Questions and Answers #3 Analytic Geometry

Questions and Answers Sheet 3 Analytic Geometry Question #1 Let 𝑄2 be the rational plane of all ordered pairs (x,y) of rational numbers with the usual interpretations of the undefined geometric terms used in analytical geometry. Show that axiom C-1 and the elementary continuity principle fail in 𝑄2 . (Hint: the segment from (0,0) to (1,1) can not be laid off on the x axis from the origin. Axiom C-1: If A, B are two points on a line a, and if A' is a point upon the same or another line a' , then, upon a given side of A' on the straight line a' , we can always find a point B' so that the segment AB is congruent to the segment A'B' . We indicate this relation by writing 𝐴𝐵 ≅ 𝐴′ 𝐵′ . Every segment is congruent to itself; that is, we always have 𝐴𝐵 ≅ 𝐴𝐵 . Answer: Step 1 Let A, B lies on Horizontal axis. Let's consider a ray r from A' at 45∘ line. The distance between A and B is a rational number, all points on r separated by rational multiples of √2 Step 2 So, 𝑄2 is an example of Hilbert -plane not Euclidean. So, 𝑄2 is not even a Hilbert Plane Hence, Axiom, C-1 and elementary continuity principle fail in 𝑄2 Question #2 A point moves so that the angle from the line joining it and the origin to the line (3, - 2) and (5, 7) is 45∘ . Find the equation of the locus. Answer: Let the coordinate of the moving point be (𝛼, 𝛽) and origine (0, 0) Then slope of the line joining this two point (𝛼, 𝛽), (0, 0) is 𝑚[1} = 𝛽−0 𝛼−0 = 𝛽 𝛼 and the slope of the line joining the point (3, -2) and (5, 7) is 7 − (−2) 9 𝑚[2} = = 5−3 2 Now given that the above two lins whose slope are 𝑚1 , 𝑚2 makes an angle 45∘ Then by angle between two lines formula 𝑚 −𝑚 tan 4 5∘ = | 1 2 | 1+𝑚1 𝑚2 𝑚1 − 𝑚2 ⇒ 𝐼 =∣ 1 + 𝑚1 𝑚2 When we remove moduls it will give + and -, (Taking + for actuale angle 45∘ ) When + Then 1 = 𝑚[1} −𝑚2 1+𝑚1 𝑚2 ⇒ 1 + 𝑚1 𝑚2 = 𝑚1 − 𝑚2 𝛽 9 𝛽 9 ⇒1+ × = − 𝛼 2 𝛼 2 ⇒ 2𝛼 + 9𝛽 = 2𝛽 − 9𝛼 ⇒ 11𝛼 + 7𝛽 = 0 ∴ sins (𝛼, 𝛽) is arbitry ∴ locus of the point will be 11x+7y=0 Question #3 a) Use Coordinate Geometry Computations to compute the intersection point of the lines AB and CD. The X and Y coordinates of stations A, B,C and D in feet are given as follows: 𝑋𝐴 = 8377.66, 𝑌𝐴 = 4340.92 𝑋𝐵 = 8719.43, 𝑌𝐵 = 4642.65 𝑋𝐶 = 84.36.76, 𝑌𝐶 = 4712.64 𝑋 − 𝐷 = 8810.45, 𝑌𝐷 = 4325.72 b. Use an alternative method (e.g. analytical geometry) to compute the intersection point. Answer: