MATH 201 Problems Sample for Exam 2, Spring 2012

1) Let f(x, y) = � 9 − x2 − y2 a) Give the domain and the range of f: Df = {(x, y) ∈ R2; x2+y2 ≤ 9}, and Range(f) = [0, 3] b) Sketch some level curves of f 1.4 1.4 1.4 1.4 1.6 1.6 1.6 1.6 1.8 1.8 1.8 1.8 2 2 2 2 2.2 2.2 2.2 2.2 2.2 2.2 2.2 2.2 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.8 2.8 2.8 2.8 2.8 The level curves of z=sqrt(9−y2–x2) x y 5 10 15 20 25 30 35 40 5 10 15 20 25 30 35 40 c) Sketch the graph of f −2 −1 0 1 2 −2 −1 0 1 2 1 1.5 2 2.5 3 x z=sqrt(9−y2–x2) y z d) find fx(0, 0) and fy(0, 0) (by using the definition) fx(0, 0) = lim x→0 f(x,0)−f(0,0) x = lim x→0 √ 9−x2−3 x = lim x→0 (9−x2)−9 x( √ 9−x2+3) = −lim x→0 x √ 9−x2+3 = 0, and fy(0, 0) = 0 (by symmetry) e) prove that f is differentiable at (0, 0) lim (x,y)→(0,0) f(x,y)−f(0,0)−fx(0,0)∆x−fy(0,0)∆y √ x2+y2 = lim (x,y)→(0,0) √ 9−x2−y2−3 √ x2+y2 = lim (x,y)→(0,0) (9−x2−y2)−9 √ x2+y2(√ 9−x2−y2+3) (multiplying by the conjugate) = − lim (x,y)→(0,0) √ x2+y2 √ 9−x2−y2+3 = 0 hence f is differentiable at (0, 0) 2) Find the directional derivative of the f(x, y) = 2xy − y2 at P = (5, 5) in the direction of u = 4i + 3j solution: ∇f = (2y, 2x − 2y) ⇒ ∇f(5, 5) = (10, 0); v = u |u| = 4i+3j √ 42+32 = 4 5i + 3 5j; hence (Df)u,P = ∇f(P) • v = 8 3) Find the direction of maximum increase of f(x, y) = x2 − 3xy + 4y2 at P(1, 2). Is there is a direction u in which the rate of change of f at P(1, 2) equals 14? Justify your answer. solution: ∇f = (2x − 3y, −3x + 8y), then ∇f(1, 2) = (−4, 13), and |∇f(1, 2)| = √16 + 169 = √ 185 The direction of maximum increase of f is the direction of ∇f(1, 2), hence (−4/ √ 185, 13/ √ 185) The rate of change in the direction of maximum increase is √ 185 < 14, then there is no direction u in which the rate of change of f at P(1, 2) equals 14. 4) The derivative of of f(x, y) at P(1, 2) in the direction of u = i + j is 2 √ 2 and in direction of v = −2j is −3. Find the derivative of f in the direction of w = −i − 2j. solution: (Df)u,P = 2 √ 2 ⇒ fx(P) × 1 √ 2 + fy(P) × 1 √ 2 = 2 √ 2 ⇒ fx(P) + fy(P) = 4 (Df)v,P = −3 ⇒ −fy(P) = −3 ⇒ fy(P) = 3, and then fx(P) = 1, hence (Df)w,P = 1 × −1 √ 5 + 3 × −2 √ 5 = −7 √ 5 5) The derivative of of f(x, y, z) at P is greatest in the direction of v = i + j + k. In this direction, the value of the derivative is 2 √ 3. Find ∇f at P. Find the derivative of f at P in the direction of w = i + j. solution: - ∇f = k. v |v| (cause ∇f and v are collinear). Moreover k = 2 √ 3, and then ∇f = 2i + 2j + 2k - (Df)w,P = ∇f(P) • w |w| = 2 × 1 √ 2 + 2 × 1 √ 2 + 2 × 0 6) Given the surface z = x2 − 4xy + y3 + 4y − 2 containing the point P(1, −1, −2). - Find an equation of the tangent plane to the surface at P. - Find an equation of the normal line to the