2008Final-Solutions

2008Final-Solutions - blue MATH 32A Final Exam LAST NAME...

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Unformatted text preview: blue MATH 32A Final Exam March 17, 2008 LAST NAME FIRST NAME ID NO. Your TA: To receive credit, you must write your answer in the space provided. DO NOT WRITE BELOW THIS LINE 1 (20 pts) 5 (20 pts) 2 (20 pts) 6 (20 pts) 3 (20 pts) 7 (20 pts) 4 (20 pts) TOTAL 2 PROBLEM 1 (20 Points) Let function f (x, y ) = x4 − x2 − y 2 . (A) Find all critical points of f (x, y ) and classify each one as a maximum, minimum, or saddle point. (B) Find the absolute maximum and minimum of f (x, y ) on the domain x2 + y 2 ≤ 4. Answer : 3 Solution: (A) First solve for the critical points: 1 fx = 4x3 − 2x = 2x(2x2 − 1) ⇒ x = 0, ± √ 2 fy = −2y = 0 ⇒ y = 0 There are three critical points: 1 (0, 0) ( √ , 0), 2 The discriminant is 1 (− √ , 0) 2 2 D(x, y ) = fxx fyy − fxy = (12x2 − 2)(−2) − 0 = 4 − 24x2 We see that D(0, 0) = 4 > 0 so (0, 0) is a local min or max. In fact, it is a local max since fx x(0, 0) = −2 < 0. We have 1 1 D(± √ , 0) = 4 − 24( ) = −8 < 0 2 2 Therefore the other two critical points are saddle points. (B) We evaluate f (x, y ) at the critical points: 1 11 1 f (0, 0) = 0, f (± √ , 0) = − − 0 = − 42 4 2 Next, we test f (x, y ) on the circle x2 + y 2 = 4. We find that on the circle f (x, y ) = x4 − x2 − y 2 = x4 − 4 (for −2 ≤ x ≤ 2) Since x4 is positive, we may conclude immediately that the maximum of f on the circle is f (2, 0) = f (−2, 0) = 24 − 4 = 12 and the minimum value is f (0, 0) = −4. Conclusion: the max of f (x, y ) on the disk x2 + y 2 ≤ 4 is 12 and the min is −4 (since the values at the critical points lie between −4 and 12). 4 PROBLEM 2 (20 Points) Suppose that f (x, y ) is a function of x and y , and that x = u2 − v 2 , y = uv . Use the table of values ∂ f and ∂ f to ∂x ∂y compute ￿ ￿ ∂f ￿ ∂f ￿ ￿ ￿ and ￿ ∂ u (u,v)=(1,1) ∂ v ￿(u,v)=(1,0) x 0 1 0 1 Answer: Answer: y 0 0 1 1 ∂f ￿ ￿ ￿ ∂ u (u,v)=(1,1) ∂f ￿ ￿ ￿ ∂ v (u,v)=(1,0) ∂f ∂ x (x, y ) ∂f ∂ y (x, y ) 2 5 -1 3 -3 1 -4 2 5 Solution: (A) When (u, v ) = (1, 1), we have x = u2 − v 2 = 0, y = uv = 1 The table gives the values ∂f ￿ ∂f ￿ ￿ ￿ = −1, = −4 ￿ ￿ ∂ x (x,y)=(0,1) ∂ y (x,y)=(0,1) By the Chain Rule: ∂f ∂x ∂f ∂y ∂f ∂f ∂f = + = (2u) + (v ) ∂u ∂x ∂u ∂y ∂u ∂x ∂y Evaluate at (u, v ) = (1, 1) and (x, y ) = (0, 1): ∂f ￿ ￿ = (−1)(2) + (−4)(1) = −6 ￿ ∂ u (u,v)=(1,1) (B) When (u, v ) = (1, 0), we have x = u2 − v 2 = 1, y = uv = 0 The table gives the values ∂f ￿ ∂f ￿ ￿ ￿ = 5, =1 ￿ ￿ ∂ x (x,y)=(1,0) ∂ y (x,y)=(1,0) By the Chain Rule: ∂f ∂f ∂x ∂f ∂y ∂f ∂f = + = (−2v ) + (u) ∂v ∂x ∂v ∂y ∂v ∂x ∂y Evaluate at (u, v ) = (1, 0) and (x, y ) = (0, 1): ∂f ￿ ￿ = (5)(0) + (1)(1) = 1 ￿ ∂ v (u,v)=(1,0) 6 PROBLEM 3 (20 Points) Let r(t) = ￿cos3 t, 0, sin3 t￿ (A) Show that ||r￿ (t)||2 = 9 cos2 t sin2 t (B) Find