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Math 241, Exam 2. 10/26/11. Name: • Read problems carefully. Show all work. No notes, calculator, or text. • The exam is roughly 15 percent of the total grade. There are 100 points total. • Write your full name in the upper right corner of page 1. • Do problem 1 on p. 1, problem 2 on p. 2,...(or at least present your solutions in numerical order). • Circle or otherwise clearly identify your final answer. x4 + 3y 3 . Does lim f (x, y) exist? If it exists, (x,y)→(0,0) 2x3 − y 3 give its value. If it does not exist, briefly explain how you reached your conclusion. 1. (10 points): Suppose that f (x, y) = Solution: Approach along x = 0 and along y = 0. • Along x = 0: We have 3y 3 lim f (0, y) = lim = −3. y→0 y→0 −y 3 • Along y = 0: We have x4 = 0. x→0 2x3 lim f (x, 0) = lim x→0 Since these limits disagree, we see that x4 + 3y 3 does not exist. (x,y)→(0,0) 2x3 − y 3 lim 2. (10 points): Find an equation of the form ax + by + cz = d for the plane tangent to the surface F (x, y, z) = y − x2 + z 2 = 0 at the point (4, 7, 3). Solution: We have ∇F (x, y, z) = h−2x, 1, 2zi, ∇F (4, 7, 3) = h−8, 1, 6i. It follows that the tangent plane has equation −8(x−4)+1(y−7)+6(z−3) = 0 ⇐⇒ −8x+y+6z = −32+7+18 = −7 ⇐⇒ 8x−y−6z = 7. 3. (20 points): (a) (12 points): Suppose that u = z sin x + zy 3 − y 2 , x = et , y = ln(t + 1), z = 3t5 + 2t + 1. Use the Chain rule to compute du/dt. You do not need to put the answer in terms of t alone; you may use the variables x, y, and z as well. Solution: We have du ∂u dx ∂u dy ∂u dz 1 t 2 = + + = z cos x·e +(3zy −2y)· +(sin x+y 3 )(15t4 +2). dt ∂x dt ∂y dt ∂z dt t+1 (b) (8 points): Suppose that z = F (x, y, t), x = G(t), y = H(s, t). Use partial derivatives of F , G, and H to write down the form of the chain rule that you would use to compute ∂z/∂t. (Use a tree diagram to show variable dependencies.) Solution: ∂F dG ∂F ∂H ∂F ∂z = + + . ∂t ∂x dt ∂y ∂t ∂t 4. (20 points): Let f (x, y, z) = xy + z 2 . (a) (13 points): Find the directional derivative of f (x, y, z) at (1, 1, 1) in the direction of (5, −3, 3). Solution: We have h4, −4, 2i 2 2 1 ~u = √ = ,− , , ∇f (x, y, z) = hy, x, 2zi, ∇f (1, 1, 1) = h1, 1, 2i. 3 3 3 36 It follows that D~u f (1, 1, 1) = h1, 1, 2i · 2 2 1 ,− , 3 3 3 = 2 2 2 2 − + = . 3 3 3 3 (b) (7 points): Give a unit vector in the direction in which f decreases most rapidly at (1, 1, 1). Solution: We have ∇f (1, 1, 1) h1, 1, 2i − =− √ = k∇f (1, 1, 1)k 6 1 1 2 −√ , −√ , −√ 6 6 6 . 5. (25 points): Let f (x, y) = x2 − 6xy + y 2 + 4x + 4y. (a) (12 points): Use the second derivative test to locate all local maxima, minima, and saddle points of f (x, y). Solution: We solve the system: fx (x, y) = 2x − 6y + 4 = 0 ⇐⇒ x − 3y + 2 = 0 ⇐⇒ x = 3y − 2 fy (x, y) = 2y − 6x + 4 = 0 ⇐⇒ y − 3x + 2 = 0 ⇐⇒ y = 3x − 2. We substitute the first equation in the second to obtain y = 3(3y − 2) − 2 = 9y − 8 ⇐⇒ y = 1, x = 1. 2 It follows that (1, 1) is the unique critical point. To apply the second derivative test, we compute fxx (x, y) = 2, fyy (x, y) = 2, fxy (x, y) = fyx (x, y) = −6, 2 −6 = 4 − 36 = −32 < 0. D = −6 2 We conclude that (1, 1) is a saddle point. (b) (13 points): Let D = {(x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 5}. Find the absolute maximum and minimum of f (x, y) on D. Justify your conclusions. Solution: It suffices to analyze f (x, y) on the boundary of D. • x = 0: f (0, y) = u1 (y) = y 2 + 4y; u01 (y) = 2y + 4 = 0 ⇐⇒ y = −2; (0, −2) 6∈ D. • y = 0: f (x, 0) = u2 (x) = x2 + 4x; (−2, 0) 6∈ D. • x = 2: f (2, y) = u3 (y) = 4 − 12y + y 2 + 8 + 4y = y 2 − 8y + 12; u03 (y) = 2y − 8 = 0 ⇐⇒ y = 4; (2, 4) ∈ D. • y = 5; f (x, 2) = u4 (x) = x2 − 26x + 45; u04 (x) = 2x − 26 = 0 ⇐⇒ x = 13; (13, 5) 6∈ D. We note that the critical point (1, 1) in the interior of D is a saddle, so f does not have an extremum there. It remains to evaluate f at (2, 4) and at the corners: (0, 0), (0, 5), (2, 0), (2, 5). We have: f (2, 4) = −4, f (0, 0) = 0, f (0, 5) = 45, f (2, 0) = 12, f (2, 5) = −3. Therefore, the absolute minimum is at (2, 4) while the absolute maximum is at (0, 5). 6. (15 points): Use Lagrange multipliers to find the point(s) (x, y, z) which maximize and minimize f (x, y, z) = x + 2y + 8z subject to the constraint g(x, y, z) = x2 + y 2 + 2z 2 = 37. Solution: We compute ∇f (x, y, z) = h1, 2, 8i , ∇g(x, y, z) = h2x, 2y, 4zi . The equation ∇f (x, y, z) = λ∇g(x, y, z) gives 1 1 2 , 2 = 2λy ⇐⇒ y = , 8 = 4λz ⇐⇒ z = . 2λ λ λ We substitute these values in g(x, y, z) = 37 to obtain 2 2 2 1 1 2 1 + 4 + 32 37 g(x, y, z) = + +2 = = 2 = 37 ⇐⇒ λ = ±1/2. 2 2λ λ λ 4λ 4λ 1 = 2λx ⇐⇒ x = We substitute λ in for x, y, and z to obtain points (1, 2, 4), (−1, −2, −4). The maximum is at f (1, 2, 4) = 37; the minimum is at f (−1, −2, −4) = −37. 3