Questions and Answers #2 Analytic Geometry

Questions and Answers Sheet 2 Analytic Geometry Question #1 Find the equation of the ellipse 𝑥 2 + 9𝑦 2 − 4𝑥 − 72𝑦 + 139 = 0 and locate the coordinates of the foci. Answer: Step 1 Ellipse is basically a plane curve around 2 focal points such that for all points on the curve, the sum of the two distances to the focal points is a constant. General equation of ellipse is given by (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 A circle is a special type of ellipse in which both the focal points are same. Step 2 Given equation of ellipse: 𝑥 2 + 9𝑦 2 − 4𝑥 − 72𝑦 + 139 = 0 Completing the squares of x and y terms: 𝑥 2 − 4𝑥 + 9𝑦 2 − 72𝑦 = −139 ⇒ 𝑥 2 − 4𝑥 + 4 + 9𝑦 2 − 72𝑦 = −139 + 4 ⇒ (𝑥 2 − 4𝑥 + 4) + 9(𝑦 2 − 8𝑦 + 16) = −139 + 4 + 144 ⇒ (𝑥 − 2)2 + 9(𝑦 − 4)2 = 9 ⇒ (𝑥−2)2 9 + (𝑦 − 4)2 = 1 Step 3 Center of the ellipse is (2, 4)=(h, k) Eccentricity of the ellipse will be: √𝑎2 − 𝑏 2 √32 − 12 √9 − 1 √8 2√2 = = = = 𝑎 3 3 3 3 Now, focii of ellipse: (h-c, k),\ (h+c, k) where, 𝑐 = √𝑎2 − 𝑏 2 = √9 − 1 = √8 = 2√2 Hence, Focii will be: (2 − 2√2, 4) 𝑎𝑛𝑑 (2 + 2√2, 4) Question #2 Using analytic geometry, find the coordinates of the point on the x axis which is equidistant 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 − 5)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 Coordinate is (4, 0) Question #3 Learn with an example If DE=7 and EF=11 , what is DF? Write your answer as a decimal or integer. Answer: Step 1 Given that DE=7 and EF=11 We have to find DF Step 2 Clearly from given picture DF=DE+EF given that, DE=7 and EF=11 , DF =? clearly, DF=DE+EF we have DE=7 and EF=11 DF=74+11=18 DF=18 ans Question #4 What is the volume of this cone? Use π = 3.14 and round your answer to the nearest hundredth. Answer: Given: Height of the cone (h)=10inch Diameter of cone (d)=14inch To find:- volume of the cone=? Solution:Height of cone (h)=10inch 𝐷 14 Radius of cone = = 2 2 =7 inch Now volume of cone (v) 1 𝑣 = π𝑟 2 ℎ 3 1 𝑣 = (3.14)(7)2 (10)(∵ π = 3.14) 3 𝑣 = (1.046)(49)(10) 𝑣 = 512.54𝑖𝑛3 Ans. Question #5 The radius of a circle is 5 miles. What is the angle measure of an arc bounding a sector with area 10π square miles? Answer: Ans. r=5 , Area 10π Sector area = π𝑟 2 × Sector area = π𝑟 2 × 10π = π ⋅ (𝑆)2 × 𝑂= 10π×360 𝑂 360 π×25 O=144 Question #6 𝑂 360 𝑂 360 What is the surface area of this cylinder? Use π ≈ 3.14 and round your answer to the nearest hundresth. What is the surface area of this rectangular pyramid? Answer: ans. Total surface area of given cylinder = 2πγℎ + 2πγ2 Radius of given cylinder = 154 2 = 77 in By