Conic sections #1 - Questions and Answers

Question 1 The current in a wire varies over time according to the relation I = 55A - (0.65A/s²)t². How many electric charges 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 flowing through a cross section in the time interval is given by q = ∫[t1,t2] It dt Substitute (55A - (0.65A/s²)t²) for I, 0s for t₁, and 8.5s for t₂ to find the charge. q = ∫[0s,8.5s] (55A - (0.65A/s²)t²)dt = [55tA - (0.65t³/3)A/s² ] [0s,8.5s] = [55(8.5) - (0.65(8.5)³/3) ]C = 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 I = 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 2 How are quadric surfaces and conic sections related? Answer The quadratic surfaces are three dimensional analogs of two dimensional conic sections. Quadratic surfaces is a term used in three dimension for surfaces which are analogs of conic sections. Question 3 The solutions of nonlinear systems of equations are points that have intersection of ? tapered sections. Answer Solution of nonlinear system of equations are the points of intersection of the curve of conic sections. The linear system is one where each equation is linear, it means that each variable in the equation has a power of 1. The graph of the linear equation is a straight line. For example 2x + 3y = 4 is linear equation. A system of two or more equations having two or more variables containing at least one equation that is not linear is known as nonlinear equations system. For example xy = 4 is a nonlinear equation because it cannot be written in the form of Ax + By + C = 0. The graph of nonlinear systems is generally have conic curves where intersection point of the curves are the solution of the system. Therefore, solutions of the nonlinear systems of equations are the points of intersection of the curve of conic sections. Question 4 For Exercise, a. Identify the equation as representing a circle, an ellipse, a hyperbola, or a parabola. b. Graph the curve. c. Identify key features of the graph. That is. If the equation represents a circle, identify the center and radius. If the equation represents an ellipse, identify the center, vertices, endpoints of the minor axis, foci, and eccentricity. If the equation represents a hyperbola, identify the center, vertices, foci, equations of the asymptotes, and eccentricity. If the equation represents a parabola, identify the vertex, focus, endpoints of the latus rectum, equation of the directrix, and equation of the axis of symmetry. x² + y² - 4x - 6y + 1 = 0 Answer Step 1 a) For the given equation simplify it and compare with standard forms of conic sections to determine which conic section or a figure it represents as. x² + y² - 4x - 6y + 1 = 0 x² - 4x + 4 - 4 + y² - 6y + 9 - 9 + 1 = 0 (x² - 4x + 4) - (y² + 6y + 9) + 1 + 9 - 4 = 0 (x - 2)² - (y + 3)² + 6 = 0 (y + 3)² - (x - 2)² = 6 (y+3)² / 6 - (x-2)² / 6 = 1 Which is a hyperbola. Step 2 Now to find the features of the given hyperbola there are standard forms where these features can be determined. Let us determine centre 'C' of this hyperbola by comparing with standard form of such hyperbola as (y-k)² / a² - (x-h)² / b² = 1 (y+3)² / 6 - (x-2)² / 6 = 1 C = (h, k) here C = (2, -3) Step 3 Vertices "V" for this hyperbola can be determined to be as. V = (h, k ± a) a² = 6 a= √6 V = (2, -3 ± √6) Step 4 Foci "F" for this hyperbola can be determined to be as F = (h, k ± c) c² = a² + b² c² = 6 + 6 c = √12 F = (2, -3 ± √12) Step 5 Equation of asymptotes for this hyperbola can be determined to be as. y = ± b/a (x - h) + k y = ± √6 / √6 (x - 2) - 3 y = ± (x - 2) - 3 Step 6 Eccentricity "e" is calculated by the formula as. e = √(a² + b²) / a = √ (6 + 6) / √6 = √12 / √6 = √2 √6 / √6 = √2 Question 5 The circle, ellipse, hyperbola, and parabola are examples of conic sections. Their quation contains x² terms, y² terms, or both. When these terms both appear, are on the same side, and have different coefficients with same signs, the equation is that of an ellipse. Answer The circle, ellipse, hyperbola, and parabola are made up of cone. Therefore the circle, ellipse, hyperbola, and parabola are examples of conic sections. Their equation contains x² terms, y² terms, or both. We define ellipse as follows. There are two given points, the foci, A ellipse is the locus of points such that the sum between the distances to each focus is constant. The standard equation of ellipse is x²/a² + y²/b² = 1 For example: x²/4 + y²/1 = 1 When these terms both appear, are on the same side, and have different coefficients with same signs, the equation is that of an ellipse. Conclusion: Ellipse is conic section that has x² terms, y² terms and they have same sign.