Step 1 a) The Coordinates are as follows: (𝑋𝐴 , 𝑌𝐴 ) = (8377.66,4340.92) (𝑋𝐵 , 𝑌𝐵 ) = (8719.43,4642.65) (𝑋𝐶 , 𝑌𝐶 ) = (8436.76,4712.64) (𝑋𝐷 , 𝑌𝐷 ) = (8810.45,4325.72) So, The equation of line AB becomes 4642.65−4340.92 (𝑥 − 8377.66) 𝑦 − 4340.92 = 8719.43−8377.66 ⇒ 𝑦 − 4340.92 = 0.877(𝑥 − 8377.66) 1) ⇒ 𝑦 = 4340.92 + 0.877(𝑥 − 8377.66) The equation of the line CD becomes 4712.64−4325.72 (𝑥 − 8436.76) 𝑦 − 4712.64 = 8810.45−8436.76 ⇒ 𝑦 − 4712.64 = −1.041(𝑥 − 8436.76) 2) ⇒ 𝑦 = 4712.64 − 1.041(𝑥 − 8436.76) Now, For the intersection point of AB and CD, equate equation (1) and (2), we get 4340.92 + 0.877(𝑥 − 8377.66) = 4712.64 − 1.041(𝑥 − 8436.76) ⇒ 0.877(𝑥 − 8377.66) + 1.041(𝑥 − 8436.76) = 4712.64 − 4340.92 ⇒ 1.918𝑥 − 16129.88 = 371.72 ⇒ 1.918𝑥 = 16501.6 ⇒ 𝑥 = 8603.54 Put the value of x in the equation (1), we get 𝑦 = 4340.92 + 0.877(8603.54 − 8377.66) ⇒ 𝑦 = 4539.02 ∴ we get Step 2 b) We have (𝑋𝐴 , 𝑌𝐴 ) = (8377.66,4340.92) (𝑋𝐵 , 𝑌𝐵 ) = (8719.43,4642.65) (𝑋𝐶 , 𝑌𝐶 ) = (8436.76,4712.64) (𝑋𝐷 , 𝑌𝐷 ) = (8810.45,4325.72) Now, The direction vector for AB is given by 𝑧𝐴𝐵 =≪ 8719.43 − 8377.66,4642.65 − 4340.92 ≫ =≪ 341.77, 299.73 ≫ ⇒ 𝑧𝐴𝐵 =≪ 341.77, 299.73 ≫ 𝑆 So, The vector equation for the line AB becomes 3) 𝑒𝐴𝐵 = (8377.66,4340.92) + 𝑎(341.77,299.73) Similarly, The direction vector for CD is given by 𝑧𝐶𝐷 =≪ 8810.45 − 8436.76, 4325.72 − 4712.64 ≫ =≪ 373.69, −388.92 ≫ ⇒ 𝑧𝐶𝐷 =≪ 373.69, − 388.92 ≫ So, The vector equation for the line CD becomes 4) 𝑒𝐶𝐷 = (8436.76,4712.64) + 𝑏(373.69, −388.92) Now, The vector equation of AB and CD will incresect if equation (3) and (4) are equal (8377.66, 4340.92) + 𝑎(341.77, 299.73) = (8436.76, 4712.64) + 𝑏(373.69, − 388.92) Break it into 2 parts as 8377.66+341.77a=8436.76+373.69b 5) ⇒ 341.77𝑎 − 373.69𝑏 = 59.1 and 4340.92+299.73a=4712.64-399.92b 6) ⇒ 299.73𝑎 + 388.92𝑏 = 371.32 By solving equation (5) and (6), we get a=0.66 b=0.44 Put the value of a in equation (4), we get (8377.66, 4340.92) + 𝑎(341.77, 299.73) = (8377.66, 4340.92) + 0.66(341.77, 299.73) = (8603.22, 4538.74) \therefore we get The intersection point of the line AB and CD is (8603.22, 4538.74) Step 3 Answer: a) The intersection point of the line AB and CD is (𝑥, 𝑦) = (8603.54,4539.02) b) The intersection point of the line AB and CD is (8603.22, 4538.74) Question #4 Using analytic geometry, find the coordinates of the point on the x axis which is equaidistant from A(5, 8) and B(-3, 4) Answer: