MATH. 2153, Spring 16, MWF 12:40 p.m. QUIZ 1 January 25, 2016 PRINT NAME A. Derdzinski Show all work. No calculators. The problem is worth 10 points.
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1 MATH. 2153, Spring 16, MWF 12:40 p.m. QUIZ 1 January 25, 2016 PRINT NAME A. Derdzinski Show all work. No calculators. The problem is worth 10 points. 1. Evaluate the area A of the triangle with the vertices P (1, 0, 1), Q(2, 1, 0), R(0, 1, 1). YOUR ANSWER: A =
2 QUIZ 1, 01/25/16 1. Evaluate the area A of the triangle with the vertices P (1, 0, 1), Q(2, 1, 0), R(0, 1, 1). Answer: We have 2A = u v, where u = P Q = i + j + k and and v = P R = i + j, so that u v = i j + 2k and 6 A = 2.
3 MATH. 2153, QUIZ 1 (make-up) Spring 16, MWF 12:40 p.m. February 2, 2016 PRINT NAME A. Derdzinski Show all work. No calculators. The problem is worth 10 points. 1. Evaluate the area A of the triangle with the vertices P (1, 1, 1), Q(1, 4, 2), R(5, 1, 0). YOUR ANSWER: A =
4 MATH. 2153, Spring 16, MWF 12:40 p.m. page 1 of 2 TEST 1 FEBRUARY 5, 2016 PRINT NAME A. Derdzinski RECITATION TIME (check one) 10:20 11:30 12:40 Show all work. No calculators. Problems worth 25 points each. 1. Determine the angle θ at the vertex A(1, 1, 2) of the triangle ABC with B(1, 0, 1) and C(1, 1, 4). Express θ both in radians and in degrees. YOUR ANSWER: (radians), (degrees). 2. Find a nonzero vector w perpendicular to a = j k and v = i + j + 2 k. YOUR ANSWER: w =
5 MATH. 2153, Spring 16, MWF 12:40 p.m., Test 1 page 2 of 2 NAME 3. Find the area A of the parallelogram in the xy-plane with vertices A(0, 0), B(1, 3), C(2, 1) and D(3, 4). YOUR ANSWER: A = 4. Let u = 3 i j k, v = 6 i j 3 k. Find a scalar c and a vector w orthogonal to u such that v = c u + w. YOUR ANSWER: w =, c =
6 MATH. 2153, Spring 16, MWF 12:40 p.m., Test 1 ANSWERS 1. Determine the angle θ at the vertex A(1, 1, 2) of the triangle ABC with B(1, 0, 1) and C(1, 1, 4). Express θ both in radians and in degrees. Answer: Set v = AB = j k, w = AC = 2 k. Hence v = 2, w = 2 and v w = 2, and cos θ = 2/2 2 = 1/ 2, i.e., θ = 3π 4 (radians), θ = 135 (degrees.) 2. Find a nonzero vector w perpendicular to u = j k and v = i + j + 2 k. Answer: We may use w = u v = 3 i j k.
7 3. Find the area A of the parallelogram in the xy-plane with vertices A(0, 0), B(1, 3), C(2, 1) and D(3, 4). Answer: Set v = AB, w = AC. Then AD = v + w, and so the vectors v and w treated as directed segments emanating from A represent two adjacent sides of the parallelogram in question. Hence, as v = i + 3 j, w = 2 i + j, we have v w = 5 k and A = v w = Let u = 3 i j k, v = 6 i j 3 k. Find a scalar c and a vector w orthogonal to u such that v = c u + w. Answer: Decomposing v into a component parallel to u and a component perpendicular to u, we obtain v = v u [ u u u + v v u ] u u u, so that c = v u u u = = 2, w = v c u = j k.