a formula for the curvature κ(t) at time t. Answer : 7 Solution: We have r￿ (t) = ￿−3 cos2 t sin t, 0, 3 sin2 t cos t￿ ￿ ￿ ||r (t)|| = 9 cos4 sin2 t + 9 sin4 t cos2 t ￿ = 3 cos2 t sin2 t(cos2 t + sin2 t) = 3| cos t sin t| r￿￿ (t) = ￿6 cos t sin t − 3 cos3 t, 0, 6 cos2 t sin t − 3 sin3 t￿ r￿ (t) × r￿￿ (t) = ￿0, 9 cos2 t sin2 t, 0￿ The curvature is ||r￿ (t) × r￿￿ (t)|| 9 cos2 t sin2 t 1 κ(t) = = = ￿ (t)||3 3 ||r |3 cos t sin t| |3 cos t sin t| 8 PROBLEM 4 (20 Points) Let P be the plane through the points P1 = (1, 2, 0), P2 = (2, 0, 1), P3 = (0, 0, 2) (A) Find an equation for the plane P . Answer : (B) Find the coordinates of the point Q where the plane P intersects the line through the two points A = (1, 0, 0), Answer : B = (0, 2, 1) 9 Solution: (A) We have P1 P2 = ￿2, 0, 1￿ − ￿1, 2, 0￿ = ￿1, −2, 1￿ P1 P3 = ￿0, 0, 2￿ − ￿1, 2, 0￿ = ￿−1, −2, 2￿ We find a normal to the plane by computing the cross product P1 P2 × P1 P3 = ￿1, −2, 1￿ × ￿−1, −2, 2￿ = ￿−2, −3, −4￿ Since P1 = (1, 2, 0) lies on the plane, the equation of the plane is −2(x − 1) − 3(y − 2) − 4z = 0 or 2x + 3y + 4z = 8 (B) The line through (1, 0, 0) and (0, 2, 1) has the parametrization r(t) = ￿1 − t, 2t, t￿ We solve for t in the equation 2(1−t)+3(2t)+4t = 8 ⇒ 2−2t+6t+4t = 8 ⇒ 8t = 6 ⇒ t = 3/4 133 3 Since r( 4 ) = ￿ 1 , 3 , 3 ￿, we find that the point Q = ( , , ) 424 424 lines on the intersection of the plane and the line. 10 PROBLEM 5 (20 Points) The gradient of pressure at the origin in r3 is ∇P = ￿4, 3, −1￿ (millibars per meter) A fly carrying a pressure gauge travels in the positive direction along the y -axis at 20 meters per second. How fast is the pressure reading on the gauge changing when the fly passes through the origin? Give your answer in the correct units. Answer : Solution: The velocity vector of the fly is v = ￿0, 20, 0￿. By the Chain Rule for Paths, the reading on the pressure gauge is changing at a rate of ∇P · v = ￿4, 3, −1￿ · ￿0, 20, 0￿ = 60 millibars per second 11 PROBLEM 6 (20 Points) Use the contour map to answer the following questions. Show and explain the computation you made to obtain the estimate. (A) Estimate De f (P ) where e is the unit vector in the direction of i + j and P is the point indicated on the map. Answer: Answer: (B) Mark the point Q on the contour map where f (Q) = 400, Answer: De f (P ) = ∂f ￿ ￿ ￿ > 0, ∂x Q Figure 1 ∂f ￿ ￿ ￿ =0 ∂y Q 12 Solution: (A) The distance from P (which lies on the level curve 400) to the level curve 200 along a segment in the i + j direction (45o angle) is approximately 50 meters. Therefore 200 − 400 De f (P ) ≈ = −4 50 (B) The conditions ∂f ￿ ∂f ￿ ￿ ￿ f (Q) = 400, ￿ > 0, ￿ =0 ∂x Q ∂y Q are satisfied at the point Q which is the left-most point of ∂f ￿ ￿ the level curve f (x, y ) = 400. Indeed, ￿ > 0 because ∂x Q f is increasing as you move horizontally to the