Given, A(5, 8), B(-3, 4) Let, Equidistent point, p(x, 0) So, using distance formula: √(𝑥 − 3)2 + (−8)2 = √(𝑥 + 3)2 + (−4)2 Both side x 2 + 25 − 10 x + 64 = x 2 + 9 + 6 x + 16 25+64-9-16=10x+6x 16x=64 x=4 Question #5 In the diagram below, a circle with centre O is given with diameter AB. The coordinates of A and B are (4 ; 4) and (2 ; -6) respectively. ya A(4; 4) B(2; -6) a) Determine the coordinates of O, the centre of the circle. b) Calculate the length of the radius of the circle. c) Write down the equation of the circle in form: (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2 d) Write down the equation of the circle in form: 𝑎𝑥 2 + 𝑏𝑥 + 𝑐𝑦 2 + 𝑑𝑦 + 𝑒 = 0 Answer: Step 1 Centre of the circle = mid Point of AB 𝑥1 + 𝑥2 𝑦1 + 𝑦2 =( , ) 2 2 4+2 4−6 ( , ) 2 2 = (3, − 1) Step 2 Length of the radius = = = = 𝐴𝐵 2 √(𝑥2 −𝑥1 )2 +(𝑦2 −𝑦1 )2 2 √(4−2)2 +(4+6)2 2 √4+100 𝑟= 2 √104 2 Question #6 Consider the case of a rotating wheel at rest and starting a clockwise rotation, meaning the negative direction of the angular velocity, and increasing (negatively) its value up to -12 rad/sec for 2 seconds. It then maintains a constant velocity for 2 seconds, and then uniformly reduces the magnitude of the velocity for 2 seconds until the wheel is momentarily stopped and restarts its rotation counter clockwise with positive angular velocity, accelerating up to 20 rad/sec in 2 seconds and remaining at a constant rotation for 2 more seconds. Finally, the wheel stops gradually in 2 seconds. Next you can see the graph of angular velocity versus time of this rotation: Slope : ) Origin intercept : Equation :  = f ( t ) Answer: Step 1 The graph in the interval [0,2] is a straight line The standard equation of a line in slope intercept form is 𝑦 = 𝑚𝑥 + 𝑐 where m is the slope of the line and c is the y-intercept. The horizontal axis is x-axis and vertical axis is y-axis. Step 2 The two points from which slope will be calculated are: (0,0) and (2, −12) which are the respective endpoints of the line. Calculate the slope. 𝑦 −𝑦 𝑚= 2 1 = 𝑥2 −𝑥1 −12−0 2−0 = −6 Hence the slope is -6. The y-intercept is the point where the line intersects the y-axis. This point is clearly the origin. Hence the origin y-intercept is 0 since the y-coordinate at this point is 0. The y-axis is represented by \omega and the x-axis is represented by The equation of the line is thus framed. 