Putting values in formula c ∴ ⇒ 2 × 3.14 × 77 × 56 + 2 × 3.14 × 77 × 77 ⇒ 27079.36 + 37234.12 ⇒ 64313.48𝑖𝑛2 ⇒ 64313𝑖𝑛2 ⇒ surface area of given square pyramid = 2𝑏𝑠 + 𝑏 2 b=length of base s=slent height By Putting values in formula c ∴ ⇒ 2 × 31 × 57 + 31 × 31 ⇒ 3534 + 961 ⇒ 4495𝑦𝑑 2 Question #7 What is the surface area of this cone? Use π ≈ 3.14 and round your answer to the nearest hundredth. What is the surface area of this cylinder? Use π ≈ 3.14 and round your answer to the nearest hundred Answer: (1) Surface Area= π𝑅(𝑙 + 𝑅) Given: R=2mm l=3mm A=(3.14)(2)(3+2) = 3.14 × 2 × 5 = 3.14 × 10 = 31.4𝑚𝑚2 𝐴 = 31.4𝑚𝑚2 Question #8 What is the volume of this triangular pyramid? What is the volume of this cylinder? Use π ≈ 3.14 and round your answer to the nearest hundredth. Answer: In this question we will use basic of geometry i.e formula of volume of triangular pyramid and volume of cylinder. 1 We will use volume of triangular pyramid= *Area of base *Height 3 Volume of cylinder = π𝑅2 ℎ Solution of the above problem is in the step below. (a) As we know the formula of Volume of triangular pyramid So, 1 Volume of pyramid = Area of base × Height 3 As we can observe that 1 1 Area of base = × 𝐵 × ℎ = × 15 × 10 = 75𝑚2 2 2 Height of triangular pyramid =8m Hence, 1 Volume of pyramid = × 75 × 8 = 200𝑚3 3 Volume of triangular pyramid = 200𝑚3 Question #9 Analytic Geometry: Trace the curve 𝑦 2 = (𝑥 − 1)(2𝑥 + 3)(3𝑥 − 1). Answer: Step 1 To trace the curve 𝑦 2 = (𝑥 − 1)(2𝑥 + 3)(3𝑥 − 1) Step 2 Consider the function 𝑓(𝑥) = (𝑥 − 1)(2𝑥 + 3)(3𝑥 − 1) has three real roots namely 3 1 𝑥 = 1, − , 2 3 The graph of the function 𝑓(𝑥) = (𝑥 − 1)(2𝑥 + 3)(3𝑥 − 1) is given below: Step 3 Now for the function 𝑦 2 = (𝑥 − 1)(2𝑥 + 3)(3𝑥 − 1) the function 𝑓(𝑥) = (𝑥 − 1)(2𝑥 + 3)(3𝑥 − 1) should be greater than or equal to zero otherwise the function 𝑦 2 = (𝑥 − 1)(2𝑥 + 3)(3𝑥 − 1) gets undefined. Now the graph of 𝑦 2 = (𝑥 − 1)(2𝑥 + 3)(3𝑥 − 1) will be symmetric about the x-axis Now the trace of the function 𝑦 2 = (𝑥 − 1)(2𝑥 + 3)(3𝑥 − 1) is given below: Question #10 Analytic Geometry: Trace the curve of 𝑥 3 − 6𝑥 2 + 11𝑥 − 𝑦 = 6 Answer: 𝑥 3 − 6𝑥 2 + 11𝑥 − 𝑦 = 6 𝑦 = 𝑥 3 − 6𝑥 2 + 11𝑥 − 6 𝑃(𝑥) = 𝑥 3 − 6𝑥 2 + 11𝑥 − 6 𝑃(𝑧) = 1 − 6 + 11 − 6 = 0 (x-l) is a talov of p(x) Diviwng p(x) by (x-l) we get 2(𝑥) = 𝑥 2 − 3𝑥 − 2𝑥 + 6 2(𝑥) = 𝑥(𝑥 − 3) − 2(𝑥 − 3) 2(𝑥) = (𝑥 − 2)(𝑥 − 3) 𝑝(𝑥) = (𝑥 − 1)2(𝑥) 𝑝(𝑥) = (𝑥 − 1)(𝑥 − 2)(𝑥 − 3) Zowes of polynaval are x=1,2,3 Question #11 What is the circumference of the circle? Use 22 7 for x. a) 44 cm b) 66 cm c) 182 cm d) 154 cm Answer: Step 1 Given: Similarto circle sphere is a two