8 MATH. 2153, Spring 16, MWF 12:40 p.m. page 1 of 2 TEST 1 (make-up) FEBRUARY 16, 2016 PRINT NAME A. Derdzinski RECITATION TIME (check one) 10:20 11:30 12:40 Show all work. No calculators. Problems worth 25 points each. 1. Determine the angle θ at the vertex A(2, 0, 0) of the triangle ABC with B(0, 2, 0) and C(0, 1, 1). Express θ both in radians and in degrees. YOUR ANSWER: θ = (radians), θ = (degrees). 2. Find a nonzero vector w perpendicular to a = i + k and v = i j 2 k. YOUR ANSWER: w =
9 MATH. 2153, Spring 16, MWF 12:40 p.m., Test 1 page 2 of 2 NAME 3. Find the area A of the parallelogram in the xy-plane with vertices A(0, 0), B(1, 4), C(2, 0) and D(3, 4). YOUR ANSWER: A = 4. Let u = i j k, v = 2 i j 3 k. Find a scalar c and a vector w orthogonal to u such that v = c u + w. YOUR ANSWER: w =, c =
10 MATH. 2153, Spring 16, MWF 12:40 p.m. QUIZ 2 February 26, 2016 PRINT NAME A. Derdzinski Show all work. No calculators. The problem is worth 10 points. 1. Evaluate the partial derivatives f x and f y for the function f given by f(x, y) = x 2 xy dt 1 + e t. YOUR ANSWER: f x (x, y) =, f y (x, y) =
11 QUIZ 2, 02/26/16 1. Evaluate the partial derivatives f x and f y for the function f given by f(x, y) = x 2 xy dt 1 + e t. Answer: Writing z = f(x, y) we have z = g(u, v), where u = xy, v = x 2 and g(u, v) = v u dt 1 + e t. The chain rule now gives z x = g u u x + g v v x and z y = g u u y + g v v y. Since g u (u, v) = e u, g v(u, v) = e v, while u x = y, u y = x, v x = 2x and v y = 0, we see that f x (x, y) = 2x 1 + e x2 y 1 + e xy, f y(x, y) = x 1 + e xy.
12 MATH. 2153, Spring 16, MWF 12:40 p.m. page 1 of 2 TEST 2 MARCH 25, 2016 PRINT NAME A. Derdzinski RECITATION TIME (check one) 10:20 11:30 12:40 Show all work. No calculators. Problems worth 25 points each. 1. Find an equation for the tangent plane at the point P = (1, 2, 3) of the surface xyz = 6. YOUR ANSWER: Equation 2. Find all the critical points of the function f(x, y) = (x 2 1) 2 + y 2 2y and decide which of them are local maxima, local minima, and saddle points. YOUR ANSWER: loc.min. at loc.max. at saddle pts. at
13 MATH. 2153, Spring 16, MWF 12:40 p.m., Test 2 page 2 of 2 NAME 3. Determine the minimum value m assumed by the function f(x, y, z) = x y on the ellipsoid 4x 2 + y 2 + z 2 + 2y + 4z + 1 = 0. YOUR ANSWER: m = 4. Use the right order of iterated integration to evaluate c = e x2 da, where R is the triangle in the xy-plane given by 0 y 2x 2. R YOUR ANSWER: c =
14 MATH. 2153, Spring 16, MWF 12:40 p.m., Test 2 ANSWERS 1. Find an equation for the tangent plane at the point P = (1, 2, 3) of the surface xyz = 6. Answer: The surface is given as the level set f = 6 for f(x, y, z) = xyz. Since f(p ) = 6i + 3j + 2k 0, the equation for the tangent plane is 6(x 1) + 3(y 2) + 2(z 3) = 0, or, equivalently, 6x + 3y + 2z = Find all the critical points of the function f(x, y) = (x 2 1) 2 + y 2 2y and decide which of them are local maxima, local minima, and saddle points. Answer: We have f x = 4x(x 2 1), f y = 2(y 1), D(x, y) = f xx f yy (f xy ) 2 = 8(3x 2 1). Thus, f has three critical points: P (0, 1), Q(1, 1), R( 1, 1), and at P we have D < 0, while at both Q and R we have D > 0 with f yy = 2 > 0. Thus, P is a saddle point, and Q, R both are local minima.
