Presentation on theme: "Slide 5- 1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley."— Presentation transcript:

Slide 5- 2 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Analytic Geometry Topics Chapter 9

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 9.1 The Parabola  Given an equation of a parabola, complete the square, if necessary, and then find the vertex, the focus, and the directrix and graph the parabola.

Slide 5- 4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Parabolas A parabola is the set of all points in a plane equidistant from a fixed line (the directrix) and a fixed point not on the line (the focus). The line that is perpendicular to the directrix and contains the focus is the axis of symmetry. The vertex is the midpoint of the segment between the focus and the directrix. Directrix Axis of symmetry Vertex Focus

Slide 5- 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Standard Equation of a Parabola with Vertex at the Origin The standard equation of a parabola with vertex (0, 0) and directrix y =  p is x 2 = 4py. The focus is (0, p) and the y-axis is the axis of symmetry. The standard equation of a parabola with vertex (0, 0) and directrix x =  p is y 2 = 4px. The focus is (p, 0) and the x-axis is the axis of symmetry.

Slide 5- 6 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find the focus, and the directrix of the parabola Then graph the parabola. Solution: We write in the form x 2 = 4py: Thus, p = 4, so the focus is (0, p), or (0, 4). The directrix is y =  p =  (4).

Slide 5- 7 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find an equation of the parabola with vertex (0, 0) and focus (3, 0). Then graph the parabola. The focus is on the x-axis so the line of symmetry is the x-axis. Thus the equation is of the type y 2 = 4px. Since the focus (3, 0) is 3 units to the right of the vertex, p = 3 and the equation is y 2 = 4(3)x, or y 2 = 12x.

Slide 5- 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Standard Equation of a Parabola with Vertex (h, k) and Vertical Axis of Symmetry

Slide 5- 9 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Standard Equation of a Parabola with Vertex (h, k) and Horizontal Axis of Symmetry

Slide 5- 10 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the parabola x 2  6x  y + 11 = 0, find the vertex, the focus, and the directrix. Then draw the graph. Solution: We first complete the square:

Slide 5- 11 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued We see that h = 3, k = 2 and p = ¼, so we have the following: Vertex (h, k): (3, 2) Focus (h, k + p): (3, 2 + ¼) or (3, 9/4) Directrix y = k  p: y = 2  ¼, or y = 7/4

Slide 5- 12 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the parabola y 2  6y  x + 4 = 0, find the vertex, the focus, and the directrix. Then draw the graph. Solution: We first complete the square:

Slide 5- 13 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued We see that h =  5, k = 3 and p = ¼, so we have the following: Vertex (h, k): (  5, 3) Focus (h + p, k): (  5 + ¼, 3) or (  19/4, 3) Directrix x = h  p: x =  5  ¼, or y =  21/4

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 9.2 The Circle and the Ellipse  Given an equation of a circle, complete the square, if necessary, and then find the center and the radius and graph the circle.  Given the equation of an ellipse, complete the square, if necessary, and then find the center, the vertices, and the foci and graph the ellipse.

Slide 5- 15 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Circle A circle is the set of all points in a plane that are at a fixed distance from a fixed point (the center) in the plane. Standard Equation of a Circle The standard equation of a circle with center (h, k) and radius r is (x  h) 2 + (y  k) 2 = r 2.

Slide 5- 16 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the circle x 2 + y 2 + 2x + 6y  15 = 0 find the center and the radius. Then graph the circle. Solution: First, we complete the square twice: Center = (  1,  3) Radius = 5

Slide 5- 17 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Ellipse An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points (the foci) is constant. The center of an ellipse is the midpoint of the segment between the foci. The major axis of an ellipse is longer than the minor axis.

Slide 5- 18 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Standard Equation of an Ellipse with Center at the Origin (Major Axis Horizontal)

Slide 5- 19 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Standard Equation of an Ellipse with Center at the Origin (Major Axis Vertical)

Slide 5- 20 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find the standard equation of the ellipse with vertices (  6, 0) and (6, 0) and foci (  4, 0) and (4, 0). Then graph the ellipse. Since the foci are on the x-axis and the origin is the midpoint of the segment between them, the major axis is horizontal and (0, 0) is the center of the ellipse. Thus the equation is of the form

Slide 5- 21 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Since the vertices are (  6, 0) and (6, 0) and the foci are (  4, 0) and (4, 0), we know that a = 6 and c = 4. These values can be used to find b 2. Thus the equation of the ellipse is To graph, we plot the vertices, and the y-intercepts are (0,  4.47) and (0, 4.47). We plot these points and connect the four points with a smooth curve.

