Questions and Answers #1 Conic Sections

Questions and Answers Sheet 1 Conic Sections Question #1 Graph the lines and conic sections 𝑟 = 8/(4 + cos 𝜃) Answer: Step 1 We have to graph the lines and conic sections 𝑟 = 8/(4 + cos 𝜃) Step 2 The given conic looks like Question #2 The type of conic sections for the nondegenerate equations given below. a) 0.1𝑥 2 + 0.6𝑥 − 1.6 = 0.2𝑦 − 0.1𝑦 2 b) 2𝑥 2 − 7𝑥𝑦 = −𝑦 2 + 4𝑥 − 2𝑦 − 1 c) 8𝑥 + 2𝑦 = 𝑦 2 + 4 Answer: a) Consider the equation 0.1𝑥 2 + 0.6𝑥 − 1.6 = 0.2𝑦 − 0.1𝑦 2 This equation can be written as: 0.1𝑥 2 + 0.1𝑦 2 + 0.6𝑥 − 0.2𝑦 − 1.6 = 0. It is known that the equation is in the form 𝐴𝑥 2 + 𝐶𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0. Here, the coefficient of 𝑥 2 and 𝑦 2 are A=0.1 and C=0.1 , that is, A=C. Therefore, the equation is of a circle. Hence, the conic section of the given equation is a circle. b) Consider the equation 2𝑥 2 − 7𝑥𝑦 = −𝑦 2 + 4𝑥 − 2𝑦 − 1. This equation can be written as: 2𝑥 2 − 7𝑥𝑦 + 𝑦 2 − 4𝑥 + 2𝑦 + 1 = 0. This is in the form of 𝐴𝑥 2 + 𝐵𝑥𝑦 + 𝐶𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 Solve for 𝐵2 − 4𝐴𝐶. 𝐵2 − 4𝐴𝐶 = (7̅ − 4(2)(1)) =49-8 =41 This implies that 𝐵2 − 4𝐴𝐶 > 0 Therefore, the equation is of a hyperbola. Hence, the conic section of the given equation is a hyperbola. c) Consider the equation 8𝑥 + 2𝑦 = 𝑦 2 + 4. This equation can be written as: −𝑦 2 + 8𝑥 + 2𝑦 − 4 = 0 It is known that the equation is in the form 𝐴𝑥 2 + 𝐶𝑦 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 In the given equation, there is no term of 𝑥 2 , which means that A=0 , that is, AC=0 Therefore, the equation is of a parabola. Hence, the conic section of the given equation is a parabola. Question #3 Solve the system and graph the curves  x 2 + 3x − y + 2 = 0   y − 5x = 1 Answer: Step 1 Given:  x 2 + 3x − y + 2 = 0  (1)   (2)  y − 5x = 1 From equation (2), we get y=1+5x Now put y=1+5x in equation (1), we get ⇒ 𝑥 2 + 3𝑥 − (1 + 5𝑥) + 2 = 0 ⇒ 𝑥 2 + 3𝑥 − 1 − 5𝑥 + 2 = 0 ⇒ 𝑥 2 − 2𝑥 + 1 = 0 ⇒ (𝑥 − 1)2 = 0 ⇒𝑥−1=0 ⇒𝑥=1 Now put x=1 in equation (2), we get y-5(1)=1 ⇒𝑦−5=1 ⇒𝑦=6 Therefore the solution is, x=1 and y=6 Step 2 Question #4 Write the following equation in standard form and sketch its graph 9𝑥 2 + 72𝑥 − 64𝑦 2 + 128𝑦 + 80 = 0 Answer: Step 1 Given equation is 9𝑥 2 + 72𝑥 − 64𝑦 2 + 128𝑦 + 80 = 0 Given equation can be written as: 9𝑥 2 + 72𝑥 − 64𝑦 2 + 128𝑦 + 80 = 0 (3𝑥)2 + 2 ⋅ 3𝑥 ⋅ 12 + (12)2 − 144 − [64𝑦 2 − 128𝑦] + 80 = 0 (3𝑥 + 12)2 − (64𝑦 2 − 128𝑦 + 64 − 64) + 80 − 144 = 0 (3𝑥 + 12)2 − (8𝑦 − 8)2 = 0 Given equation is pair of straight lines. Step 2 Graph of the given equation is: Question #5 1 Graph the lines and conic sections 𝑟 = (1+2 sin 𝜃) Answer: Step 1 Consider the given polar equation, 1 𝑟= (1 + 2 sin(𝜃)) Compare the given with standard equation of the hyperbola, 𝑟= 𝑘𝑒 1+2 sin(𝜃) we get, 1 e=2 and 𝑘 = 2 so the given equation is the equation of the hyperbola. Step 2 So, the graph of the hyperbola is, Question #6 1 Graph the lines and conic sections 𝑟 = (1+cos 𝜃) Answer: Step 1 To identify the conic section, write the given equation in the form of 𝑟 = 𝑒𝑝 1+𝑒 cos(𝜃) where e is the eccentricity. 