15 3. Determine the minimum value m assumed by the function f(x, y, z) = x y on the ellipsoid 4x 2 + y 2 + z 2 + 2y + 4z + 1 = 0. Answer: The ellipsoid is given by g(x, y, z) = 0 with g(x, y, z) = 4x 2 +y 2 +z 2 +2y+4z+1. The partial derivatives f x = 1, f y = 1, f z = 0, g x = 8x, g y = 2y + 2, g z = 2z + 4 exist everywhere and f has no critical points, while the only critical point of g is (0, 1, 2), which is not on the ellipsoid surface (in fact, it is the center of the ellipsoid). The minimum value m thus can be found using the method of Lagrange multipliers. The system 4x 2 + y 2 + z 2 + 2y + 4z + 1 = 0, 1 = 8λx, 1 = 2λ(y + 1), 0 = 2λ(z + 2) has two solutions (x, y, z, λ), given by ( ± 1, 4 ) 5 1, 2, ± They yield two values for f, namely, 1 ± 5, the smaller of which is m = Use the right order of iterated integration to evaluate c = e x2 da, where R is the triangle in the xy-plane given by 0 y 2x 2. R Answer: With u = x 2, c = 1 2x e x2 dy dx = 1 2xe x2 dx = 1 e u du = e
16 MATH. 2153, Spring 16, MWF 12:40 p.m. page 1 of 2 TEST 2 (make-up) APRIL 1, 2016 PRINT NAME A. Derdzinski RECITATION TIME (check one) 10:20 11:30 12:40 Show all work. No calculators. Problems worth 25 points each. 1. Find an equation for the tangent plane at the point P = (3, 1, 2) of the surface xy z = 5. YOUR ANSWER: Equation 2. Find all the critical points of the function f(x, y) = (y 2 1) 2 + x 2 + 2x and decide which of them are local maxima, local minima, and saddle points. YOUR ANSWER: loc.min. at loc.max. at saddle pts. at
17 MATH. 2153, Spring 16, MWF 12:40 p.m., Test 2 page 2 of 2 NAME 3. Determine the minimum value m assumed by the function f(x, y, z) = x z on the ellipsoid 4x 2 + y 2 + z 2 + 4y + 2z + 1 = 0. YOUR ANSWER: m = 4. Use the right order of iterated integration to evaluate c = e x2 da, where R is the triangle in the xy-plane given by 0 y x 1. R YOUR ANSWER: c =
18 MATH. 2153, Spring 16, MWF 12:40 p.m. QUIZ 3 April 8, 2016 PRINT NAME A. Derdzinski RECITATION TIME (check one) 10:20 11:30 12:40 Show all work. No calculators. The problem is worth 10 points. 1. Evaluate the triple integral c = z dv, where D is the region consisiting of all (x, y, z) with D x 2 + y 2 + z 2 1 and 0 z x 2 + y 2. Reminder: for integration in cylindrical/spherical/arbitrary coordinates, dx dy dz = r dr dθ dz = ρ 2 sin ϕ dρ dϕ dθ = (x, y, z) (u, v, w) du dv dw. YOUR ANSWER: c =
19 QUIZ 3, 04/08/16 1. Evaluate the triple integral c = z dv, where D is the region consisiting of all (x, y, z) with D x 2 + y 2 + z 2 1 and 0 z x 2 + y 2. Reminder: for integration in cylindrical/spherical/arbitrary coordinates, dx dy dz = r dr dθ dz = ρ 2 sin ϕ dρ dϕ dθ = (x, y, z) (u, v, w) du dv dw. Answer: In spherical coordinates, z = ρ cos ϕ, while D is given by ρ 1 and π/4 ϕ π/2, so c = 2π π/2 1 0 π/4 0 ρ 3 cos ϕ sin ϕ dρdϕdθ = π 8.
