14.1 PARAMETRIC EQUATIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Programming a Robot Using a Parameter Example 1 If t is time in minutes, describe the path followed by a robot given by x = 2t, y = t for 0 ≤ t ≤ 5. Solution At time t = 0, the robot’s position is given by x = 2 · 0 = 0, y = 0, so it starts at the point (0, 0). One minute later, at t = 1, its position is given by x = 2 · 1 = 2, y = 1, so it has moved to the point (2, 1). At time t = 2, its position is given by x = 2 · 2 = 4, y = 2, so it has moved to the point (4, 2). Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally The path followed by the robot is given by (0, 0) → (2, 1) → (4, 2) → (6, 3) → (8, 4) → (10, 5). At time t = 5, the robot stops at the point (10, 5) because we have restricted the values of t to the interval 0 ≤ t ≤ 5. In the figure we see the path followed by the robot; it is a straight line. Robot stops here starts here t=0 t=1 t=2 t=3 t=4 t=5

Programming a Robot Using a Parameter Example 2 (a) A robot begins at the point (1, 0) and follows the path given by the equations x = cos t, y = sin t where t is in minutes, 0 ≤ t ≤ 6. (a)Describe the path followed by the robot. Solution (a) At time t = 0, the robot’s position is given by x = cos 0 = 1, y = sin 0 = 0, so it starts at the point (1, 0). One minute later, at t = 1, its position is given by (cos1,sin1) ≈ (.54,.84) Replacing t by 2, 3, 4, 5, and 6, the path followed by the robot is (1,0) → (0.54,0.84) → (−0.42,0.91) → (−0.99,0.14) → (−0.65,−0.76) → (0.28,−0.96) → (0.96,−0.28) Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally In the figure we see the path followed by the robot; it is circular with a radius of one meter. At the end of 6 minutes the robot has not quite returned to its starting point at (1, 0). Robot stops here t=0 t=1 t=2 t=3 t=4 t=5 t=6

Example 2 (b) A robot begins at the point (1, 0) and follows the path given by the equations x = cos t, y = sin t where t is in minutes, 0 ≤ t ≤ 6. (b) What happens when you try to eliminate the parameter t ? Solution (b) One way to eliminate t from this pair of equations is to use the Pythagorean identity, cos 2 t + sin 2 t = 1. Since x = cos t and y = sin t, we can substitute x and y into this equation: x 2 + y 2 = 1. This is the equation of a circle, centered at the origin, with radius 1. Solving for y 2, we have y 2 = 1 – x 2 and, solving for y, we get: The first solution represents the upper half of the circle, while the second solution represents the bottom half of the circle. Eliminating the Parameter t Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Example 4 Describe the motion of the robot that follows the path given by: (a) x = cos(−t), y = sin(−t) for 0 ≤ t ≤ 6 (b) x = cos t, y = sin t for 0 ≤ t ≤ 10 Solution Different Motions Along the Same Path Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally (a) The robot travels around the circle in the clockwise direction, opposite to that in Examples 2 and 3. (b) The robot travels around the circle more than once but less than twice, coming to a stop southwest of the landing site. t=0 t=1 t=2 t=3 t=4 t=5 t=6 t=7 t=8 t=9 t=10 Robot stops here Robot stops here

In polar coordinates, the Archimedean spiral is the graph of the equation introduced on page 352: r = θ. Since the relationship between polar coordinates and Cartesian coordinates is x = r cos θ and y = r sin θ, we can write the Archimedean spiral r = θ as x = θ cosθ and y = θ sinθ. Replacing θ (an angle) by t (a time), we obtain the parametric equations for the spiral in the figure: x = t cos t y = t sin t. The Archimedean Spiral Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally t = 4π t = 3π t = 2π t = π t = π/2 t = 3π/2 t = 5π/2 t = 7π/2 t = 0

Example 5 The first two graphs show functions, f(t) and g(t). Describe the motion of the particle whose coordinates at time t are given by x = f(t), y = g(t). Solution As t increases from 0 to 1, the x-coordinate increases from 0 to 1, while the y-coordinate stays fixed at 0. The particle moves along the x-axis from (0, 0) to (1, 0). As t increases from 1 to 2, the x-coordinate stays fixed at x = 1, while the y-coordinate increases from 0 to 1. Thus, the particle moves along the vertical line from (1, 0) to (1, 1). Between times t = 2 and t = 3, it moves horizontally backward to (0, 1), and between times t = 3 and t = 4 it moves down the y-axis to (0, 0). Thus, it traces out the square. Using Graphs to Parameterize a Curve Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Solution Curve f(t)g(t)g(t) t t x x y y