1 𝑟= = 1+cos(𝜃) 1(1) 1+(1) cos(𝜃) Therefore, e=1 and p=1. Hence, the given conic is a parabola. Step 2 1 To graph the conic substitute √𝑥 2 + 𝑦 2 for r and x for 𝑟 cos 𝜃 in the given equation 𝑟 = , and find 1+cos(𝜃) the cartesian equation of the parabola. The lines present in the graph when 𝜃 = ±𝜋, ±3𝜋, ±5𝜋,⋅ 1 𝑟= 1+cos 𝜃 𝑟(1 + cos 𝜃) = 1 𝑟 + 𝑟 cos 𝜃 = 1 √𝑥 2 √𝑥 2 + 𝑦2 𝑦2 +𝑥 =1 + = 1−𝑥 𝑥 2 + 𝑦 2 = (1 − 𝑥)2 𝑥 2 + 𝑦 − (1 − 𝑥)2 = 0 𝑥 2 + 𝑦 2 − 1 + 2𝑥 − 𝑥 2 = 0 𝑦 2 = 1 − 2𝑥 Question #7 𝜋 Graph the lines and conic sections 𝑟 = 4 sec (𝜃 + ) 6 Answer: Step 1 We first make a table for the given function  r 0   2 4.6188 −8 −4.6188 Step 2 3  2 8 2 4.6188 Then we plot the points and join using curve Question #8 Graph the lines and conic sections 𝑟 = Answer: 8 4+sin 𝜃 Step 1 To Graph: 8 𝑟= 4+sin 𝜃 Explanation: 8 𝑟= 4+sin 𝜃 when 𝜃 = 0 8 𝑟= =2 4+0 𝜋 when 𝜃 = 𝑟= 2 8 4+sin(𝜋) 8 = = 1.6 5 when 𝜃 = 𝜋 𝑟= 8 8 4+sin(𝜋) when 𝜃 = 𝑟= 8 4+sin( = =2 4 3𝜋 2 3𝜋 ) 2 = 8 4−1 when 𝜃 = 2𝜋 𝑟= 8 4+sin(2𝜋) Step 2 8 = =2 4 8 = = 2.67 3 Question #9 To identify: The conic represented by the following polar equation and graph it by using the graphing utility: 8 𝑟= 4+3 sin 𝜃 Answer: Formula used: The conics sections of polar form 𝑟 = 𝑒𝑝 1±cos 𝜃 or 𝑟 = 𝑒𝑝 1±𝑒 sin 𝜃 are said to be: Parabola if e=1 Ellipse if e<1 Hyperbola if e>1 The given polar equation is: 8 𝑟= 4+3 sin 𝜃 Compared it with the standard equation of the polar form of a conic 𝑟 = 3 Get that: 𝑒 = < 1 4 𝑒𝑝 1±𝑒 sin 𝜃 So, here the eccentricity is equal to one e<1 , so the polar equation represents ellipse. Graph for the given function is: Question #10 Instructions: Graph the conic section and make sure to label the coordinates in the graph. Give the standard form (SF) and the general form (GF) of the conic sections. CIRCLES: Center is at (2, -4); the diameter's length is 6. The endpoints of the diameter is at (-1, -4) and (3, 6). Answer: Step 1 We have to find standard form and general form of conic section and also we have to make the graph of these conic. Step 2 Centre (2, -4) and diameter's length is 6. We have given diameter =6 6 So, radius of circle (𝑟) = = 3 2 Concept: If centre of a circle is (h, k) and radius of circle is r, then equation of circle in standard form is: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 And we can find general form of circle after expanding standard form: We have centre of circle (2, -4) radius 3 So, equation of circle in standard form is: 2 (𝑥 − 2)2 + (𝑦 − (−4)) = 32 𝑜𝑟(𝑥 − 2)2 + (𝑦 + 