right. And since the tangent line to the level curve is vertical, we see ∂f ￿ ￿ that ￿ = 0. ∂y Q 13 PROBLEM 7 (20 Points) Find the coordinates of the point (a, b) lying on the surface 2x2 − 2xy + y 2 + 4x − 2y = 7 with the largest x-coordinate. Hint: use Lagrange multipliers. Answer : 14 Solution: We must find the maximum value of the function f (x, y ) = x subject to the constraint g (x, y ) = 2x2 − 2xy + y 2 + 4x − 2y − 7 = 0 The Lagrange equations are ￿1, 0￿ = λ￿4x − 2y + 4, −2x + 2y − 2￿ The equation 1 = λ(4x − 2y + 4) shows that λ ￿= 0, and the second equation yields 0 = −2x + 2y − 2 ⇒ y = x + 1 Substitute back in the constraint equation: 2x2 − 2xy + y 2 + 4x − 2y − 7 == 0 2x2 − 2x(x + 1) + (x + 1)2 + 4x − 2(x + 1) − 7 = 0 2x2 − 2x2 − 2x + x2 + 2x + 1 + 4x − 2x − 2 − 7 = 0 x2 + 2x − 8 = 0 (x + 4)(x − 2) = 0 Therefore, x = 2 (and y = x + 1 = 3) or x = −4 (and y = x + 1 = −4). The maximum value of x occurs at the point (2, 3). blue MULTIPLE CHOICE QUESTIONS ENTER ALL ANSWERS ON YOUR SCANTRON FORM 882-ES (in pencil) If you change any answers on the scantron form, please make sure to erase completely and leave no smudges. 1 2 PROBLEM 1 (10 Points) Find the equation of the tangent plane to the surface x2 y + y 3 + z 4 = 3 at the point P = (1, 1, 1). (A) 2x + 3y + 4z = 9 (B) 3x + 2y + 4z = 0 (C) 2x + 4y + 4z = 10 (D) 2x + 5y + 4z = 11 (E) 3x + 3y + 4z = 10 Solution: C . The gradient provides a normal vector: ∇F(1,1,1) = ￿2xy, x2 + 3y 2 , 4z 3 ￿ = ￿2, 4, 4￿ The equation of the plane is or 2(x − 1) + 4(x − 1) + 4(x − 1) = 0 2x + 4y + 4z = 10 3 PROBLEM 2 (10 Points) Let f (x, y ) = x3 y 4 − x − y Find a vector v pointing in a direction along which the directional derivative of f (x, y ) at P = (2, −1) is equal zero. (A) ￿33, 11￿ (B) ￿12, 4￿ (C) ￿12, −32￿ (D) ￿12, 32￿ (E) ￿11, −33￿ Solution: Both A and B are correct (and both answers will receive full credit). We have and ∇f = ￿(3x2 y 4 − 1, 4x3 y 3 − 1￿ ∇f (2, −1) = ￿(11, −33￿ The vector v = ￿33, 11￿ is perpendicular to ∇f (2, −1) and so the directional derivative in the direction of v = ￿33, 11￿ is zero. But 4 ￿12, 4￿ = ￿33, 11￿ 11 so ￿12, 4￿ is also correct. 4 PROBLEM 3 (10 Points) Find the minimum value taken by f (x, y ) = xy − x2 y − 4xy 2 on the hyperbola xy = 1. (A) −4 (B) −3 (C) − 3 2 (D) −2 (E) − 1 2 Solution: The correct answer is B . The Lagrange condition is This yields ∇f = ￿y − 2xy − 4y 2 , x − x2 − 8xy ￿ = λ￿y, x￿ y − 2xy − 4y 2 x − x2 − 8xy = y x Notice that x and y are both non-zero if (x, y ) satisfies xy = 1, so it is legitimate to divide by x and y . Thus we obtain λ= 1 − 2x − 4y = 1 − x − 8y ⇒ 4y = x The constraint yields xy = 4y 2 = 1, so y = ± 1 and the critical points 2 are (2, 1 ) and (−2, − 1 ). We have 2 2 1 1 f (2, ) = xy − x2 y − 4xy 2 = 1 − 2 − 4( ) = −3 2 2 1 1 f (−2, − ) = 1 + 2 − 4(− ) = 5 2 2 We conclude that the