Math 241, Midterm 1 A Solutions May 4, 2009 Part I. 1. Circle T for true and F for false. 2xy = −1 . + y2 2xy b) lim = 0. (x,y)→(0,0) x2 + y 2 a) T lim (x,y)→(1,−1) x2 F T F c) If f (x, y) = ln(y), then ∇f (x, y) = 1/y . p d) The domain of the function f (x, y) = z − x2 − y 2 consists of all points (x, y, z) that lie above the cone z = x2 + y 2 . p dx + y dy e) The differential of u = 2x + y 2 is du = p . 2x + y 2 T F f) The set of points { (x, y, z) | x2 + y 2 = 1 } is a circle. T F g) The graph of the surface y 2 + z 2 = 1 + x2 is an ellipsoid. T F h) The maximum rate√of change of the function f (x, y, z) = tan(x + 2y + 3z) at the point (−5, 1, 1) is 3 3 . T F T F T F i) The critical point (0,0) of the function f (x, y) = x sin(y) is a saddle point. T F j) The linear approximation of the function f (x, y) = sin(x − 3y) at (6,2) is T F T (x, y) = x − 3y . 2. Fill in the blanks. Simplify all expressions. a) An equation for the surface obtained by rotating the curve y = x2 + 1 about the y-axis is y − 1 = x2 + z 2 This an elliptical paraboloid centered at (0,1,0) b) Write the equation x2 + y 2 + z 2 = 2z in spherical coordinates: c) If f (x, y, z) = cos(4x+3y +2z), then fxyz (x, y, z) = d) The linear approximation of the function f (x, y, z) = ρ2 = 2ρ cos(φ) or ρ = 2 cos(φ) 24 sin(4x + 3y + 2z) p − x2 + y 2 + z 2 at the point (3, 2, 6) is xz + sin(x + y + z) xy + sin(x + y + z) . . f) If g(s, t) = s2 et , then the directional derivative of g at the point (2,0) in the direction of v = i + j is √ ∇g(2, 0) · √12 h1, 1i = 4 2 . 2 2 2 g) The critical points √ of f (x, y) = x + y + x y + 4 are (0,0) and (± 2, −1). Classify each one as a local max, local min, or saddle point. The point √(0,0) ( √2, −1) (− 2, −1) . . T (x, y, z) = 7 + 73 (x − 3) + 72 (y − 2) + 76 (z − 6) = 3x/7 + 2y/7 + 6z/7 e) If xyz = cos(x + y + z), then ∂z/∂y = . is classified as a local min saddle saddle Part II. 1. Find the volume of the rectangular box with largest volume if it lies in the first octant (x ≥ 0, y ≥ 0, z ≥ 0) with three of its faces in the coordinate planes, one vertex at (0,0,0), and the opposite vertex in the plane x + 2y + 3z = 6. Let V be the volume of the box and (x, y, z) be the point at its vertex that is opposite to (0, 0, 0. We wish to maximize V = xyz when x > 0, y > 0, z > 0, and x + 2y + 3z = 6. Since x = 6 − 2y − 3z the problem can be reduced to two dimensions by substituting for x. So, we will maximize V = (6 − 2y − 3z)yz on the open set y > 0 and z > 0. The maximum of V will be attained at one of its critical points. To find the critical points simplify to V = 6yz − 2y 2 z − 3yz 2 so Vy = 6z − 4yz − 3z 2 = z(6 − 4y − 3z) Vz = 6y − 2y 2 − 6yz = 2y(3 − y − 3z) . Since y and z are not 0, the critical point equations Vy = 0 and Vz = 0 are equivalent to the following system of linear equations. 4y + 3z = 6 y + 3z = 3 Subtracting the second equation from the first we see that 3y = 3 so y = 1. From this it follows that z = 2/3, and x = 6 − 2y − 3z = 2. The maximum volume is 2 · 1 · (2/3) = 4/3. 