4)2 = 32 Now, Expland (𝑥 − 2)2 + (𝑦 + 4)2 = 32 we get 𝑥 2 + 𝑦 2 − 4𝑥 + 8𝑦 + 4 + 16 = 9 𝑥 2 + 𝑦 2 − 4𝑥 + 8𝑦 + 11 = 0 So, general form of circle is: 𝑥 2 + 𝑦 2 − 4𝑥 + 8𝑦 + 11 = 0 Now, graph of this circle: Step 3 End points of the diameter are (-1, -4) and (3, 6) Centre of circle lies on the mid-point of the diameter So, centre of coordinate (𝑥, 𝑦) = ( 3−1 6−4 2 , 2 ) = (1,1) And distance between (-1, -4) and (3, 6) 2 2 = √(6 − (−4)) + (3 − (−1)) = √(6 + 4)2 + (3 + 1)2 = √100 + 16 = √116 = 2√29 Distance of diameter = 2√29 So, radius of circle (𝑟) = 2√29 2 = √29 We have centre (1, 1) and radius (𝑟) = √29 So, equation of circle in standard form is: 2 (𝑥 − 1)2 + (𝑦 − 1)2 = (√29) Now, 2 Expand (𝑥 − 1)2 + (𝑦 − 1)2 = (√29) we get 𝑥 2 + 𝑦 2 − 2𝑥 − 2𝑦 + 1 + 1 = 29 𝑥 2 + 𝑦 2 − 2𝑥 − 2𝑦 − 27 = 0 So, general form of circle is: 𝑥 2 + 𝑦 2 − 2𝑥 − 2𝑦 − 27 = 0 Graph of this function: Question #11 Give the eccentricities of conic sections with one focus at the origin along with the directrix corresponding to that focus. Find a polar equation for each conic section. e=1, y=2 Answer: Step 1 Given: e=1 and y=2 Now directrix is, y=p ⇒𝑝=2 Step 2 Therefore equation of the conic is, 𝑒𝑝 𝑟= 1+𝑒 sin 𝜃 1⋅2 ⇒𝑟= 1 + 1 sin 𝜃 2 ⇒𝑟= 1 + sin 𝜃 Question #12 Give the eccentricities of conic sections with one focus at the origin along with the directrix corresponding to that focus. Find a polar equation for each conic section. e=1/5, y=-10 Answer: Step 1 Given: e=1/5, y=-10 Step 2 For a conic with a focus at the origin, if the directrix is 𝑦 = ±𝑝 , where p is a positive real number, and the eccentricity is a positive real number e, the conic has a polar equation 𝑒𝑝 𝑟= 1±𝑒 sin 𝜃 For 𝑒 = 1/5, 𝑦 = −10 𝑟= = = 1 ×(−10) 5 1 1− sin 𝜃 5 −2 5−sin 𝜃 5 −10 5−sin 𝜃 Question #13 To identify: The conic represented by the following polar equation and graph it by using the graphing utility: −4 𝑟= −1+cos 𝜃 Answer: The conics sections of polar form 𝑟 = Parabola if e=1 Ellipse if e<1 Hyperbola if e>1 𝑒𝑝 1±𝑒 cos 𝜃 or 𝑟 = 𝑒𝑝 1±𝑒 sin 𝜃 are said to be: The given polar equation is: −4 𝑟= −1+cos 𝜃 Compared it with the standard equation of the polar form of a conic 𝑟 = 𝑒𝑝 1±𝑒 cos 𝜃 Get that: e=1 So, here the eccentricity is equal to one e=1, so the polar equation represents parabola. Hence, the given conic section (𝑟 = −4 −1+cos 𝜃 ) is the parabola. For parabola: Vertex lies at (𝑟, 𝜃) = (2, 𝜋) with a horizontal axis. That means this parabola opens on upward direction. Now, confirm by using the graphing utility as: From graph, it confirms that the graph represents a parabola. Thus the result verified successfully. Question #14 To identify: The conic represented by the following polar equation and graph it by using the graphing utility: 𝑟= 5 1−sin 𝜃 Answer: The conics sections of polar form 𝑟 = 𝑒𝑝 1±𝑒 cos 𝜃 or 𝑟 = 𝑒𝑝 