20 MATH. 2153, Spring 16, MWF 12:40 p.m. QUIZ 3 (make-up) April 11, 2016 PRINT NAME A. Derdzinski RECITATION TIME (check one) 10:20 11:30 12:40 Show all work. No calculators. The problem is worth 10 points. 1. Evaluate the triple integral c = z dv, where D is the region consisiting of all (x, y, z) with D x 2 + y 2 + z 2 1 and z x 2 + y 2. Reminder: for integration in cylindrical/spherical/arbitrary coordinates, dx dy dz = r dr dθ dz = ρ 2 sin ϕ dρ dϕ dθ = (x, y, z) (u, v, w) du dv dw. YOUR ANSWER: c =
21 MATH. 2153, Spring 16, MWF 12:40 p.m. QUIZ 3 (make-up) April 13, 2016 PRINT NAME A. Derdzinski RECITATION TIME (check one) 10:20 11:30 12:40 Show all work. No calculators. The problem is worth 10 points. 1. Evaluate the triple integral c = z dv, where D is the region consisiting of all (x, y, z) with D x 2 + y 2 + z 2 1 and 3 z x2 + y 2. Reminder: for integration in cylindrical/spherical/arbitrary coordinates, dx dy dz = r dr dθ dz = ρ 2 sin ϕ dρ dϕ dθ = (x, y, z) (u, v, w) du dv dw. YOUR ANSWER: c =
22 MATH. 2153, Spring 16, MWF 12:40 p.m. page 1 of 2 TEST 2 (make-up) APRIL 22, 2016 PRINT NAME A. Derdzinski RECITATION TIME (check one) 10:20 11:30 12:40 Show all work. No calculators. Problems worth 25 points each. 1. Find an equation for the tangent plane at the point P = (3, 1, 2) of the surface xy z = 5. YOUR ANSWER: Equation 2. Find all the critical points of the function f(x, y) = (y 2 1) 2 + x 2 + 2x and decide which of them are local maxima, local minima, and saddle points. YOUR ANSWER: loc.min. at loc.max. at saddle pts. at
23 MATH. 2153, Spring 16, MWF 12:40 p.m., Test 2 page 2 of 2 NAME 3. Determine the minimum value m assumed by the function f(x, y, z) = x z on the ellipsoid 4x 2 + y 2 + z 2 + 4y + 2z + 1 = 0. YOUR ANSWER: m = 4. Use the right order of iterated integration to evaluate c = e x2 da, where R is the triangle in the xy-plane given by 0 y x 1. R YOUR ANSWER: c =
24 MATH. 2153, Spring 16, MWF 12:40 p.m. page 1 of 2 FINAL EXAM (early) APRIL 27, 2016 PRINT NAME A. Derdzinski RECITATION TIME (check one) 10:20 11:30 12:40 Show all work. No calculators. 1. Evaluate the derivative f (x) of the function f given by f(x) = sin x 3x e t2 dt. YOUR ANSWER: f (x) = 2. Determine the maximum value M and the minimum value m assumed by the function f(x, y) = x + 2y on the region R in the xy-plane given by x 2 + y 2 10 and y x. Find all points at which the extremum values are assumed. YOUR ANSWER: m =, M = 3. Find a polar-coordinate equation for a circle of radius 10 whose center C has the Cartesian coordinates (6, 8). (34 points) YOUR ANSWER: the equation
25 MATH. 2153, Spring 16, MWF 12:40 p.m., Final Exam page 2 of 2 NAME 4. Set up and solve a double integral to find the volume V of the solid bounded by the planes x = 0, y = 0, z = 0, and x + y + z = 2. (The solid in question is the tetrahedron with the vertices at (0, 0, 0), (2, 0, 0), (0, 2, 0) and (0, 0, 2).) YOUR ANSWER: V = = 5. Evaluate the line integral W = F d r, where C is any oriented curve in space C with the initial point P (1, 0, 0) and terminal point Q(2 + 3, 1, 1), and F denotes the gradient of the function ϕ(x, y, z) = arctan (x 2y) + ln(z 2 + 1). YOUR ANSWER: W = 6. Evaluate the triple integral c = z dv, where D is the region consisiting of all (x, y, z) with x 0, y 0, 0 z 1, x 2 + y 2 1. Reminder: for integration in cylindrical/spherical/arbitrary coordinates, dx dy dz = r dr dθ dz = ρ 2 sin ϕ dρ dϕ dθ = (x, y, z) (u, v, w) du dv dw. D YOUR ANSWER: c =
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