EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, 2000

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1 EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, 2000 This examination has 30 multiple choice questions. Problems are worth one point apiece, for a total of 30 points for the whole examination. You may use a scientific calculator, such as the TI-83, but not any calculator that has a CAS. You may also use both sides of a 3 5 notecard which you have prepared beforehand with notes and formulas. An answer key appears on the last page. Solutions are given for problems 2, 25, 26 and Find a Cartesian equation for the curve described by the polar equation r =cosθ. (A) (x 1) 2 + y 2 =1 (B) (x 1 2 )2 + y 2 = 1 (C) x 2 +(y 1) 2 =1 (D) x 2 +(y 1 2 )2 =1 (E) (y 1) 2 =2x (F) y =1 2x (G) y 2 x 2 =1 (H) xy =1 (I) x 2 =2y (J) 2x 2 + y 2 =1 1

2 2 EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, Find the area enclosed by the polar curve r =2 sin θ. (A) 7π 2 (B) 15 π (C) π (D) 17π (E) 9 2 π (F) 19π (G) 5π (H) 21π (I) 11π 2 (J) 23π Solution: Curve is traced out once over 0 θ 2π. Area = R 2π R 0 2π ( sinθ 0 +sin2 θ) dθ = 1(8π 0+π) = 9π (2 sin θ)2 dθ = 3. Find the length of the polar curve r = e θ,over0 θ π. (A) (B) 30.1 (C) (D) (E) (F) (G) (H) (I) (J) 33.81

3 EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, Find the surface whose equation in cylindrical coordinates is r 2 + z 2 = 25. (A) Circular cylinder of radius 5, axis is the z-axis. (B) Circular cone, axis is the z-axis, vertex at the origin. (C) Sphere of radius 5, center at the origin. (D) Sphere of radius 1, center at (5, 5, 5). (E) Circular cylinder of radius 5, axis is the x-axis. (F) Circular cylinder of radius 25, axis is the z-axis. (G) Sphere of radius 5, center at (0, 0, 1). (H) Vertical plane containing the z-axis and the line x = y. (I) Horizontal plane through (0, 0, 5). (J) Circular cone, axis is the z-axis, vertex at (5, 5, 5). 5. Find the equation in spherical coordinates for x 2 + y 2 + z 2 =2z. (A) ρ =2sinθ (B) ρ =2cosθ (C) θ = π (D) ϕ = π (E) ρ = 1 2 (F) ρ =1 (G) ρ =cosϕ (H) ρ =sinϕ (I) ρ =2cosϕ (J) ρ =2sinϕ

4 EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, For r(t) =ht 2 2t, sin 3t, e t i, find lim t 0 r( t) r(0) t. (A) h6, 1, 7i (B) h5, 1 2, 6i (C) h, 0, 5i (D) h3, 1 2, i (E) h2, 1, 3i (F) h1, 3 2, 2i (G) h0, 2, 1i (H) h 1, 5 2, 0i (I) h 2, 3, 1i (J) h 3, 7 2, 2i 7. If u(t) andv(t) are vector functions, if u v 0 = ht 2,t 3 cos t t, t 2 cos ti and if v u 0 = ht 2,t 3t 2 sin t, 2t sin t 2i then find (u v) 0 (t). (A) ht 2,t 3 cos t +2t +3t 2 sin t, 2 t 2 cos t 2t sin ti (B) h0, t 3 cos t + t +3t 2 sin t, 1 t 2 cos t 2t sin ti (C) h t 2,t 3 cos t +3t 2 sin t, t 2 cos t 2t sin ti (D) h 2t 2,t 3 cos t t +3t 2 sin t, 1 t 2 cos t 2t sin ti (E) h 3t 2,t 3 cos t 2t +3t 2 sin t, 2 t 2 cos t 2t sin ti (F) h t 2,t 3 cos t 3t +3t 2 sin t, 3 t 2 cos t 2t sin ti (G) h 5t 2,t 3 cos t t +3t 2 sin t, t 2 cos t 2t sin ti (H) h 6t 2,t 3 cos t 5t +3t 2 sin t, 5 t 2 cos t 2t sin ti (I) h 7t 2,t 3 cos t 6t +3t 2 sin t, 6 t 2 cos t 2t sin ti (J) h 8t 2,t 3 cos t 7t +3t 2 sin t, 7 t 2 cos t 2t sin ti

5 EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, Find the unit tangent vector T at t =0ofthecurver(t) =3t i +costj +sintk. (A) 3j + k 5 5 (B) 5 i 3 5 j (C) 1i + 2j k q (D) 2i 3j +2 3 k (E) 3i + k 5 5 (F) i 3k 5 5 (G) 5 j 3 5 k (H) 3i j 5 5 (I) 2 5 i j k q (J) 2 3 i + 3j + 2k Find the length of the helix r(t) =sint i +3tj +costk, over0 t 2π. (A) 19π 2 (B) 39π (C) 10π (D) 1π (E) 21π 2 (F) 3π (G) 11π (H) 5π (I) 23 2 π (J) 7π

