (∂/∂x)g(f(x, y)) = g0(f(x, y))fx(x, y), (∂/∂y)g(f(x, y)) =g0(f(x, y))fy(x, y).CRITICAL POINT. ∇f(x, y) = (0, 0). Is also called stationary point.DOUBLE INTEGRAL.RbaRg(x)f (x)f(x, y) dydx is an example of a double integral.2D POLAR INTEGRAL.R RRf(r, φ) rdrdφ in polar coordinates.GRADIENT. f(x, y) function of two variables, ∇f(x, y) = (∂xf(x, y), ∂yf(x, y)) = (fx(x, y), fy(x, y)).HESSIAN MATRIX f(x, y) function of two variables. The Hessian is the matrix H(x, y) =fxx(x, y) fxy(x, y)fyx(x, y) fyy(x, y).HESSIAN DETERMINANT. D = fxx(x, y)fyy(x, y) − fxy2(x, y). LEVEL SURFACE. f (x, y, z) = 0 has gra-dients ∇f(x, y, z) as normals.LINEAR APPROXIMATION. L(x, y) = f(x0, y0) + ∇f(x0, y0) ·(x − x0, y − y0).LOCAL MAXIMUM. A critical point for which det(H(x, y)) > 0, Hxx(x, y) < 0 is a local maximum.LOCAL MINIMUM. A critical point for which det(H(x, y)) > 0, Hxx(x, y) > 0 is a local minimum.PARTIAL DIFFERENTIAL EQUATION. Equation for a function. Involves partial derivatives of the function.Example: ftt= fxxwave equation, ft= fxxheat equation.SADDLE POINT. A critical point for which det(H(x, y)) < 0.SECOND DERIVATIVE TEST. D < 0 ⇒ saddle, D > 0, fxx> 0 ⇒ min, D > 0, fyy< 0 ⇒ max.II) AlgorithmsINTEGRATION OVER A DOMAIN R.1) Eventually chop the region into pieces which can be parametrized.2) Start with one variable, say x and find the smallest x-interval [a, b] which contains R.3) For fixed x, intersect the line x = const with R to determine the y-bounds [f (x), g(x)].4) Evaluate the integralRbahRg(x)f (x)f(x, y)idydx.5) Solve the double integral by 1D integration starting from inside.6) In case of problems with the integral, try to switch the order of integration. (Go to 2) and start with y).EXAMPLE. Integrate x2y2over the triangle x + y/2 ≤ 3, x > 0, y > 1. The triangle is contained in the strip0 ≤ x ≤ 3. The x-integration ranges over the interval [0, 3]. For fixed x, we have y ≥ 1 and y ≤ 2(3 − x) whichmeans that the y bounds are [0, 2(3 − x)]. The double integral isR30R6−2x1x2y2dydx.FINDING THE MAXIMUM OF A FUNCTION f(x, y) OVER A DOMAIN R with boundary g(x, y) = c.1) First look for all stationary points ∇f(x, y) in the interior of R.2) Eventually classify the points in the interior by looking at det(H)(x, y), Hxx(x, y) at the critical points.3) Locate the critical points at the boundary by solving ∇F (x, y) = λ∇G(x, y), G(x, y) = c.4) List the values of F evaluated at all the points found in 1) and 3) and compare them.EXAMPLE. Find the maximum of F (x, y) = x2− y2− x4− y4on the domain x4+ y4≤ 1.∇F (x, y) = (2x − 4x3, −2y − 4y3). The critical points inside the domain are obtained by solving 2x − 4x3=0, −2y − 4y3= 0 which means x = 0, x = 1/4, y = 0, y = −1/4. We have four points P1= (0, 0), P2=(1/√2, 0), P3= (0, −1/√2), P4= (1/√2, −1/√2).The critical points on the boundary are obtained by solving the Lagrange equations (2x − 4x3, −2y − 4y3) =λ(4x3, 4y3), x4+ y4= 1. Solutions (see below) are P5= (0, −1), P6= (0, 1), P7= (−1, 0), P8= (1, 0). A listof function values F (P1) = 0, F(P2) = 1/2 − 1/4, F (P3) = −1/2 − 1/4, F (P4) = −2/4, F(P5) = −2, F(P6) =−2, F (P7) = 0, F(P8) = 0 shows that P2in the interior is the maximum. Indeed, the Hessian at this point isH = diag(−1, −2) which has positive determinant and negative H11.SOLVING THE LAGRANGE EQUATIONS.1) Write down the equations neatly.2) See whether some variable can be eliminated easily. λ can always be eliminated.3) If some variable can be eliminated easily, go back to 1) using one variable less and repeat.4) Try to combine, rearrange, simplify the equations. The system might not have an algebraic solution.EXAMPLE.2x − 4x3= 4λx3−2y − 4y3= 4λy3x4+ y4= 12x − (4 + 4λ)x3= 0−2y − (4 + 4λ)y3= 0x4+ y4= 1x = 0 or x = 1/√2 + 2λy = 0 or y = −1/√2 + 2λx4+ y4= 1If x=0, then y=1, or y=-1, if y = 0 then x = 1 or x = −1. There are 4 critical points.III) ”Understanding”. Try to answer questions like:• What do the Lagrange equations mean geomet-rically? Explain the method to somebody witha drawing.• Explain why only critical points can be can-didates for maxima or minima of a functionF (x, y).• What can happen at a critical point if the dis-criminant fxxfyy− f2xyis 0 at this point?• What is the geometric meaning of the entry Hxyin the Hessian?• Discuss the chain rule F (r(t)) for F (x, y) =√x2, r(t) = (x(t), y(t)) = (t, t2).• In which situations is the Lagrange multiplyerλ = 0?• If a function f (x, y) is replaced by its linear ap-proximation L(x, y), what do you expect the er-ror L(0.01, 0.01) − f(0.01, 0.01) to be?• How can the chain rule be used to find zx(x, y)if f(x, y, z(x, y)) = 0?• Can we have functions which contain two saddlesas critial points and no other critical points?• Find the second partial derivatives of g(f (x, y)).IV) Apply to new situations.Problem solving and creativity skills are acquired best by ”doing” it, by pondering over new questions, workingon specific problems. Nevertheless, there is also theoretical help: for example, G. Polya
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