2. Find the absolute maximum and minimum values of the function f (x, y) = xy 2 on the set D = { (x, y) | x ≥ 0 , y ≥ 0 , x2 + y 2 ≤ 3 } . Exact values, please. The region D is shown on the right. Note that f is non-negative on D and f (x, y) = 0 at all points on the x and y axes. Therefore, the absolute minimum of f on D is 0. Since fx = y 2 and fy = 2xy, all of the points on the x axis are critical points, and there are no more. Consequently, the absolute maximum of f must occur on the circular part of its boundary where y 2 = 3 − x2 and z = f (x, y) = x(3 − x2 ) . 2 Thus z = 3x −√ x3 and dz/dx = 3 − 3x2 = 3(1 − √x ) implying that the maximum value of z is attained when x = 1 and y = 2. The maximum value is f (1, 2) = 2. 3. Do both parts. a) If ρ = p x2 + y 2 + z 2 , then show that 2 ∂2ρ ∂2ρ ∂2ρ + 2+ 2 = . ∂x2 ∂y ∂z ρ (Student solution) Start with ρ2 = x2 + y 2 + z 2 and differentiate with respect to x to obtain, 2ρρx = 2x. Therefore, ρx = x . ρ Differentiate again with respect to x to obtain ρxx = ρ − xρx ρ − x · (x/ρ) 1 x2 = = − 3. 2 2 ρ ρ ρ ρ A similar calculation will show that ρyy = 1 y2 − ρ ρ3 and ρzz = 1 z2 − . ρ ρ3 Therefore, 3 x2 + y 2 + z 2 − ρ ρ3 2 3 ρ 3 1 2 = − 3 = − = . ρ ρ ρ ρ ρ ρxx + ρyy + ρzz = b) If z = cos(xy) + y cos(x) , where x = u2 + v and y = u − v 2 , use the Chain Rule to find ∂z/∂u and ∂z/∂v. According to the Chain Rule, ∂z ∂z ∂x ∂z ∂y = + . Therefore, ∂u ∂x ∂u ∂y ∂u ∂z = (−y sin(xy) − y sin(x)) · 2u + (−x sin(xy) + cos(x)) · 1 . ∂u Similarly, ∂z ∂z ∂x ∂z ∂y = + , so ∂v ∂x ∂v ∂y ∂v ∂z = (−y sin(xy) − y sin(x)) · 1 + (−x sin(xy) + cos(x)) · (−2v) . ∂v 4. The surface on the right is the graph of the function f (x, y) = sin(πx) sin(πy) . The point on the surface is P = (0.2, 1.2, f (0.2, 1.2)) . a) In the space below the surface make a rough sketch of a contour map of this function for 0 ≤ x ≤ 2 , 0 ≤ y ≤ 2 . b) Based upon the surface, at the point P , (circle i, ii, iii, or iv) i. fx (P ) > 0 and fy (P ) > 0. ii. fx (P ) > 0 and fy (P ) < 0. iii. fx (P ) < 0 and fy (P ) > 0. iv. fx (P ) < 0 and fy (P ) < 0. c) If a bug is at the point P , in what direction (from the point (0.2, 1.2) in the xy-plane) should it move in order to go up the surface at the fastest rate? Show your work in the space below. Draw the contour plot on these axes. Let z = f (x, y). Then z increases the fastest when the bug moves in the direction of the gradient of f at the point (0.2, 1.2). Since ∇f (x, y) = hπ cos(πx) sin(πy), π sin(πx) cos(πy)i the bug should move in the direction of the vector ∇f (0.2, 1.2) = hπ cos(0.2π) sin(1.2π), π sin(0.2π) cos(1.2π)i ≈ h−1.494, −1.494i . d) What will be the rate that the bug moves upward if it walks from the point P in the direction you found in part c)? Show your work below. When the bug walks in the direction of the gradient the rate of change of z is p |∇f (0.2, 1.2)| ≈ 1.4942 + 1.4942 ≈ 2.113 .
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