18.02 Problem Set 3Due Thursday, March 5 at 1:00 PM in 2-106ReadingThe material for this problem set is covered in section SD of the 18.02 Notes, Exercises andSolutions, and sections 19.4, 19.5, 19.6, 19.7 of Simmons, Calculus with Analytic Geometry, 2ndEdition.Part A (10 points)Exercises from the Supplementary Notes and Exercises. Please turn in the 10 underlined problemsfor 1 point each, the others are for more practice:• (2C): 2C-1, 2C-2, 2C-3, 2C-4• (2D): 2D-1, 2D-2, 2D-3, 2D-4, 2D-6, 2D-7• (2E): 2E-1, 2E-2, 2E-4• (2F): 2F-1, 2F-2, 2F-5• (2H): 2H-1, 2H-3, 2H-4, 2H-5Part B (25 points)1. (8 points, lecture 10) Consider a triangle in the plane, with angles α, β, γ. Assume that thatthe radius of its incircle (the circle tangent to all three sides) is 1.(a) By decomposing the triangle into six right triangles having the incenter as a commonvertex, express the area A of the triangle in terms of α, β, and γ. The use your result toshow that A can be expressed as a function of the two variables α and β by the formulaA = cotα2+ cotβ2+ tanα + β2(b) What is the set of possible values for α and β? Find all the critical points of the functionA in this region.(c) By computing the values of A at the critical points and its behavior on the boundary ofthe region where it is defined, find the maximum and the minimum of A (justify youranswer). Describe the shapes of the triangles corresponding to these two situations.(d) Use the second derivative test to confirm the nature of the critical points you found in(b).2. (3 points, lecture 11) The electrical resistance of a piece of wire is given by the formulaR = ρL/S, where the resistivity ρ is a constant depending only on the material, L is thelength and S is the cross-section area.(a) Show thatdR = R1LdL −1SdS1(b) The electrical resistance of a copper wire of length 100 m and cross-section 1 mm2isequal to 1.5 ohm. Using the differential, approximate the resistance when the lengthbecomes 105 m and the cross-section becomes 1.1 mm2. Compare your approximationwith the actual value.3. (5 points, lecture 11) Let x = uv, y =12(v2− u2), and f = f(x, y).(a) Use the chain rule to derive the change of variables formula in matrix form:fufv= Afxfy(b) Invert the matrix A to obtain a formula expressing fxand fyin terms of fuand fv.(c) If f(u, v) = ln(u2+ v2), express fxand fyinterms of u and v.4. (5 points, lecture 12) Consider the functionf(x, y) = x(x − 1)(x − 2) + (y − 1)(x − y)(a) Find the maximum and minimum values of the direction derivativedfdsˆuat (1, 3/2) asˆu varies.(b) Say in which direction ˆu the maximum and minimum occur.(c) Find the direction(s) ˆu for whichdfdsˆu= 0.5. (5 points, lecture 12)(a) Find the direction from (4, 2, 3) in which g(x, y, z) = x2+ y2− 6z decreases fastest.(b) Follow the line in the direction you found in part (a) to estimate, using linear approxima-tion, the location of the point closest to (4, 2, 3) at which g = 0. Do not use a calculator.Express your answer using fractions. Next use a calculator to evaluate g at your point.(The value should be reasonably
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