14.2 IMPLICITLY DEFINED CURVES AND CIRCLES Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

The curves known as conic sections include two we are already familiar with, circles and parabolas. They also include ellipses and hyperbolas. Conic sections are so-called because, as was demonstrated by the Greeks, they can be constructed by slicing, or sectioning, a cone. Conic sections arise naturally in physics, since the path of a body orbiting the sun is a conic section. We will study them in terms of parametric and implicit equations. As we have already studied parabolas, we now focus on circles, ellipses and hyperbolas. Conic Sections Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Example 2 Graph the parametric equations x = cos t and y = sin t. Solution We see that x varies between 3 and 7 with a midline of 5 while y varies between 1 and 5 with a midline of 3. The figure gives a graph of this function. It appears to be a circle of radius 2 centered at the point (5, 3). Circles Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally center is (5, 3) radius = 2

Parametric and Nonparametric Equations of a Circle For r > 0, the parametric equations of a circle of radius r centered at the point (h, k) are: x = h + r cos t y = k + r sin t 0 ≤ t ≤ 2π. For r > 0, the parametric equations of a circle of radius r centered at the point (h, k) are: x = h + r cos t y = k + r sin t 0 ≤ t ≤ 2π. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally An implicit equation for the circle of radius r centered at the point (h, k) is: (x − h) 2 + (y − k) 2 = r 2. This is called the standard form of the equation of a circle.

Example 3 + Write the equation of this circle in standard form. x = cos t and y = sin t. Solution We have h = 5, k = 3, and r = 2. This gives (x − 5) 2 + (y − 3) 2 = 2 2 = 4. Equations for Circles Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally center is (5, 3) radius = 2 If we expand the equation for the circle in Example 3, we get a quadratic equation in two variables: x 2 − 10x y 2 − 6y + 9 = 4, or x 2 + y 2 − 10x − 6y + 30 = 0. Note that the coefficients of x 2 and y 2 (in this case, 1) are equal. Such equations often describe circles; we put them into standard form by completing the square.

Example 4 Describe in words the curve defined by the equation x x + 20 = 4y − y 2. Solution Rearranging terms gives x x + y 2 − 4y = − 20 We complete the square for the terms involving x and (separately) for the terms involving y: (x x + 25) + (y 2 − 4y + 4) − 25 − 4 = −20 (x + 5) 2 + (y – 2) 2 − 29 = −20 (x + 5) 2 + (y – 2) 2 = 9 = 3 2 This equation is a circle of radius r = 3 with center (h, k) = (−5, 2). Equations for Circles Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

14.3 ELLIPSES Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Parametric Equations for an Ellipse The parametric equations of an ellipse centered at (h, k) are: x = h + a cos t y = k + b sin t 0 ≤ t ≤ 2π. We usually take a > 0 and b > 0. The parametric equations of an ellipse centered at (h, k) are: x = h + a cos t y = k + b sin t 0 ≤ t ≤ 2π. We usually take a > 0 and b > 0. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally In the special case where a = b, these equations give a circle of radius r = a. Thus, a circle is a special kind of ellipse. The parametric equations of an ellipse are transformations of the parametric equations of the unit circle, x = cos t, y = sin t. These transformations have the effect of shifting and stretching the unit circle.

Example 1 Graph the ellipse given by x = cos t, y = sin t, 0 ≤ t ≤ 2π. Solution The Graph of an Ellipse Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally The center of the ellipse is (h, k) = (7, 4). The value a = 5 determines the horizontal “radius” of the ellipse, while b = 2 determines the vertical “radius”. The value of x varies from a maximum of 12 to a minimum of 2 about the vertical midline x = 7. Similarly, the value of y varies from a maximum of 6 to a minimum of 2 about the horizontal midline y = 4. The ellipse is symmetric about the midlines x = 7 and y = 4. Graph of the ellipse and midlines (7,4) x=7 y=4