Slide 5- 22 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the ellipse find the vertices and the foci. Then draw the graph. a = 7 and b = 6. The major axis is vertical. Vertices: (0, 7) and (0,  7)

Slide 5- 23 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued The foci are (0, 3.61) and (0,  3.61). The x-intercepts are (  6, 0) and (6, 0). We plot the vertices and the x-intercepts and connect the four points with a smooth curve.

Slide 5- 24 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the ellipse find the center, the vertices, and the foci. Then draw the graph. First, we complete the square twice to get standard form.

Slide 5- 25 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Center is (  3, 1) a = 3 and b = 2 The major axis is vertical so the vertices are 3 units above and below the center. (  3, 1 + 3) (  3, 1  3) or (  3, 4) and (  3,  2). We know that c 2 = a 2  b 2, so c 2 = 3 2  2 2 = 5 and c =. Then the foci are 2.24 units above and below the center:

Slide 5- 26 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued To graph, we plot the vertices. Note also that since b = 2, two other points on the graph are the endpoints of the minor axis, 2 units right and left of center.

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 9.3 The Hyperbola  Given an equation of a hyperbola, complete the square, if necessary, and then find the center, the vertices, and the foci and graph the hyperbola.

Slide 5- 28 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Hyperbola A hyperbola is the set of all points in a plane for which the absolute value of the difference of the distances from two fixed points (the foci) is constant. The midpoint of the segment between the foci is the center of the hyperbola.

Slide 5- 29 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Standard Equation of a Hyperbola with Center at the Origin -- Horizontal

Slide 5- 30 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Standard Equation of a Hyperbola with Center at the Origin -- Vertical

Slide 5- 31 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find an equation of the hyperbola with vertices (0,  3) and (0, 3) and foci (0,  5) and (0, 5). Solution: We know that a = 3 and c = 5. We find b 2. Since the vertices and foci are on the y-axis, we know that the transverse axis is vertical. We can now write the equation of the hyperbola:

Slide 5- 32 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the hyperbola given by 16x 2  25y 2 = 400, find the vertices, the foci, and the asymptotes. Then graph the hyperbola. First, we find standard form:

Slide 5- 33 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued The hyperbola has a horizontal transverse axis, so the vertices are (  5, 0) and (5, 0). For the standard form of the equation, we know that a 2 = 5 2, or 25, and b 2 = 4 2, or 16. We find the foci: Thus, the foci are

Slide 5- 34 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Next, we find the asymptotes: To draw the graph, sketch the asymptotes first.

Slide 5- 35 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Standard Equation of a Hyperbola with Center at (h, k) -- Horizontal

Slide 5- 36 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Standard Equation of a Hyperbola with Center at (h, k) -- Vertical

Slide 5- 37 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the hyperbola given by 9y 2  4x 2  54y  8x + 41 = 0, find the center, the vertices, and the foci. Then draw the graph. First, we complete the square to get standard form:

Slide 5- 38 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Center = (1, 3) a = 2b = 3 Vertices: (1, 3  2) or (1, 1) (1, 3 + 2) or (1, 5) We know that c 2 = a 2 + b 2, so c 2 = 4 + 9 = 13 and The foci are units below and above the center:

Slide 5- 39 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued The asymptotes are: We sketch the asymptotes, plot the vertices, and draw the graph.

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 9.4 Nonlinear Systems of Equations  Solve a nonlinear system of equations.  Use nonlinear systems of equations to solve applied problems.  Graph nonlinear systems of inequalities.

Slide 5- 41 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Nonlinear Systems of Equations The graphs of the equations in a nonlinear system of equations can have no point of intersection or one or more points of intersection. The coordinates of each point of intersection represent a solution of the system of equations. When no point of intersection exists, the system of equations has no real-number solution. We can solve nonlinear systems of equations by using the substitution or elimination method.

Slide 5- 42 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the following system of equations: We use the substitution method. First, we solve equation (2) for y.

Slide 5- 43 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Next, we substitute y = 2x  3 in equation (1) and solve for x: Now, we substitute these numbers for x in equation (2) and solve for y. y = 2x  3 y = 2(0)  3 y =  3 (0,  3) and Check each solution.