1±𝑒 sin 𝜃 are said to be: Parabola if e=1 Ellipse if e<1 Hyperbola if e>1 The given polar equation is: 5 𝑟= 1−sin 𝜃 Compared it with the standard equation of the polar form of a conic 𝑟 = 𝑒𝑝 1±𝑒 sin 𝜃 Get that: e=1 So, here the eccentricity is equal to one e=1, so the polar equation represents parabola. Hence, the given conic section (𝑟 = 5 1−sin 𝜃 ) is the parabola. For parabola: Vertex lies at (𝑟, 𝜃) = (2.5, 3𝜋 2 ) with a horizontal axis. That means this parabola opens on upward direction. Now, confirm by using the graphing utility as: From graph, it confirms that the graph represents a parabola. Thus the result verified successfully. Question #15 Sketch the following conic sections: 42 − 4𝑦 + 8 − 4𝑥 2 = 0 Answer: Step 1 Given Equation: 𝑦 2 − 4𝑦 + 8 − 4𝑥 2 = 0 Converting the above equation in Standard hyperbola form standard form (𝑥−ℎ)2 𝑎2 − (𝑦−𝑘)2 𝑏2 =1 The above equation in Standard form 𝑥2 − (𝑦−2)2 4 =1 comparing with the standard form we have h=0, a=1, k=2, b=2 Step 2 The graph of the above equation is shown below Question #16 Give the eccentricities of conic sections with one focus at the origin along with the directrix corresponding to that focus. Find a polar equation for each conic section. 1 1) 𝑒 = , 𝑥 = −2 4 1 2) 𝑒 = , 𝑦 = 6 3 Answer: Step 1 Problem 1: Recall the following fact. Equation of conic with eccentricity e and x=-k corresponding to focus at origin is given by 𝑟 = 𝑘𝑒 1−𝑒 cos 𝜃 Obtain the polar equation as follows. 1 𝑒 = and k=2 4 1 1 2× 4 2 𝑟= = 1 4 − cos 𝜃 1 − cos 𝜃 4 4 2 ⇒𝑟= 4−cos 𝜃 Step 2 Problem 2: Recall the following fact. Equation of conic with eccentricity e and y=-k corresponding to focus at origin is given by 𝑟 = 𝑘𝑒 1−𝑒 sin 𝜃 Obtain the polar equation as follows. 1 𝑒 = and k=-6 3 𝑟= 1 3 −6× 1 3 1− sin 𝜃 ⇒𝑟=− = −2 3−sin 𝜃 3 6 3−sin 𝜃 Question #17 The current in a wire varies with time according to the relation I=55A?(0.65A/s2)t2 . How many coulombs of charge pass a cross section of the wire in the time interval between t=0 and t = 8.5s ? Express your answer using two significant figures. What constant current would transport the same charge in the same time interval? Express your answer using two significant figures. Answer: Step 1 The charge passing through a cross section in the time interval is given by 𝑡 𝑞 = ∫𝑡 2 𝐼𝑡𝑑 1 Substitute (55𝐴 − (0.65𝐴/𝑠 2 )𝑡 2 ) for I, 0s for 𝑡1 , and 8.5s for 𝑡2 to find the charge. 8.5𝑠 𝑞=∫ (55𝐴 − (0.65𝐴/𝑠 2 )𝑡 2 )𝑑𝑡 0𝑠 = [55𝑡𝐴]8.5𝑠 0𝑠 − [ 0.65𝑡 3 2 8.5𝑠 ] 0𝑠 =467.5C-133.1C =334.1C The amount of charge that pass a cross section in the given time interval is 334.1C. Step 2 The constant current is given by the equation 𝑄 𝐼= △𝑡 𝑄 = 𝑡2 −𝑡1 Substitute 334.1 C for Q and 8.5s for 𝑡2 and 0s for 𝑡1 to find I. 