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Ma 116 / Prof. Miller Name: Exam 3 April 18, 2002 Solutions ID: x2 + 3y 3 . 2 Determine which are local maxima, which are local minima, and which are saddle points. 1. [9 pts] Find all critical points of f (x, y) = 2 + 3x y + Solution: f x (x, y) = 3y + x = 0 ⇒ x = −3y 2 f y (x, y) = 3x + 9y = 0 ⇒ −9y + 9y 2 = 0 ⇒ y(y − 1) = 0 y = 0 ⇒ x = 0, y = 1 ⇒ x = −3 Two critical points: (0, 0) and (−3, 1) f x x (x, y) = 1, f yy (x, y) = 18y, f x y (x, y) = 3 D(x, y) = f x x f yy − [ f x y ]2 = 18y − 9 D(0, 0) = −9 < 0 ⇒ (0, 0) is a saddle point D(−3, 1) = 18 − 9 > 0, f x x (−3, 1) > 0 ⇒ (−3, 1) is a local minimum 2. [8 pts] For the function f (x, y) = p 4 − 4x 2 − y 2 , (a) determine the domain and range of f (x, y). (b) calculate the partial derivatives Solution: Part (a): Domain and Range Domain: 4 − 4x 2 − y 2 ≥ 0 ∂f ∂2 f and . ∂x ∂ y∂ x or f min = 0 (when 4x 2 + y 2 = 4), Range: 0 ≤ f (x, y) ≤ 2. Part (b): x 2 + (y/2)2 ≤ 1. f max = 2 (when x = y = 0) Partial derivatives ∂f ∂x = −1/2 2 2 1/2 4 − 4x − y −8x ∂2 f ∂ y∂ x = −3/2 −4x −1/2 4 − 4x 2 − y 2 −2y = = p −4x 4 − 4x 2 − y 2 −4x y 4 − 4x 2 − y 2 3/2 3. [8 pts] For the the function f (x, y) = x 2 + 2y 2 + x y, (a) find the directional derivative of f at the point (1, 1) in the direction h −1, −1 i. (b) in which direction does the function f decrease most rapidly at the point (1, 1). Solution: Part (a): Directional derivative ∇ f (x, y) = h2x + y, 4y + xi −→ ∇ f (1, 1) = h3, 5i h−1, −1i (unit vector in direction of h−1, −1i) uE = √ 2 √ h3, 5i ◦ h−1, −1i −3 − 5 Du f (1, 1) = ∇ f (1, 1) ◦ uE = = √ = −4 2 √ 2 2 Part (b): The gradient vector points in the direction of steepest ascent. Therefore the function is decreasing most rapidly in the direction opposite the gradient. Direction of steepest descent = −∇ f (1, 1) = h−3, −5i. 4. [8 pts] For the function f (x, y) = √ x y, determine the equation of the tangent plane √ at P = (2, 8). Use this plane to estimate (2.02) · (7.96) . Solution: r 1 y f x (x, y) = (1/2)(x y) y= −→ f x (2, 8) = 1 2 x r 1 x f y (x, y) = (1/2)(x y)−1/2 x = −→ f x (2, 8) = 1/4 2 y −1/2 Tangent Plane: z = f (2, 8) + f x (2, 8)(x − 2) + f y (2, 8)(y − 8) z = 4 + (x − 2) − (y − 8)/4 f (2.02, 7.96) ≈ 4 + (2.02 − 2) − (7.96 − 8)/4 = 4 + 0.02 − 0.01 = 4.01 √ 5. [8 pts] Consider the surface obtained by rotating the curve y = xe−x about the x-axis on the interval 0 ≤ x ≤ 3. Determine a parametric representation for this surface. [ Be sure to include the parameter domain. ] Solution: Cross-sections at x = xo are circles in the yz-plane with radius √ r (xo ) = xo e−xo . Let u parametrize the axis of rotation (x = u) and v be the angle of rotation about the x-axis, the polar angle in the yz-plane. √ √ x = u, y = u e−u cos v, z = u e−u sin v 0 ≤ u ≤ 3, 0 ≤ v ≤ 2π 6. [9 pts] For the surface defined by the equation F(x, y, z) = x 4 + z 2 + x yz = 3, (a) determine the equation of the tangent plane to this surface at the point P = (1, −1, 2). (b) assuming z to be a function of x and y, determine the value of the partial derivative ∂z at the point P = (1, −1, 2). ∂x Solution: Part (a): Tangent Plane ∇ F(x, y, z) = 4x 3 + yz, x z, 2z + x y ∇ F(1, −1, 2) = h4 − 2, 2, 4 − 1i = h2, 2, 3i Tangent Plane: ∇ F(1, −1, 2) ◦ hx − 1, y + 1, z − 2i = 0 2(x − 1) + 2(y + 1) + 3(z − 2) = 0 −→ 2x + 2y + 3z = 6 Part (b): Assume z = z(x,y) − ∂∂ Fx P ∂z 2 = − = ∂F ∂x P 3 ∂z P