Mathematics 123.3: Solutions to Lab Assignment #1

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1 Mathematics 123.3: Solutions to Lab Assignment #1 2x 2 1 if x<= 1/2 orifx>= 1/2 (A12) 1 2x 2 = 1 2x 2 if 1/2 <= x<= 1/2 2x 2 1 if x (, 1/2] [ 1/2, ) = 1 2x 2 if x ( 1/2, 1/2) (A36) Solve ()(x 2)(x + 3) <= 0 First, rewrite as (x ( 3))(x ( 1))(x 2) <= 0, so that the number line is partitioned by 3, 1, and 2. Next, make a table: (, 3) 3 ( 3, 1) 1 ( 1, 2) 2 (2, ) x ( 3) x ( 1) x ()(x 2)(x + 3) Thus the solution set is (, 3] [ 1, 2] (A40) Solve 3 < 1 x <= 1 We need to look at two cases: x>0 and x<0. Case I: x > 0. We may multiply the given inequalities by x without changing their direction: x( 3) <x 1 x <= x(1) simplifies to: 3x <1 <= x. Thus, for positive x the given inequalities are satisfied if x>= 1. Case II: x<0. We may multiply the given inequalities by x, changing their direction: x( 3) > x 1 x >= x(1) simplifies to: 3x > 1 >= x. Thus, for negative x the given inequalities are satisfied if 3x > 1. Now we multiply this inequality by 1/3, and in so doing change its direction: ( 1/3)( 3x)<( 1/3)1, which simplifies to x< 1/3. Combining the two cases, our solution set is (, 1/3) [1, ) (A54) Solve for x: 2x 1 = 3 First perform long division to get 2x 1 = We then look for values where this expression equals 3 or3. It will equal 3 when 3 1 = 5, or = 3/5, or x = 2/5. It will equal 3 when 3 1 = 1, or = 1, or x = 4.

2 (B6) (a b) 2 + (b a) 2 = 2 b a (B10) Find the slope of the line through P( 1, 4) and Q(6, 0). 0 ( 4) 6 ( 1) = 4 7 (B14) (a) Show that the points A( 1, 3), B(3, 1), and C(5, 15) are collinear by showing that AB + BC = AC AB = (3 ( 1)) 2 + (11 3) 2 = = = 80 = BC = (5 3) 2 + (15 11) 2 = = = 20. AC = (5 ( 1)) 2 + (15 3) 2 = = = 180 = (b)use slopes to show that A, B, and C are collinear. Slope of AB = ( 1) = = 2, Slope of AC = 4 5 ( 1) = 12 6 = 2 (B36) Find an equation of the line through (1/2, 2/3) perpendicular to the line 4x 8y = 1. The given line has slope 2, so its perpendiculars have slope 1/2. Using the point slope form, the required line has equation y ( 2/3) = ( 1/2)(x 1/2), which simplifies to y + 2/3 = 1/2/4, or 62y = 5 (B62) A car leaves Detroit at 2:00 P.M., travelling at a constant speed west along I-0. It passes Ann Arbor, 40 mi from Detroit, at 2:50 P.M. (a) Express the distance travelled in terms of time elapsed. D = 40/50t mi/min = 4/5t mi/min Thus, if t = 50 min, we have D = 40/50(50 min) mi/min = 40 mi. (b) Draw the graph of the equation in part (a). Your graph should a line through the origin with slope 4/5. (c)what is the slope of this line? 4/5 What does it represent? velocity (in miles per minute)