Slide 5- 44 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Check: (0,  3) Check: Visualizing the Solution

Slide 5- 45 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the following system of equations: xy = 4 3x + 2y =  10 Solve xy = 4 for y. Substitute into 3x + 2y =  10.

Slide 5- 47 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Substitute values of x to find y. 3x + 2y =  10 x =  4/3 x =  2 The solutions are (  4/3,  3) and (  2,  2). Visualizing the Solution

Slide 5- 48 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the system of equations: Solve by elimination. Multiply equation (1) by 2 and add to eliminate the y 2 term.

Slide 5- 49 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Substituting x =  1 in equation (2) gives us: The possible solutions are (1, 3), (  1, 3), (  1,  3) and (1,  3).

Slide 5- 50 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued All four pairs check, so they are the solutions. Visualizing the Solution

Slide 5- 51 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Modeling and Problem Solving Example: For a student recreation building at Southport Community College, an architect wants to lay out a rectangular piece of land that has a perimeter of 204 m and an area of 2565 m 2. Find the dimensions of the piece of land. 1. Make a drawing and label it. 2. Translate. We now have the following: Perimeter: 2l + 2w = 204 Area: lw = 2565 Area = lw = 2565 m 2 w w l l

Slide 5- 52 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Modeling and Problem Solving continued 3. Carry out. Solve the system. Solving the second equation for l gives us l = 2565  w. We then substitute into equation (1) and solve for w.

Slide 5- 53 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Modeling and Problem Solving continued If w = 57, then l = 2565  w = 2565  57 = 45. If w = 45, then l = 2565  w = 2565  45 = 57. Since the length is generally considered to be longer than the width, we have the solution l = 57 and w = 45. 4. Check. If l = 57 and w = 45, the perimeter is 2(45) + 2(57) = 204. The area is 57(45) = 2565. The numbers check. 5. State. The length of the piece of land is 57 m and the width is 45 m.

Slide 5- 54 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Nonlinear Systems of Inequalities Example: Graph the solution set of the system.

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 9.5 Rotation of Axes  Use rotation of axes to graph conic sections.  Use the discriminant to determine the type of conic represented by a given equation.

Slide 5- 56 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rotation of Axes When B is nonzero, the graph of is a conic section with an axis that is not parallel to the x– or y– axis. Rotating the axes through a positive angle  yields an coordinate system.

Slide 5- 57 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rotation of Axes Formulas If the x– and y– axes are rotated about the origin through a positive acute angle  then the coordinates of and of a point P in the and coordinates system are related by the following formulas.

Slide 5- 58 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Suppose that the axes are rotated through an angle of 45º. Write the equation in the coordinate system. Solution: Substitute 45º for  in the formulas

Slide 5- 59 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Now, substitute these into the equation This is the equation of a hyperbola in the coordinate system.

Slide 5- 60 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Eliminating the xy- Term To eliminate the xy- term from the equation select an angle  such that and use the rotation of axes formulas.

Slide 5- 61 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph the equation Solution: We have

Slide 5- 62 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Substitute these into and simplify. Parabola with vertex at (0, 0) in the coordinate system and axis of symmetry

Slide 5- 63 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Discriminant The expression is, except in degenerate cases, is the discriminant of the equation The graph of the the equation 1. an ellipse or circle if 2. a hyperbola if and 3. a parabola if

Slide 5- 64 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph the equation so Solution: We have The discriminant is negative so it’s a circle or ellipse. Determine  Substitute into the rotation formulas

Slide 5- 65 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued After substituting into the equation and simplifying we have:

Slide 5- 66 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued with vertices is an ellipse,

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 9.6 Polar Equations of Conics  Graph polar equations of conics.  Convert from polar to rectangular equations of conics.  Find polar equations of conics.

Slide 5- 68 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley An Alternative Definition of Conics Let L be a fixed line (the directrix); let F be a fixed point (the focus), not on L; and let e be a positive constant (the eccentricity). A conic is the set of all points P in the plane such that where PF is the distance from P to F and PL is the distance from P to L. The conic is a parabola if e = 1, an ellipse if e 1.

Slide 5- 69 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Polar Equations of Conics To derive equations position focus F at the pole, and the directrix L either perpendicular or parallel to the polar axis. In this figure L is perpendicular to the polar axis and p units to the right of the pole.