𝐼= 334.1𝐶 8.5𝑠−0𝑠 =39.3A . The constant current that would transport the charge is 39.3A. Answer: The amount of charge that pass a cross section in the given time interval is 334.1C. The constant current that would transport the charge is 39.3A. Question #18 Determine the current flowing through an element if the charge flow is given by a) 𝑞(𝑡) = (3𝑡 + 8)𝑚𝐶 b) 𝑞(𝑡) = (8𝑡 2 + 4𝑡 − 2)𝐶 c) 𝑞(𝑡) = (3𝑒 −1 − 5𝑒 −2𝑡 )𝑛𝐶 d) 𝑞(𝑡) = 10 sin 1 20𝜋𝑡𝑝𝐶 e) 𝑞(𝑡) = 20𝑒 −4𝑡 cos 5 0𝑡𝜇𝐶 Answer: Step 1 The Current Flow through an element is given by: 𝑖= 𝑑𝑞 𝑑𝑡 Step 2 Solving for a: 𝑞(𝑡) = (3𝑡 + 8)𝑚𝐶 𝑖= 𝑑 𝑑𝑡 (3𝑡 + 8) 𝑖 = 3𝑚𝐴 Step 3 Solving for b: 𝑞(𝑡) = (8𝑡 2 + 4 − 2)𝐶 𝑖= 𝑑 𝑑𝑡 (8𝑡 2 + 4𝑡 − 2) 𝑖 = (16𝑡 + 4)𝐴 Step 4 Solving for c: 𝑞(𝑡) = (3𝑒 −𝑡 − 5𝑒 −2𝑡 )𝑛𝐶 𝑑 𝑖 = (3𝑒 −𝑡 − 5𝑒 −2𝑡 ) 𝑑𝑡 𝑖 = (−3𝑒 −𝑡 + 10𝑒 −2𝑡 )𝑛𝐶 Step 5 Solving for d: 𝑞(𝑡) = (10 sin 1 20𝜋𝑡)𝑝𝐶 𝑖= 𝑑 𝑑𝑡 (10 sin 1 20𝜋𝑡) 𝑖 = (10 cos 1 20𝜋𝑡)(120𝜋) 𝑖 = (1200𝜋 cos 1 20𝜋𝑡)𝑝𝐶 Step 6 Solving for d: 𝑞(𝑡) = (20𝑒 −4𝑡 cos 5 0𝑡)𝜇𝐶 𝑖= 𝑑 𝑑𝑡 (20𝑒 −4𝑡 cos 5 0𝑡) 𝑖 = (50)(20𝑒 −4𝑡 (𝑠𝑖𝑛)50𝑡) + (−80𝑒 −4𝑡 cos 5 0𝑡) 𝑖 = (−1000𝑒^{−4𝑡 sin 50𝑡 − 80𝑒^{−4𝑡} cos 50𝑡)𝜇𝐶 Question #19 Find an equation of the tangent line to the curve at the given point. 𝑦 = 𝑥3 − 3𝑥 + 1, (2,3) Answer: Evaluate the derivative using the limit definition, because the easy way is in a later section. 𝑓(𝑥) = 𝑥 3 − 3𝑥 + 1, (2,3) 𝑓`(𝑎) = lim = lim ℎ⇒0 = lim ℎ⇒0 = lim 𝑓(𝑎+ℎ)−𝑓(𝑎) ℎ ℎ⇒0 (𝑎+ℎ)3 −3(𝑎+ℎ)+1−(𝑎3 −3𝑎+1) ℎ 𝑎3 +3ℎ𝑎2 +3ℎ2 𝑎+ℎ3 −3𝑎−3ℎ+1−𝑎3 +3𝑎−1 ℎ 3ℎ𝑎2 +3ℎ2 𝑎+ℎ3 −3ℎ ℎ⇒0 ℎ = lim (3𝑎2 + 3ℎ𝑎 + ℎ2 − 3) ℎ⇒0 = 3𝑎2 + 0 + 0 − 3 = 3𝑎2 − 3 Evaluate the derivative at a=2 𝑓`(2) = 3(2)2 − 3 = 9 f`(x) is the slope of the tangent line to f(x) at x, so we have m=9 at point (2, 3). Plug into the point-slope line equation 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) 𝑦 = 𝑚(𝑥 − 𝑥1 ) + 𝑦1 = 9(𝑥 − 2) + 3 = 9𝑥 − 18 + 3 = 9𝑥 − 15 Result = 9𝑥 − 15 Question #20 At one point in a pipeline the water’s speed is 3.00 m/s and the gauge pressure is 5.00 \times 104 𝑃𝑎. Find the gauge pressure at a second point in the line, 11.0 m lower than the first, if the pipe diameter at the second point is twice that at the first. Answer: Step 1 The diameters of the two pipe cross-sections are 𝑑1 and 𝑑2 = 2𝑑1 . The area of the first is 𝐴1 = 𝜋𝑑12 and of the the second 𝐴2 = 𝜋𝑑22 = 𝜋4𝑑12 = 4𝐴1 Step 2 Let's first determine the relation between the velocities based on the continuity equation 𝑣1 𝐴1 = 𝑣2 𝐴2 and the fact that 𝐴2 = 4𝐴1 from which we can write 𝑣1 𝐴1 𝑣2 = 𝐴2 𝑣1 𝑣2 = 4 Step 3 Since the atmospheric pressures are the same we can write Bernoulli's equation as 1 1 𝑝1 + 𝜌𝑔𝑦1 + 𝜌𝑣12 = 𝑝2 + 𝜌𝑔𝑦2 + 𝜌𝑣22 where 𝑝1 and 𝑝2 are gauge pressures. We can now insert 𝑣2 = 2 2 𝑣1 /4 to get 1 𝑝2 = 𝑝1 + 𝜌𝑔(𝑦1 − 𝑦2 ) + 𝜌(𝑣12 − 𝑣12 /16) 𝑝2 = 𝑝1 + 𝜌𝑔(𝑦1 − 𝑦2 ) + 2 15 32 𝜌𝑣12 Step 4 At this point we can insert all the given values to get the following 15 𝑝2 = 5 × 104 + 103 × 9.81 × 11 + 103 × × 32 32 𝑝2 = 162𝑘𝑃𝑎