𝜔 = 𝑚𝑡 + 𝑐 𝜔 = −6𝑡 + 0 Hence the equation of the line is 𝜔 = −6 Question #7 Find the area of the surface generaled by revolving the curve 𝑥 = 𝑦3 5 , 0 ≤ 𝑦 ≤ 3 about the y-axis. Answer: Step 1 The curve 𝑥 = 𝑔(𝑦), 𝑐 ≤ 𝑦 ≤ 𝑑 is revolved about the x-axis, then area of the surface is 𝑑 𝐴 = 2𝜋 ∫𝑐 𝑔(𝑦)√1 + [𝑔′ (𝑦)]2 𝑑𝑦 Step 2 3 Now differentiate 𝑔(𝑦) = 𝑦 2 , with resprct to y, we get 3𝑦 2 𝑔′ (𝑦) = 5 Step 3 Now substitute the value of 𝑔(𝑦) and 𝑔′ (𝑦) in above formula, we get 2 2 3 𝑦3 𝐴 = 2𝜋 ∫0 √1 + (3𝑦 ) 𝑑𝑦 5 5 Step 4 Simplify the above integral, we get 2 2 3 𝑦3 𝐴 = 2𝜋 ∫0 √1 + (3𝑦 ) 𝑑𝑦 5 5 2𝜋 3 3√1+9𝑦4 25 𝑑𝑦 𝐴= ∫ 𝑦 5 0 Put 1 + 9𝑦 4 25 = 𝑢 ⇒ 𝑦 3 𝑑𝑦 = 25 36 𝑑𝑦 When 𝑦 = 0, 𝑢 = 1 and when 𝑦 = 3, 𝑢 = 2 25 745 25 udu So, A = 5 36 1 𝐴= 3 𝑢2 754 25 5𝜋 [ ] 18 3 2 1 3 𝐴= 5𝜋 𝐴= 18 [( 5𝜋 27 754 2 ) 25 − 1] [165.63 − 1] 𝐴 = 30.49𝜋 754 25 Question #8 Find 𝑚∠𝐹 Answer: Step 1 Given, a right angle triangle ∠𝐸 = 90∘ 𝐸𝐷 = 6 𝐸𝐹 = 3 Step 2 Let ∠𝐹 = 𝜃 𝑝 𝐷𝐸 then tan 𝜃 = = ⇒ tan 𝜃 = 6 𝑏 𝐸𝐹 3 ⇒ tan 𝜃 = 2 ⇒ 𝜃 = tan−1 2 ⇒ 𝜃 = 63.434 ≅ 63.4 ∴ on ∠𝐹 = 63.4 Question #9 Write the equation of the circle with Center (-1, -8) and Radius: √7 a) 𝑥 2 + 2𝑥 + 𝑦 2 + 16𝑦 + 58 = 0 b) 𝑥 2 + 2𝑥 + 1 + 𝑦 2 + 16𝑦 + 64 = 7 c) 𝑥 2 + 2𝑥 + 𝑦 2 + 16𝑦 + 72 = 0 d) 𝑥 2 − 2𝑥 + 𝑦 2 − 16𝑦 + 58 = 0 Answer: Step 1 Given point on circle ⇒ (𝑎, 𝑏) = (−1, −8) and radius ⇒ 𝑟 = √7 As we know given equation of circle is i.e (𝑥 − 𝑎)2 + (𝑦 − 𝑏}2 = 𝑟 2 2 or (𝑥 + 1)2 + (𝑦 + 8)2 = (√7) = 7 𝑥 2 + 1 + 2𝑥 + 𝑦 2 + 64 + 16𝑦 = 7 𝑥 2 + 𝑦 2 + 2𝑥 + 16𝑦 + 58 = 0 Hence option (a) is correct Question #10 2 3𝑛 Consider the sequence {𝐴𝑛 } , whose nth term is given by 𝐴𝑛 = (1 + ) 𝑛 Use limits to analyticlly show how we can determine whether the sequence is convergent or divergent. If it is convergent, find the limit of the sequence. Write a sentence to summariae your findings. Answer: Step 1 If the limit of the sequence is finite then the series is convergent otherwise divergent. 2 3𝑛 lim (𝐴𝑛 ) = 𝐿 = lim (1 + ) 𝑛⇒∞ 𝑛⇒∞ 𝑛 taking log both sides    2 3 n   log ( L ) = lim log  1 +    n    n    2 log(𝐿) = lim [3𝑛 log (1 + )] 𝑛⇒∞ 𝑛 2 3    2 2        2 n n = lim 3n  −   +   − сs   n  n  2 3       2 8   lim 6 − 2 + 3 − сs  n  n 3n  log(𝐿) = 6 𝐿 = 𝑒6 thus, lim (𝐴𝑛 ) = 𝐿 = 𝑒 6 (finite) 𝑛⇒∞ Thus, the limit of the sequence is finite and hence the given sequence is convergent. Question #11 x is an acuate. Find the value of x in degrees cos(𝑥) = √3 3 𝑥 =? Answer: Step 1 Given ⇒ cot(𝑥) = ⇒ cot(𝑥) = ⇒𝑥= √3 3 1 √3 −1 1 cot ( } √3 ∘ 𝑎𝑟𝑟𝑤𝑜𝑥 = 60 Question #12 As viewed from above, a swimming pool has the shape of the ellipse 𝑥2 𝑦2 + = 1, 2500 900 where x and y are measured in feet. The cross sections perpendicular to the x-axis are squares. Find the total volume of the pool. 