Example 2 Graph the ellipse given by x = cos t, y = sin t, 0 ≤ t ≤ 2π. How is this ellipse similar to the one in Example 1? How is it different? Solution The Major and Minor Axes of an Ellipse Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally This ellipse has the same center (h, k) = (7, 4) as the ellipse in Example 1. However, as we see in the figure, the vertical axis of this ellipse is the longer one and the horizontal axis is the shorter one. The longer axis of an ellipse is called the major axis and the shorter axis is called the minor axis. In Example 1, the horizontal axis was the major axis, but in this example the vertical axis is the major axis. Graph of the ellipse and its midlines (7,4) x=7 y=4

Eliminating the Parameter t in the Equations for an Ellipse The implicit equation for an ellipse centered at (h, k) and with horizontal axis 2a and vertical axis 2b is: Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally As with circles, we can eliminate t from the parametric equations for an ellipse by first rewriting the equations for x and y as follows: x = h + a cos t → x – h = a cos t → (x – h)/a = cos t y = k + b sin t → y – k = b sin t → (y – k)/b = sin t Then the result follows by applying the Pythagorean Identity.

14.4 HYPERBOLAS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

We now use another form of the Pythagorean identity to parameterize a curve. With tan t = sin t/cos t and sec t = 1/cos t, on page 370 we showed that sec 2 t − tan 2 t = 1. We use this identity to investigate the curve described by the parametric equations x = sec t and y = tan t. By eliminating the parameter t, we get x 2 − y 2 = 1. This implicit equation looks similar to the equation for a unit circle x 2 + y 2 = 1. However, the curve it describes is very different. Solving for y 2, we find that y 2 = x 2 − 1. Introduction to Hyperbolas Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

The hyperbola with the equation y 2 = x 2 − 1 or x 2 − y 2 = 1 is called the Unit Hyperbola by analogy to the unit circle. The graphs of other hyperbolas will be viewed as transformations of the unit hyperbola. It has two branches, one to the right of the y-axis and one to the left. As x grows large in magnitude, either toward + ∞ or − ∞, the graph approaches the asymptotes, y 2 = x 2. Graphing the Unit Hyperbola Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally A graph of x 2 − y 2 = 1 The X-shaped graph of the asymptotes y 2 = x 2 (y = ± x) has been dashed in. Note that, in order for y 2 to be positive (and y real), |x| ≥ 1.

A General Formula for Hyperbolas The parametric equations for a hyperbola centered at (h, k) and opening to the left and right are x = h + a sec t y = k + b tan t 0 ≤ t ≤ 2π. We usually take a > 0 and b > 0. The parametric equations for a hyperbola centered at (h, k) and opening to the left and right are x = h + a sec t y = k + b tan t 0 ≤ t ≤ 2π. We usually take a > 0 and b > 0. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Example 2 Graph the equation Solution Graphing Hyperbolas Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Since h = 4 and k = 7, we have shifted the unit hyperbola 4 units to the right and 7 units up and it will open up and down. Since we have a 2 = 9 (a = 3) and b 2 = 25 (b = 5), we have stretched the unit hyperbola horizontally by a factor of 3 and vertically by a factor of 5. The equations of the asymptotes are Note the red box with labels, demonstrating the roles of a and b. Graph of the hyperbola opening up and down and its asymptotes 2b=10 2a=6 (4,7)

Eliminating the Parameter t in the Equations for a Hyperbola The implicit equation for a hyperbola, describes a hyperbola that opens left and right. Its asymptotes are diagonal lines through the corners of a rectangle of width 2a and height 2b centered at the point (h, k). The graph with equation has a similar shape, except that it opens up and down. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

14.5 GEOMETRIC PROPERTIES OF CONIC SECTIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Geometric Definitions of Circles and Ellipses A circle of radius r is the set of points in the plane a distance r from a given point. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally An ellipse is the set of points in the plane for which the sum of the distances to the foci is constant. Each of the two points is called a focus or focal point of the ellipse. The two focal points lie on the major axis, and are equally spaced about the minor axis.The constant sum of distances is the length of the major axis. An ellipse is the set of points in the plane for which the sum of the distances to the foci is constant. Each of the two points is called a focus or focal point of the ellipse. The two focal points lie on the major axis, and are equally spaced about the minor axis.The constant sum of distances is the length of the major axis.