3 (C8) 16x y 2 + 8x + 32y + 1 = 0 Rearrange and complete squares: 16x y y = 1 16(x 2 + 1/2x) + 16(y 2 + 2y) = 1 16(x 2 + 2(1/4)x + (1/4) 2 (1/4) 2 ) + 16(y 2 + 2y ) = 1 16((/4) 2 1/16) + 16((y + 1) 2 1) = 1 16(/4) ((y + 1) 2 16 = 1 16(/4) (y + 1) 2 = 16 (/4) 2 + (y + 1) 2 = 1 (x ( 1/4)) 2 + (y 1) 2 = 1 Thus the centre of the circle is at ( 1/4, 1) and the radius is 1. (C10) Under what conditions on the coefficients a, b, and c does the equation x 2 + y 2 + ax + by + c = 0 represent a circle? When that condition is satisfied, find the centre and radius of the circle. Rearrange and complete squares: x 2 + ax + y 2 + by = c x 2 + 2(a/2)x + y 2 + 2(b/2)y = c x 2 + 2(a/2)x + (a/2) 2 (a/2) 2 + y 2 + 2(b/2)y + (b/2) 2 (b/2) 2 = c (x + a/2) 2 + (y + b/2) 2 = c + (a/2) 2 + (b/2) 2 (x ( a/2)) 2 + (y ( b/2)) 2 = c + (a/2) 2 + (b/2) 2 The circle will only exist if the right-hand side is positive, so the requirement is that the square of the distance from ( a/2, b/2) to the origin must exceed c. The centre will then be ( a/2, b/2) and the radius will be c + (a/2) 2 + (b/2) 2 (C28) Identify and plot x 2 y 2 4x + 3 = 0. Rearrange and complete squares: x 2 4x y = 0 x 2 4x y = 0 (x 2) 2 y 2 = 1 Thus the centre is at (2,0), the vertices are at (1,0) and (3,0), and the asymptotes are the lines with slopes -1 and 1 passing through (2,0). Ask your instructor to draw the graph. (C30) y 2 2x + 6y + 5 = 0. Rearrange and complete squares: y 2 + 6y + 5 = 2x y 2 + 6y = 2x (y + 3) 2 4 = 2x (y ( 3)) 2 = 2x + 4 (y ( 3)) 2 = 2(x + 2) (y ( 3)) 2 = 4(1/2)(x ( 2)) Thus the vertex is at (-2,-3) Ask your instructor to draw the graph.

4 (C36) Find anequation of the ellipse with centre at the origin that passes through the points 2 5 (1, 10 ) and ( 2, ). The ellipse has equation of the form x 2 a 2 + y2 Substituting the coordinates of the given points, we get which simplify to 1 2 a 2 + ( 10 b 2 1 a )2 = 1 and ( 2)2 a 2 + (5 5 3 )2 200 and 4 a 2 + Let A = 1 a 2 and B = 1 so that our two equations are b2 125 A B = 1 and 4A + B = 1 Multiplying the first equation by 4 and adding it to the second yields ( ) B = 3or 675 B = 3, so B = = Substituting this into the first equation we get A Thus we have a = 3 and b = 5 and so the required equation is x y2 5 2 = = 1orA + 8 = 1, so A = 1.

5 (D14) If a circle has radius 10 cm, what is the length of arc subtended by a central angle of 72? In radians, the central angle is 72 π 180 = 2π 5, so the length of the required arc is 10 cm 2π 5 = 4π cm (D16) Find the radius of a circular sector with angle 3π 4 We have 6 cm = r 3π 4 (D58) Prove the identity cos 3θ = 4 cos 3 θ 3 cos θ,sor = 6cm 4 3π = 8 π cm and arc length 6 cm. Starting with the left hand term, we have: cos 3θ = cos 2θ + θ = cos 2θ cos θ sin 2θ sin θ = (2 cos 2 θ 1) cos θ 2 sin θ cos θ sin θ = cos θ(2 cos 2 θ 1 2 sin 2 θ) = cos θ(2 cos 2 θ 1 2(1 cos 2 θ)) = cos θ(4 cos 2 θ 3) = 4 cos 3 θ 3 cos θ which is the right-hand side. (D64) If sin x = 1/3 and sec y = 5/4, where x and y lie between 0 and π 2, evaluate cos 2y. cos 2y = 2 cos 2 y 1 = 2 sec 2 y 1 = 2 25/16 1 = 32/25 1 = 7/25 (D84) See text for question. By the Law of Cosines, AB 2 = AC 2 + BC 2 2 AC BC cos 103 = m m 2 2(820m)(10m) cos 103 = 10 2 ( (82)(1) cos 103 )m 2 = 10 2 ( cos 103 )m 2 = 10 2 ( cos 103 )m ( ( ))m 2 = 10 2 ( )m 2 = so AB 1355m