𝑉 = cubic feetask Answer: Step 1 Suppose that the x and y axes are places so that the origin is at the center of the pool. Observe that each vertical strip perpendicular to the x-axis is in the shape of a square. Because we have the origin of our coordinate system in the center of the pool, the y-coordinate for each value of x equals half of the width of the pool at that location. Observe the following. The dimension of the square shaped slice is width = 2 𝑦 and depth = 2𝑦 . The thickness of each slice is dx. Area ofcross section = 2𝑦 × 2𝑦 = 4𝑦 2 The volume of each cross section is 𝑑𝑉 = 𝐴(𝑥)𝑑𝑥 = 4𝑦 2 𝑑𝑥 The given equation is 𝑥2 ⇒ 𝑦 2 = 𝑏 2 (1 − 𝑎2 𝑥2 𝑎2 + 𝑦2 𝑏2 = 1 with 𝑎2 = 2500 and 𝑏 2 = 900 . ) ⇒The volme of cross section, 𝑑𝑉 = 4𝑦 2 𝑑𝑥 = 4𝑏 2 (1 − 𝑥2 𝑎2 ) 𝑑𝑥 Step 2 The volume of the pool is the sum of all of the slices, which can be computed by integrating dV. Due to symmetry, the volume of half of the pool can be computed, and then multiply by 2 to get the total volume. This means that integration is from the origin (𝑥 = 0)to the edge of the pool (𝑥 = 𝑎) . 𝑎 Half volume, 𝑉ℎ = ∫ 𝑑𝑉 = ∫0 4𝑏 2 (1 − = 4𝑏 𝑥2 𝑎2 ) 𝑑𝑥 𝑎 𝑥2 2 ∫0 (1− 2 )𝑑𝑥 𝑎 = 4𝑏 2 [𝑥 − 𝑥3 3𝑎2 𝑎 ] 0 𝑎 = 4𝑏 2 [𝑎 − ] 3 8 = 𝑎𝑏 2 3 8 Total volume, 𝑉 = 2𝑉ℎ = 2 ( 𝑎𝑏 2 ) = 3 16 (50)(900) = 3 = 240000 cubic feetask 16 3 𝑎𝑏 2 Question #13 Evaluate the following expression: 20!/(13! (20 − 13)!) Answer: 20!/(13! (20 − 13)!) = 20!/(13! 7!) (20𝑥𝑥19𝑥𝑥18𝑥𝑥17𝑥𝑥16𝑥𝑥15𝑥𝑥14𝑥𝑥13!)/(13! 7!) = (20𝑥𝑥19𝑥𝑥18𝑥𝑥17𝑥𝑥16𝑥𝑥15𝑥𝑥14)/(7𝑥𝑥6𝑥𝑥5𝑥𝑥4𝑥𝑥3𝑥𝑥2𝑥𝑥1) = (20𝑥𝑥19𝑥𝑥18𝑥𝑥17𝑥𝑥16𝑥𝑥15𝑥𝑥14)/((7𝑥𝑥2)(5𝑥𝑥4)(6𝑥𝑥3) = (20𝑥𝑥19𝑥𝑥18𝑥𝑥17𝑥𝑥16𝑥𝑥15𝑥𝑥14)/(14𝑥𝑥20𝑥𝑥18) = 19𝑥𝑥18𝑥𝑥16𝑥𝑥15 = 77520 Question #14 Multiply 1/9 and 24/25 Answer: 1/9𝑥𝑥24/25 = 24/225 = (24−: 3)/(225−: 3) = 8/75 Question #15 Find 𝑥𝑃(𝑥), 𝑥 2 and 𝑥 2 𝑃(𝑥) when: 1. 𝑥 = 1; 𝑃(𝑥) = 0.25; 2. 𝑥 = 2; 𝑃(𝑥) = 0.28; 3. 𝑥 = 3; 𝑃(𝑥) = 0.41; 4.𝑥 = 4; 𝑃(𝑥) = 0.06; Answer: 1. 𝑥𝑃(𝑥) = 1𝑥𝑥0.25 = 0.25 𝑥 2 = 12 = 1 𝑥 2 𝑃(𝑥) = 12 𝑥𝑥0.25 = 0.25 2. 𝑥𝑃(𝑥) = 2𝑥𝑥0.28 = 0.56 𝑥 2 = 22 = 4 𝑥 2 𝑃(𝑥) = 22 𝑥𝑥0.28 = 1.12 3. 𝑥𝑃(𝑥) = 3𝑥𝑥0.41 = 1.23 𝑥 2 = 32 = 9 𝑥 2 𝑃(𝑥) = 32 𝑥𝑥0.41 = 3.69 4. 