Exam 3 with Answer Key - Calculus and Analytic Geometry | MATH 222, Exams of Analytical Geometry and Calculus

Download Exam 3 with Answer Key - Calculus and Analytic Geometry | MATH 222 and more Exams Analytical Geometry and Calculus in PDF only on Docsity! Math 222, Exam III, April 20, 2001 Answers I. (40 points.) The coordinate axes are to be rotated through an angle α so that the new equation for the curve x2 + xy + y2 = 3 has no cross-product term. Find α and the new equation. Sketch the curve. Your sketch must show both the x, y axes (as indicated) and the rotated axes. Answer: The coordinates (x′, y′) of a point in the rotated system and the coordinates (x, y) of that same point in the original system are related by x = x′ cosα− y′ sinα y = x′ sinα + y′ cosα. The transformed equation is 3 = x2 + xy + y2 = (x′ cosα− y′ sinα)2 +(x′ cosα− y′ sinα)(x′ sinα + y′ cosα) +(x′ sinα + y′ cosα)2. This expands to 3 = (x′)2(cos2 α + cosα sinα + sin2 α) +x′y′(−2 cosα sinα + cos2 α− sin2 α + 2 cosα sinα) +(y′)2(sin2 α− sinα cosα + cos2 α) which simplifies to 3 = (x′)2(1 + 1 2 sin 2α) + x′y′ cos 2α + (y′)2(1− 1 2 sin 2α) To kill the x′y′ term take 2α = π/2 so α = π/4 so sin 2α = sin(π/2) = 1 and the equation becomes 3 = 3 2 (x′)2 + 1 2 (y′)2 which is an ellipse. (This is Example 2 page 428 of the text.) 1 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -3 -2 -1 0 1 2 3 x y x′y′ The graph was drawn as follows. The equation of the ellipse in the x′y′- plane is (x′)2 2 + (y′)2 6 = 1. This can be parameterized via x′ = √ 2 cos(t), y′ = √ 6 sin(t). Hence the equations x = x′ cosα− y′ sinα = √ 2 cos(t) cos(π/4)− √ 6 sin(t) sin(π/4) y = x′ sinα + y′ cosα = √ 2 cos(t) sin(π/4) + √ 6 sin(t) cos(π/4) are parametric equations for the ellipse in the xy-plane. The x′-axis is given parametrically by x′ = t, y′ = 0 or x = t cosα, y = t sinα. 2 IV. (40 points.) Let P (a, b) lie on the parabola 4py = x2 and L be the tangent line to the parabola at P . Show that the line from the focus F (0, p) to the point P and the vertical line x = a through P make equal angles with the tangent line L to the parabola at P . Draw a diagram and be sure to define any symbols that you use. Answer: Let D be the directrix x = −p (it is parallel the x axis), V be the vertical line through P , L be the tangent line to the parabola at P , α be angle from D to FP , β be the angle from FP to L, β′ be the angle from L to V , and γ be the angle from D to L. We must show that β = β′. Then tanα = slope of FP = b− p a− 0 = a2/(4p)− p a = a2 − 4p2 4ap and tan γ = slope of L = dy dx ∣∣∣∣ x=a = 2a 4p = a 2p . F • P • D V L 5 From the picture (and the fact that the angles of a triangle sum to π) we get α + β + (π − γ) = π, π 2 − γ = β′ so β = γ − α and tan β′ = cot γ = 2p a . Now tan β = tan(γ − α) = tan γ − tanα 1 + tan γ tanα = a 2p − a2−4p2 4ap 1 + ( a 2p )( a2−4p2 4ap ) · 8ap2 8ap2 = = 4a2p− 2p(a2 − 4p2) 8ap2 + a(a2 − 4p2) = 2p(2a2 − a2 + 4p2) a(8p2 + a2 − 4p2) = 2p(a2 + 4p2) a(4p2 + a2) = 2p a = tan β′ as required. (This is problem 36 on page 410.) __________ Mon Apr 23 16:13:03 2001 There are 137 scores grade range count percent A 135...150 18 13.1% AB 120...134 19 13.9% B 105...119 29 21.2% BC 90...104 25 18.2% C 75... 89 16 11.7% D 50... 74 15 10.9% F 0... 49 15 10.9% Mean score = 98.4. Mean grade = 2.45. -- 6