surface at P. solution: Let w(x, y, z) = x2−4xy+y3+4y−2−z; ∇w = (2x−4y, −4x+3y2+4, −1) ⇒ ∇w(1, −1, −2) = (6, 3, −1), and the - (T) : 6(x − 1) + 3(y + 1) − (z + 2) = 0 - (N) : x = 6t + 1, y = 3t − 1, z = −t − 2 , t ∈ R 7) Find parametric equation for the line tangent to the curve of intersection of the surfaces xyz = 1 and x2 + 2y2 + 3z2 = 6 at the point P(1, 1, 1). solution: ∇f = (yz, xz, xy) ⇒ ∇f(1, 1, 1) = (1, 1, 1) , and ∇g = (2x, 4y, 6z) ⇒ ∇g(1, 1, 1) = (2, 4, 6) ������� i j k 1 1 1 2 4 6 ������� = 2i − 4j + 2k ⇒ tangent line T : x = 1 + 2t, y1 − 4t, z = 1 + 2t 8) By about how much will f(x, y, z) = ln � x2 + y2 + z2 change if the point p(x, y, z) moves from P(3, 4, 12) a distance of 0.1 unit in the direction of u = 3i + 6j − 2k ? solution: the change df is given by df = (Df)u,P × ds; ∇f = ( x x2+y2+z2 , y x2+y2+z2 , z x2+y2+z2 ), and ∇f(P) = ( 3 169, 4 169, 12 169) (Df)u,P = ∇f(P) • u |u| = 3 169 × 3 7 + 4 169 × 6 7 − 12 169 × 2 7 = 9 1183; then df = 0.9/1183 ≃ 0.0008 9) Locate all relative extrema and saddle points of f(x, y) = x3 − y3 − 2xy + 6. solution: critical points: (0, 0) (saddle point); (−2/3, 2/3) (local maximum) and the maximum is f(−2/3, 2/3) = 120/27. 10) Locate all relative extrema and saddle points of f(x, y) = 4xy − x4 − y4. solution: The critical points are (0, 0) (saddle point); (1, 1) (local maximum) and f(1, 1) = 2; (−1, −1) (local maximum) and f(−1, −1) = 2. 11) Find the absolute minimum and maximum values of f(x, y) = 3xy −6x−3y +7 on the closed triangular region R whose vertices are (0, 0), (3, 0) and (0, 5). solution: critical point: ∂f/∂x = 3y − 3 and ∂f/∂y = −3x − 6, then the only critical point is (−2, 1) on the path x = 0: g(y) = f(0, y) = −3y + 7, g′(y) = −3 ̸= 0, on the path y = 0: h(x) = f(x, 0) = −6x + 7, h′(x) = −6 ̸= 0, on the path y = 5 − 5 3x: k(x) = f(x, x) = −5x2 + 14x − 8, k′(x) = −10x + 14, k′(x) = 0 → x = 7/5, and then y = 8/3 point value (-2,1) 9 (0,0) 7 (3,0) -11 (0,5) -8 (7/5,8/3) 81/5 The minimum is then equals to −11 at (3, 0); the maximum is equals to 81/5 at (7/5, 8/3) 12) Find the absolute minimum and maximum of the function f(x, y) = x2 + 2y2 − y − 1 over the region R = {(x, y); x2 + y2 ≤ 1 , y ≥ 0}. solution: critical point: ∂f/∂x = 2x and ∂f/∂y = 4y − 1, then the only critical point is (0, 1/4) on the path y = 0: g(x) = f(x, 0) = x2 − 1, g′(x) = 2x; g′(x) = 0 ⇒ x = 0, and then y = 0; (0, 0) is also a critical point on the semi circle x2 + y2 = 1 ⇒ x2 = 1 − y2: h(y) = f(1 − y2, y) = y2 − y, h′(y) = 2y − 1, h′(y) = 0 → y = 1/2, and then x = ± √ 3/2 point value (0,1/4) -9/8 (-1,0) 0 (1,0) 0 (0,0) -1 ( √ 3/2, 1/2) 1/2 (− √ 3/2, 1/2) 1/2 The minimum is then equals to −9/8 at (0, 1/4); the maximum is equals to 1/2 at ( √ 3/2, 1/2) and (− √ 3/2, 1/2).