𝑥𝑃(𝑥) = 4𝑥𝑥0.06 = 0.24 𝑥 2 = 42 = 16 𝑥 2 𝑃(𝑥) = 42 𝑥𝑥0.06 = 0.96 Question #16 What is the least common multiple of {120, 124, 165} ? Answer: To find the least common multiple, let's do a ' factorization on these numbers: 120 = 2 × 2 × 2 × 3 × 5 124 = 2 × 2 × 31 165 = 3 × 5 × 11 To find the lowest common multiple, we take the largest group of 's we can choose from. For instance, the first ' is 2. The 120 has 3 of them and that is the most that any of our other numbers have. So we need 3 2s. 2 ×2 ×2 ×… For 3, the next ', two of the numbers have a 3, so we need 1 also. 2 ×2 ×2 ×3 ×… We also have two numbers with a 5 each: 2 ×2 ×2 ×3 ×5 ×… And there is an 11 and a 31: 2 × 2 × 2 × 3 × 5 × 11 × 31 = 40,920 120 × 11 × 31 = 40,920 124 × 2 × 3 × 5 × 11 = 40,920 165 × 2 × 2 × 2 × 31 = 40,920 Question #17 How do you find the Least common multiple of 28𝑏 2 , 20𝑎𝑏 3 , 16𝑏 4 ? Answer: 𝐿𝐶𝑀 = 2 × 2 × 2 × 2 × 5 × 7 × 𝑎 × 𝑏 × 𝑏 × 𝑏 × 𝑏 = 560𝑎𝑏 4 Explanation: To find LCM, I like first to do a ' factorization of the numbers in question: 28𝑏 2 = 2 × 14 × 𝑏 × 𝑏 = 2 × 2 × 7 × 𝑏 × 𝑏 20𝑎𝑏 3 = 2 × 10 × 𝑎 × 𝑏 × 𝑏 × 𝑏 = 2 × 2 × 5 × 𝑎 × 𝑏 × 𝑏 × 𝑏 16𝑏 4 = 2 × 8 × 𝑏 × 𝑏 × 𝑏 × 𝑏 = 2 × 2 × 4 × 𝑏 × 𝑏 × 𝑏 × 𝑏 = 2 × 2 × 2 × 2 × 𝑏 × 𝑏 × 𝑏 × 𝑏. The LCM will include all the 's listed (and all the variables too) in the biggest quantity from any given number we're working with. For instance, we have 2s. There are 4 of them in the 16𝑏 4 , so our LCM will have 4 as well: 𝐿𝐶𝑀 = 2 × 2 × 2 × 2 ×? We also have a 5 in the 20𝑎𝑏 3 . 𝐿𝐶𝑀 = 2 × 2 × 2 × 2 × 5 ×? There's a 7 in the 28𝑏 2 . 𝐿𝐶𝑀 = 2 × 2 × 2 × 2 × 5 × 7 ×? That's it for numbers. We have an a in 20𝑎𝑏 3 and we have 4 b in 16𝑏 4 . 𝐿𝐶𝑀 = 2 × 2 × 2 × 2 × 5 × 7 × 𝑎 × 𝑏 × 𝑏 × 𝑏 × 𝑏 All of this gives us: 𝐿𝐶𝑀 = 2 × 2 × 2 × 2 × 5 × 7 × 𝑎 × 𝑏 × 𝑏 × 𝑏 × 𝑏 = 560𝑎𝑏 4 . Question #18 Multiply (−6𝑥 − 5)(𝑥 3 − 𝑥 + 2) Answer: Step 1 Given: (−6𝑥 − 5)(𝑥 3 − 𝑥 + 2) Step 2 = −6𝑥(𝑥 3 − 𝑥 + 2) − 5(𝑥 3 − 𝑥 + 2) = −5𝑥 4 + 6𝑥 2 − 12𝑥 − 5𝑥 3 + 5𝑥 − 10 = −6𝑥 4 − 5𝑥 3 + 6𝑥 2 − 7𝑥 + 10 Question #19 How to factor 𝑥 2 − 2𝑥 − 24 ? Answer: Then 4 − 6 = 2 and 4 ∗ −6 = −24 we can factor this equation as: (𝑥 + 4)(𝑥 − 6) Question #20 What is 89.1 ∗ 0.47 ? Answer: Stack the numbers in a straight line. 89.1 × .47 Use multiplication to multiply as if there are no decimals. 38493 Then count how many decimal places there are total in the equation. For example, 3.5 has one decimal place. Place a decimal point in your answer counting from right to left. 38.493