Slide 5- 70 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Polar Equations of Conics For an ellipse and a hyperbola, the eccentricity e is given by e = c/a, where c is the distance from the center to a focus and a is the distance from the center to a vertex. Simplified:

Slide 5- 71 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Describe and graph the conic Solution: Since e < 1, it’s an ellipse with vertical directrix 6 units to the right of the pole. The major axis lies along the polar axis. Let  = 0, and π to find vertices: (2, 0) and (6, π ). The center is (2, π ). The major axis = 8, so a = 4. Since e = c/a, then 0.5 = c/4, so c = 2 The length of the minor axis is given by b: b =

Slide 5- 72 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Sketch the graph

Slide 5- 73 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Polar Equations of Conics A polar equation of any of the four forms is a conic section. The conic is a parabola if e = 1, an ellipse if 0 1.

Slide 5- 74 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Polar Equations of Conics

Slide 5- 75 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Converting from Polar to Rectangular Equations Use the relationships between polar and rectangular coordinates. Remember:

Slide 5- 76 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Convert to a rectangular equation: Solution: We have This is the equation of a parabola.

Slide 5- 77 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Finding Polar Equations of Conics We can find the polar equation of a conic with a focus at the pole if we know its eccentricity and the equation of the directrix.

Slide 5- 78 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find a polar equation of a conic with focus at the pole, eccentricity 1/3 and directrix Solution: Choose an equation for a directrix that is a horizontal line above the polar axis and substitute e = 1/3 and p = 2

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 9.7 Parametric Equations  Graph parametric equations.  Determine an equivalent rectangular equation for parametric equations.  Determine parametric equations for a rectangular equation.  Determine the location of a moving object at a specific time.

Slide 5- 80 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing Parametric Equations We have graphed plane curves that are composed of sets of ordered pairs (x, y) in the rectangular coordinate plane. Now we discuss a way to represent plane curves in which x and y are functions of a third variable t. One method will be to construct a table in which we choose values of t and then determine the values of x and y.

Slide 5- 81 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph the curve represented by the equations The rectangular equation is

Slide 5- 82 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Parametric Equations If f and g are continuous functions of t on an interval I, then the set of ordered pairs (x, y) such that x = f(t) and y = g(t) is a plane curve. The equations x = f(t) and y = g(t) are parametric equations for the curve. The variable t is the parameter.

Slide 5- 83 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determining a Rectangular Equation for Given Parametric Equations Solve either equation for t. Then substitute that value of t into the other equation. Calculate the restrictions on the variables x and y based on the restrictions on t.

Slide 5- 84 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find a rectangular equation equivalent to Solution The rectangular equation is:

Slide 5- 85 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determining Parametric Equations for a Given Rectangular Equation Many sets of parametric equations can represent the same plane curve. In fact, there are infinitely many such equations. The most simple case is to let either x (or y ) equal t and then determine y (or x ).

Slide 5- 86 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find three sets of parametric equations for the parabola Solution

Slide 5- 87 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Applications The motion of an object that is propelled upward can be described with parametric equations. Such motion is called projectile motion. It can be shown that, neglecting air resistance, the following equations describe the path of a projectile propelled upward at an angle  with the horizontal from a height h, in feet, at an initial speed v 0, in feet per second: We can use these equations to determine the location of the object at time t, in seconds.

Slide 5- 88 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A baseball is thrown from a height of 6 ft with an initial speed of 100 ft/sec at an angle of 45º with the horizontal. a) Find parametric equations that give the position of the ball at time t, in seconds. b) Find the height of the ball after 1 sec, 2 sec and 3 sec. c) Determine how long the ball is in the air. d) Determine the horizontal distance that the ball travels. e) Find the maximum height of the ball.

Slide 5- 89 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Solution a) b)The height of the ball at time t is represented by y.

Slide 5- 90 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Solution c) The ball hits the ground when y = 0. The ball is in the air for about 4.5 sec.

Slide 5- 91 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Solution d) Substitute t = 4.5 So the horizontal distance the ball travels is 318.2 ft. e)Find the maximum value of y (vertex). So the maximum height is about 84.1 ft.

Slide 5- 92 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Applications The path of a fixed point on the circumference of a circle as it rolls along a line is called a cycloid. For example, a point on the rim of a bicycle wheel traces a cycloid curve.

Slide 5- 93 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Applications The parametric equations of a cycloid are where a is the radius of the circle that traces the curve and t is in radian measure.

Slide 5- 94 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example The graph of the cycloid described by the parametric is shown below. equations