dimensional space where the set of points that are at the same distance r from agiven point in a three dimensional space. In analytical geometry with a center and radius is the locus of all points is called sphere Step 2 Given: Step 3 Circumference The distance around a circle is called its circumference. The circumference of the circle is 𝐶 = 2π𝑟 22 Taking π ≈ 7 Step 4 The object is to find the circumference for the given circle as follows, The radius of the circle is r=21 Step 5 Then the circumference of the circle is, 𝐶 = 2π𝑟 C = 2( 22 )(21) 7 C = 132in . So the circumference of the circle is approximately 132 in. Question #12 What is the volume of this cone? Use π ≈ 3.14 and round your answer to the nearest hundredth. What is the volume of this cylinder? Use π ≈ 3.14 and round your answer to the nearest hundredth. Answer: 1 Volume of the cone 𝑣 = π𝑟 2 ℎ 3 r=13 f h=16 f 1 × (3.14) × (13)2 × (16) 3 v=2830.19 cubic ft volume of the cylinder = π𝑟 2 ℎ 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 8𝑐𝑚 𝑟= = = 4𝑐𝑚 2 2 h=3.3cm 𝑣 = (3.14)(4)2 (3.3) =165.79 cubic cm 𝑣= Question #13 What is the Volume of this sphere? Use π ≈ 3.14 and round your answer to the nearest hundredth, Maple syrup in a bottle has a mass of 918 grams and a density of 1.36 grams per cubic centimeter. What is its volume? Answer: Soin given 4 Volume of sphere = π𝑟 3 4 = × 3.14 × (5)3 3 3 4 = × 3.14 × 125 = 3 1570 3 = 523.33𝑚𝑚3 𝑎𝑚 Soln given Mass=918 gram density = 1.36𝑔𝑟𝑎𝑚/𝑐𝑚3 We know thed 𝑚𝑎𝑠𝑠 Volyme = 𝑣= 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 918𝑔𝑟𝑎𝑚 1.36𝑔𝑟𝑎𝑚/𝑐𝑚3 3 𝑣 = 675𝑐𝑚 ans Question #14 An inverted cone has a height of 15 mm and a radius of 16mm. The volume of the inverted cone is decreasing a rate of 534 cubic mm per second, with the height begin held constant. What is the rate of change of the radius, in mm per second, when the radius is 6 mm? Round your answer to the nearest hundreth. (Do not include any units in your answer Remember that the volume of a cone is = 𝑉 = 1/3π𝑟 2 ℎ Answer: Step 1 The volume of a cone is = 𝑉 = 1/3\𝑝𝑖𝑟 2 ℎ The volume of the inverted cone is decreasing at a rate of 534 cubic mm per second, with the height begin held constant. = 𝑑𝑉/𝑑𝑡 = 1/3πℎ. 𝑑(𝑟 2 )/𝑑𝑡 as h is held constant and hence = 𝑑ℎ/𝑑𝑡 = 0 Step 2 Find the formula for the rate of change of the volume, in mm per second. 𝑑𝑉 1 𝑑 = = πℎ (𝑟 2 ) 𝑑𝑡 1 3 = πℎ(2𝑟) 3 2 𝑑 3 𝑑𝑡 = π𝑟ℎ 𝑑𝑡 𝑑 𝑑𝑡 (𝑟) (𝑟) Step 3 To find the rate of changt of the radius, in mm per second, when the radius is 6 mm. Here =dV/dt=534 cubic mm per second, =h=15mm and =r=6mm . 