6 6 EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, Find the curvature, κ(x), at x = 6oftheparabolay = x 2. (A).010 (B).011 (C).012 (D).013 (E).01 (F).015 (G).016 (H).017 (I).018 (J) Find the principal normal vector N(t) ofthecurver(t) = 3costi + 3sintj. (A) k 1 (B) 2 (i + j) (C) 3sinti +3costj (D) sin t i +costj (E) sin t i cos t j (F) cos t i +sintj (G) cos t i sin t j (H) cos t i +sintj (I) 3 cos t i +3sintj (J) cos t i +sintj +3tk

7 EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, The position function of a particle is r(t) =ht 3 3, ti. Find its speed at t =1. (A) 3 (B) 3.25 (C) 3.5 (D) 3.75 (E) (F).25 (G).5 (H).75 (I) 5 (J) The position vector of a particle is r(t) =(1 sin t)i +(t cos t)j. Find the tangential component of the acceleration vector at t = π. 6 (A) 1 16 (B) 1 8 (C) 3 16 (D) 1 (E) 5 16 (F) 3 8 (G) 7 16 (H) 1 2 (I) 9 16 (J) 5 8

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10 10 EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, 2000 x 18. Find lim 2 (x,y) (1,2). x 2 +y 2 (A) This limit doesn t exist. (B) 1 2 (C) 1 3 (D) 1 (E) 1 5 (F) 5 (G) 7 (H) 9 (I) 3 (J) If f(x, y) = x2 y 3 2x 2 +y 2 for (x, y) 6= (0, 0), and if f is continuous at (0, 0), find f(0, 0). (A).8 (B).7 (C).6 (D).5 (E). (F).3 (G).2 (H).1 (I) 0 (J) 1 3

11 EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, If f(x, y) =9 x 2 y 2, find lim x 0 f(2+ x,1) f(2,1) x. (A) 0 (B) 2 (C) (D) 6 (E) 8 (F) 10 (G) 12 (H) 1 (I) 16 (J) If z = f(x, y) satisfies the equation xy + yz = xz, find z x (A) 2 (B) 3 (C) (D) 5 (E) 6 (F) 7 (G) 8 (H) 9 (I) 10 (J) 5 8 at x =1,y =2.

12 12 EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, Find the tangent plane to the elliptic paraboloid z =2x 2 + y 2 at the point (1, 1, 3). (A) z =3 (B) z =3+8(y 1) (C) z =3+9(x 1) + 7(y 1) (D) z =3+8(x 1) + 6(y 1) (E) z =3+7(x 1) + 5(y 1) (F) z =3+6(x 1) + (y 1) (G) z =3+5(x 1) + 3(y 1) (H) z =3+(x 1) + 2(y 1) (I) z =3+3(x 1) + 1(y 1) (J) z =3+2(x 1) 23. For f(x, y) =x 2 y + x y, find f xy (3, ). (A) 5 (B) 5.25 (C) 5.5 (D) 6 (E) 6.25 (F) 6.5 (G) 6.75 (H) 7 (I) 7.25 (J) 7.5

13 EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, Find the slope of the tangent line to the polar curve r = θ at θ =2π. (A) 1 3 π (B) π 2 (C) π (D) 3π 2 (E) 2π (F) 5 π (G) 7π 6 (H) 7π (I) 5 3 π (J) 3π 25. Find the area of the region enclosed by one loop of the curve r =sin3θ. (A) π 16 (B) π 1 (C) π 12 (D) π 10 (E) π 8 (F) π (G) π 3 (H) π 2 (I) π (J) 3π Solution: The curve goes through 1 loop over 0 θ π/3. Area = R π/3 1 2 R π/ (1 cos 6θ) dθ = 1 ( π 3 0) = π sin2 3θ dθ =

14 1 EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, Find the vector function that represents the curve of intersection of the cone z = p x 2 + y 2 and the plane z =1+y. (A) ht, 1 2 (t2 1), 1 2 (1 + t2 )i (B) h 1 2 (t2 1),t, 1(1 + 2 t2 )i (C) h 1 2 (t2 1),t,1+ti (D) hcos t, sin t, 1+sinti (E) hcos t, sin t, 1(1 + 2 t2 )i (F) hcosh t, t, sinh ti (G) hcosh t, sinh t, 1+sinhti (H) ht, s, 1+si (I) ht, t 2 +1, 1+t 2 i (J) ht 2 1,t, 1(1 + 2 t2 )i Solution: On the intersection, 1 + y = z = p x 2 + y 2. Square both sides to get 1+2y + y 2 = x 2 + y 2,whichgivesy = x2 1. Therefore, we can use x as the parameter. 2 Put x = t, y = t2 1 and z =1+y =1+ t2 1 = 1+t

15 EXAM 2 ANSWERS AND SOLUTIONS, MATH 233 WEDNESDAY, OCTOBER 18, For a curve r(t), a table of values for the arclength function s(t) and the curvature function κ(t) are given, respectively, by the tables t s t k Find the value of the curvature, κ, whenthearclengths =1.50. (A) (B) (C) (D) (E) (F) (G) (H) (I) (J) 0.150

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