minimum value is −3 5 PROBLEM 4 (10 Points) Find a parametrization of the line through P = (1, 0, 5) that is perpendicular to the plane 12x − 6y − 24z = 7 (A) L(t) = ￿12t, −t, −24t￿ (B) L(t) = ￿5 − 24t, −t, 1 + 12t￿ (C) L(t) = ￿1 + 4t, −t, 5 − 12t￿ (D) L(t) = ￿1 + 2t, −t, 5 − 4t￿ (E) L(t) = ￿7 + 12t, −t, 24t￿ Solution: The correct answer is D . The vector n = ￿12, −6, −24￿ is normal to the plane and therefore it is a direction vector for the line. We may also divide n by 6 and use v = ￿2, −1, −4￿ as a direction vector. Therefore, the following is a parametrization of the desired line: L(t) = ￿1, 0, 5￿ + t￿2, −1, −4￿ = ￿1 + 2t, −t, 5 − 4t￿ 6 PROBLEM 5 (10 Points) The figure below shows a cube. The point Q is located at the center of the top face of the cube. Calculate the cosine of the angle between → → the vectors v = P R and w = P Q Figure 1 (A) (B) √ 2 2 1 4 1 (C) √ 3 √ 3 (D) 2 (E) 1 2 Solution: The correct answer is E . If we place the origin at the vertex directly below P , then we have 11 P = (0, 0, 1), Q = ( , , 1), R = (1, 0, 0) 22 and → → 11 v = P R = ￿1, 0, −1￿, w = P Q = ￿ , , 0￿ 22 The cosine of the angle is cos θ = 1 v·w 1 2 =√￿ = ||v|| ||w|| 2 2 1/2 7 PROBLEM 6 (10 Points) One moon of the Vulcan Planet has a circular orbit of radius 1000 km and completes one revolution of its orbit is 50 hours. A second moon completes its circular orbit in 400 hours. According to Kepler’s Third Law, the radius of the second moon’s orbit is (A) 250 km (B) 125 km (C) 500 km (D) 2000 km (E) 4000 km Solution: The correct answer is E . We have or 10003 r3 = 4002 502 r3 = 10003 Therefore 4002 = 10003 (82 ) 502 r = 1000(22 ) = 4000 km 8 PROBLEM 7 (10 Points) Calculate the directional derivative of f (x, y, z ) = xy 2 z 3 at P = (2, 1, 1) in the direction of the vector v = ￿1, 2, 2￿. 3 (A) 2 (B) 8 3 (C) 4 (D) 7 (E) 8 Solution: The correct answer is D . We have ∇f (2, 1, 1) = ￿y 2 z 3 , 2xyz 3 , 3xy 2 z 2 ￿ = ￿1, 4, 6￿ The unit vector in the direction of v is e = ￿ 1 , 2 , 2 ￿ and the direction 333 derivative is 122 122 1 8 12 ∇f (2, 1, 1) · ￿ , , ￿ = ￿1, 4, 6￿ · ￿ , , ￿ = + + =7 333 333 33 3 9 PROBLEM 8 (10 Points) A particle moves to the right along the parabola y = x2 (units in centimeters). At time t = 1 sec, the particle passes the origin, at which time the velocity and acceleration vectors of the particle are v = ￿4, 0￿, a = ￿3, 6￿ How fast is the particle’s speed changing at t = 1? (A) 3 cm/s2 (B) 12 cm/s2 (C) 2 cm/s2 (D) 6 cm/s2 (E) 4 cm/s2 Solution: The correct answer is A . The speed is changing at a rate of ￿v￿ ￿ ￿4, 0￿ ￿ v￿ = a · = ￿3, 6￿ · = 3 cm/s2 ||v|| 4 ...
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This note was uploaded on 10/26/2011 for the course MATH 32A taught by Professor Gangliu during the Spring '08 term at UCLA.

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