𝑑𝑟 2 𝑑 = = π𝑟ℎ (𝑟) 𝑑𝑡 =⇒ = 3 𝑑𝑟 𝑑𝑡 3 = 3 𝑑𝑡 𝑑𝑉 2π𝑟ℎ 𝑑𝑡 2π×6×15 (𝑟) × 534 = 2.8329 Question #15 Find the points on the ellipse 4𝑥 2 + 𝑦 2 = 4 that are the farthest from the point (1;0) Answer: Let (x;y) be the points on the ellipse 4𝑥 2 + 𝑦 2 = 4 Then, 4𝑥 2 + 𝑦 2 = 4 ⇒ 𝑦 2 = 4 − 4𝑥 2 𝑟𝑖𝑔ℎ𝑡 ⇒ 𝑦 = ±2√1 − 𝑥 2 𝑑(𝑥) = √(𝑥 − 1)2 + 𝑦 2 If 𝑦 2 = 4 − 4𝑥 2 , then we have: 𝑑(𝑥) = √(𝑥 − 1)2 + 4 − 4𝑥 2 √−3𝑥 2 − 2𝑥 + 5 Then maximise 𝑓(𝑥) = −3𝑥 2 − 2𝑥 + 5 1 𝑓 ′ (𝑥) = −6𝑥 − 2 = 0 ⇒ − (the only critical value) 𝑓 ′′ (𝑥) = −6 ⇒ 𝑥 = − 3 1 3 1 2 Since 𝑦 = ±2√1 − (− ) = ± 4√2 3 1 3 4√2 3 3 The farthest points are (− ; ± ) Question #16 Let v=zk be the velocity field (in meters per second) of a fluid in R3. Calculate the flow rate (in cubic meters per second) through the upper hemispere (z>0) of the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 1 Answer: Step 1 Similar to circle sphere is a two dimensional space where the set of points that are at the same distance r from a given point in a three dimensional space. In analytical geometry with a center and radius is the locus of all points is called sphere Given: The upper hemisphere of the sphere is 𝑥 2 + 𝑦 2 + 𝑧 2 = 1 Formula used: Volume = ∫ ∫ ∫(𝑑𝑖𝑣𝐹)𝑑𝑣 Step 2 V=(0, 0, z) 𝑑 ÷, 𝐹 = (0) + 𝑑𝑥 𝑑 𝑑𝑦 (0) + 𝑑 𝑑𝑧 (𝑧) On solving the value, ÷ F=0+0+1=1 Step 3 Volume = ∫ ∫ ∫(÷ 𝐹)𝑑𝑣 Where, 4 𝑑𝑣 = π𝑟 3 3 𝑥 2 + 𝑦2 + 𝑧2 = 1 = 𝑟2 Where, r=1 4 π𝑟 2 3 Substituting the value of r, then the required value is Question #17 The lesson is analytic geometry standartize the conic (7𝑥 2 + 4𝑥𝑦 + 4𝑦 2 − 2𝑥 − 6𝑦 + 1 = 0) Answer: Step 1 Comparing the equation with standard form 𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0 ⇒ 𝑎 = 7, ℎ = 2, 𝑏 = 4, 𝑔 = −1, 𝑓 = −3, 𝑐 = 1 △= 𝑎𝑏𝑐 + 2𝑓𝑔ℎ − 𝑎𝑓 2 − 𝑏𝑔2 − 𝑐ℎ2 △= 28 + 2(−3)(−1)(2) − 7(−3)2 − 4(−1)2 − 1(2)2 △= 40 − 63 − 4 − 2 ≠ 0 Now ℎ2 − 𝑎𝑏 = (2)2 − 7 × 4 4π 3 -24<0 and 𝑎 ≠ 𝑏 So, curve represents an Elipse 1) 7𝑥 2 + 4𝑥𝑦 + 4𝑦 2 − 2𝑥 − 6𝑦 + 1 = 0 Now find the angle by which the axes are rotated 𝑎−𝑏 So, cot 2 θ = 7−4 cot 2 θ = 2ℎ 3 = 4 So, cos 2 θ = 3 4 5 Step 2 Now, sin θ = √ So, cos θ = √ 1−cos 2θ 2 1+cos 2θ 2 1 = = √5 2 √5 Substitute 𝑠𝑖𝑛 θ of 𝑐𝑜𝑠 θ in rotation formulae 𝑥 = 𝑥 ′ cos θ − 𝑦 ′ sin θ 𝑎𝑛𝑑 𝑦 = 𝑥 ′ sin θ + 𝑦 ′ 𝑐𝑜𝑠 θ 2𝑥 ′ − 𝑦 ′ 𝑥 ′ + 2𝑦 ′ 𝑥= 𝑜𝑓 𝑦 = √5 √5 Substituting x of y in eqn (1) 2 2 (2𝑥 ′ − 𝑦 ′ )(𝑥; +2𝑦 ′ ) 2𝑥 ′ − 𝑦 ′ 𝑥 ′ + 2𝑦 ′ 7( ) +4 + 4( ) 5 √5 √5 2𝑥 ′ − 𝑦 ′ 𝑥 ′ + 2𝑦 ′ = −2 ( )− 6( )+1=0 √5 √5 ⇒ 7(4𝑥 ′2 + 𝑦 ′2 − 4𝑥 ′ 𝑦 ′ ) + 4(2𝑥;2 + 4𝑥 ′ 𝑦 ′ − 𝑥 ′ 𝑦 ′ − 2𝑦 ′ )2 + 4(𝑥 ′2 + 4𝑦 ′2 + 4𝑥 ′ 𝑦 ′ ) = −2√5(2𝑥 ′ − 𝑦 ′ ) − 6√5(𝑥 ′ + 2𝑦 ′ ) + 5 = 0 ⇒ 40𝑥 ′2 + 15𝑦 ′2 − 10√5𝑥 ′ − 10√5𝑦 ′ + 5 = 0 8𝑥 ′2 + 3𝑦 ′2 − 2√5𝑥 ′ − 2√5𝑦 ′ + 1 = 0 8 (𝑥 ′2 − √5 4 𝑥 ′ ) + 3 (𝑦 ′2 − 2√5 3 2 2 2 2 2 2 𝑦′) + 1 = 0 5 5 √5 √5 8 (𝑥 − ) + 3 (𝑦 ′ − ) = −1 + 8 ( ) + 3 ( ) 8 3 64 9 ′ 2 5 √5 √5 8 (𝑥 − ) + 3 (𝑦 ′ − ) = + 8 3 3 8 ′ Step 3 31 √5 √5 8 (𝑥 − ) + 3 (𝑦 ′ − ) = 8 3 24 ′ √5 (𝑥 ′ − 8 ) 2 31 ) 192 ( √5 + (𝑦 ′ − 3 ) 2 31 ) 72 ( It is of term =1 (𝑥−ℎ)2 𝐴2 + (𝑦−𝑘)2 𝐵2 = 1 Ellipse Where x' of y' are axes after rotation through angle \theta when tan θ = Question #18 Find the Amplitude, Midline, Midline, period and Asymptotes for 1 𝑓(𝑥) = − cot (𝑥) + 2 3 Answer: Step 1 1 2 𝑥 𝑓(𝑥) = − cot ( ) + 2 2 𝑓(𝑥) = 𝐴 cot (𝐵𝑥) + 𝐷 Midline is y=2 line the vertical shift of 2 units up. A=-1 Amplitude =|-A|=|-1|=1 Amplitude =1 Step 2 𝐵= 1 2 period = = 2π 2π 𝐵 1 2 Period = 4π In analytic geometry, an asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity. 𝑥 Now, finding values of x for which cot ( ) goes to infinity; 𝑥 2 2 = 𝑛 π where, 𝑛 ∈ 𝑍 Asymptotes: 𝑥 = 2𝑛π where 𝑛 ∈ 𝑍 Question #19 The Laplace equation ∂2 ∂𝑥 2 𝑓+ ∂2 ∂𝑦 2 𝑓=0 Answer: The steady sate temparature distribution is a one diamensional slab of thermal conductivity 50 thickness 50 mm is found to be 𝑇 = 𝑎 + 𝑏𝑥 2 Where 𝑎 = 2000𝑐, 𝑏 = 20000 𝑐 𝑚2 𝑤 𝑚 × 𝑘 and . T is in degrees celsius and x in meters. Given: 𝑓(𝑥, 𝑦) = ln 𝑥 2 + 𝑦 2 Now, ∂𝑓 1 2𝑥 𝑥 = × = 2 2 2 2 2 ∂𝑥 √𝑥 + 𝑦 𝑥 + 𝑦2 2√𝑥 + 𝑦 ∂2 𝑓 𝑥 2 + 𝑦 2 − 2𝑥 2 𝑦2 − 𝑥 2 = = (𝑥 2 + 𝑦 2 )2 (𝑥 2 + 𝑦 2 )2 ∂𝑥 2 ∂𝑓 1 2𝑦 𝑦 𝑦 = × = = ∂𝑦 √𝑥 2 + 𝑦 2 2√𝑥 2 + 𝑦 2 (√𝑥 2 + 𝑦 2 )2 𝑥 2 + 𝑦 2 ∂2 𝑓 𝑥 2 + 𝑦 2 − 2𝑦 2 𝑥 2 − 𝑦2 = = (𝑥 2 + 𝑦 2 )2 (𝑥 2 + 𝑦 2 )2 ∂𝑦 2 2 f 2 f y 2 − x2 x2 − y 2  2 + 2 = 2 + x y ( x + y 2 )2 ( x 2 + y 2 ) = 𝑦 2 −𝑥 2 +𝑥 2 −𝑦 2 (𝑥 2 +𝑦 2 )2 2 =0 Therefore, 𝑓(𝑥, 𝑦) = 𝑙𝑚√𝑥 2 + 𝑦 2 is the solution of the equation ∂2 𝑓 ∂𝑥 2 + ∂2 𝑓 ∂𝑦 2 =0 Question #20 Let E2 be the new system {𝑂, 𝑥 ′ , 𝑦 ′ } obtained by rotating the (O, x, y} perpendicular coordinate system arounf the point O in the Euclidean plane. Accordingly, the point {𝑂, Whic of the following is the coordinate according to the, 𝑥 ′ , 𝑦 ′ } system? Analytic geometry. Answer: Step 1 Given that 𝐸2 be the new system {𝑂, 𝑥 ′ , 𝑦 ′ } obtained by rotating the old system {𝑂, 𝑥, 𝑦} Which gives perpendicular coordinate system around point O in the euclidean plane Step 2 If the axes of old co-ordinate be rotates through an angle O in anti-clockwise direction keeping the origin O fixed. Then the new coordinate in terms of old co-ordinate can be written as x = x cos  + y sin  y = − x sin  + y cos  For the given problem we have θ = 90∘ So, 𝑥 ′ = 𝑥 cos 9 0∘ + 𝑦 sin 9 0∘ = 𝑦 𝑦 ′ = −𝑥 sin 9 0∘ + 𝑦 cos 9 0∘ = −𝑥 So, the new co-ordinate system is {𝑂, 𝑦, −𝑥}