MATH 201 Exam 2 with Solution, Fall 2009

American University of Beirut MATH 201 Calculus and Analytic Geometry III Fall 2009-2010 quiz # 2 - solution 1. for each of the following functions, find the domain and the range. a) f(x, y, z) = √ 9 − x2 − y2 − z2 Df = {(x, y, z) ∈ R3/x2 + y2 + z2 ≤ 9}; Range: [0, 3]; boundary: {x2 + y2 + z2 = 9}; the domain is closed cause it contains its boundary, and bounded. b) g(x, y) = x x2−y Dg = {(x, y) ∈ R2/y ̸= x2}; Range: R; boundary: {y = x2}; the domain is open cause it does not contain its boundary, and unbounded. 2. remark: take to the same denominator, then use series representation of cos lim t→0 ( 1 2 − 2 cos t − 1 t2 ) = lim t→0 (t2 − 2 + 2 cos t t2(2 − 2 cos t) ) = lim t→0 ( t2 − 2 + 2(1 − t2 2 + t4 24 + o(t6)) t2(2 − 2(1 − t2 2 + t4 24 + o(t6))) ) = lim t→0 ( t4 12 + o(t6) t4 − t6 12 + o(t6) ) = lim t→0 ( 1 12 + o(t2) 1 − t2 12 + o(t4) ) = 1 12 3. give the Taylor series expansion of f(x) = 2+x (1−x)(1+2x) at x = −1, then find f(n)(−1) remark: do not derive f directly! use a substitution Let u = x+1, then g(u) = f(u−1) = u + 1 (2 − u)(2u − 1) = 1 2 − u + 1 2u − 1 = 1 2. 1 1 − u 2 − 1 1 − 2u The Taylor series of f at x = −1 is the Maclaurin series of g, hence g(u) = 1 2 ∞ ∑ n=0 (u 2 )n − ∞ ∑ n=0 (2u)n = ∞ ∑ n=0 ( 1 2n+1 − 2n ) un , and f(x) = ∞ ∑ n=0 ( 1 2n+1 − 2n ) (x + 1)n is the Taylor series of f at x = −1 f(x) = ∞ ∑ n=0 f(n)(−1) n! (x + 1)n , then by comparison, f(n)(−1) = n! × ( 1 2n+1 − 2n ) method 2 : write f in fractions then derive ! f(x) = 2 + x (1 − x)(1 + 2x) = 1 1 − x + 1 1 + 2x f′(x) = 1 (1 − x)2 − 2 (1 + 2x)2 f′′(x) = 2 (1 − x)3 + 4 (1 + 2x)3 , and hence f(n)(x) = n! (1 − x)n+1 + (−1)nn!2n (1 + 2x)n+1 , and then f(n)(−1) = n! 2n+1 − n!2n 4. find the area inside the circle r = −2 cos θ and outside the circle r = 1 remark: sketch the curves; find the points of intersection, then find the area. method 1 : Area= ∫ 4π 3 2π 3 1 2((−2 cos θ)2−1)dθ = π 3 + √ 3 2 (by using the rule cos(2θ) = 2 cos2 θ− 1) the area can also be calculated by another way 6 - x y r1 = 1 r2 = −2 cos θ −2 0 1 R method 2 : (longer!) Area(R)= π − 2×Area(shaded region) = π − 2 × (∫ 2π 3 π 2 1 2(−2 cos θ)2dθ + ∫ π 2π 3 1 2dθ ) = π − 2 × (∫ 2π 3 π 2 (cos(2θ) + 1)dθ + ∫ π 2π 3 1 2dθ ) = π − 2 × [ sin(2θ) 2 + θ ] 2π 3 π 2 + π 3 = π 3 + √ 23 θ = 4π 3 θ = 2π 3 5. a. ���� x2y 2x2 + y2 ���� ≤ ���� x2y 2x2 ���� ≤ |y| 2 −→ 0 as (x, y) → (0, 0), then by sandwich theorem, f(x, y) −→ 0 as (x, y) → (0, 0) b. consider the path y = m(x − 1); note that (1, 0) belongs to this path! f(x, m(x − 1)) = m(x − 1)2 x2 + m2(x − 1)2 = m 1 + m2 −→ m 1 + m2 as (x, y) → (1, 0); the limit depends on m, then by the two path test, f has no limit at (1, 0) 6. a. a0 = 1 2π (∫ π 0 x dx + ∫ 2π π dx ) = π 4 + 1 2 b. f(x) = π 4 + 1 2 + +∞ ∑ n=1 1 π (−1)n − 1 n2 cos(nx) + +∞ ∑ n=1 bn sin(nx) x = π is a point of discontinuity of f; then at x = π the series converges to f(π+)+f(π−) 2 π 2 + 1 2 = π 4 + 1 2 + +∞ ∑ n=1 1 π (−1)n − 1 n2 cos(nπ) split the sum into even and odd gives π 4 = +∞ ∑ n=1 1 π. (−1)n − 1 n2 . (−1)n = +∞ ∑ n=0 1 π. (−1)2n+1 − 1 (2n + 1)2 . (−1)2n+1+ +∞ ∑ n=1 1 π. (−1)2n − 1 (2n)2 . (−1)2n the second term is equal to 0, then π 4 = +∞ ∑ n=0 1 π. 2 (2n + 1)2 and finally π2 8 = +∞ ∑ n=0 1 (2n + 1)2