Example 2 By finding a formula for c in terms of a and b, determine the focal points (±c, 0) for the ellipse where a and b are positive and a > b: x 2 /a 2 + y 2 /b 2 = 1 Solution Locating the Focal Points of an Ellipse Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally -a a -b b -cc a a From the geometric definition of an ellipse, the sum of the distances to the two focal points is the same at every point on the curve. Thus, if the point is (a, 0), the sum of the distances to (c, 0) and (-c, 0) = (a – c) + (a + c) = 2a. Similarly, if the point is (0, b), the sum of the distances to (c, 0) and (-c, 0) is the same: making the distance from (0, b) to (c, 0) “a.” Using the Pythagorean Theorem a 2 = b 2 + c 2 or Graph of x 2 /a 2 + y 2 /b 2 = 1 with focal points

The Focal Points of an Ellipse Location of the Focal Points of an Ellipse The focal points of the ellipse x 2 /a 2 + y 2 /b 2 = 1 with a > b > 0 are (±c, 0) where The focal points lie on the major axis. If b > a, the focal points are (0, ±c) with Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Reflective Properties of Ellipses Reflective property of an ellipse A ray that originates at one focus of an ellipse is reflected off the ellipse to the other focus. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally ● Focus 2: End point of reflection Focus 1: Point of origination Point of reflection This reflective property, is used in applications involving light, sound, and shock waves

Example 3 An elliptical whispering gallery has a major axis of 100 m and minor axis of 40 m. Where should two speakers position themselves so as to most easily hear each other? Solution Imagine the room centered at the origin with major axis on the x-axis. Then the positive x-intercept is a = 50 and the positive y-intercept is b = 20. The focal points are located at The speakers should position themselves approximately 4.2 m in from the ends of the major axis. Reflective Property of an Ellipse Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Geometric Definition of Hyperbolas Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally A hyperbola is the set of points in the plane for which the difference of the distances to two points is constant. Each of the two points is called a focus or focal point of the hyperbola. Location of the Focal Points of a Hyperbola The focal points of the hyperbola given by equation x 2 /a 2 − y 2 /b 2 = 1 are (±c, 0) where Location of the Focal Points of a Hyperbola The focal points of the hyperbola given by equation x 2 /a 2 − y 2 /b 2 = 1 are (±c, 0) where F1F1 F2F2 ● A

Example 4 (a) Find the focal points and the asymptotes for the hyperbola given by x 2 /9 − y 2 /16 = 1.(b) Graph the hyperbola. (c) For each of the two x-intercepts, find the difference of the distances to the two focal points. Solution Since a = 3 and b = 4, So the focal points are located at (5,0) and (-5,0) and the asymptote equations are y = ±4/3 x Locating the Focal Points of a Hyperbola Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Plot of the hyperbola with asymptotes The intercept (3, 0) is distance 2 from one focus and distance 8 from the other focus. The difference of the distance to the two focal points is 6. Similarly, the difference of the distances to the intercept (−3, 0) is also 8 − 2 = 6.

Reflective property of hyperbolas: A beam aimed at the far focus behind the hyperbola is reflected off the curve to hit the focus in front of the hyperbola. Alternately, a beam originating from the focus in front of the hyperbola is reflected off the curve so that it appears to have originated from the focus behind the hyperbola. Reflective Properties of Hyperbolas Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally This reflective property, is used in navigational systems and is associated with sonic booms. ● Point of reflection

Example 5 A hyperbolic mirror has equation x 2 /16 − y 2 /9 = 1. Find the equation of the line along which a beam of light could be sent out from point A = (15, 50) to reflect off the hyperbola and arrive at the focus closest to point A. Solution The focal points for this hyperbola are (±c, 0) with The focal point closest to A is (5, 0) and the focal point farthest away is (−5, 0). Since we want the beam of light to reflect off the hyperbola and arrive at (5, 0), we aim the beam at the other focal point, (−5, 0). The line from point A to (−5, 0) has slope m where: m = (50 – 0)/(15 – (-5)) = 50/20 = 2.5 and its equation is y = 2.5(x − (−5)) = 2.5x A beam of light sent along the line y = 2.5x toward the point (−5, 0) is reflected off the hyperbola to the point (5, 0). Reflective Property of a Hyperbola Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Geometric Definition of Parabolas A parabola is the set of points in the plane for which the distance from a point is equal to the distance from a fixed line. The point is called the focus and the line is called the directrix. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Location of the Focus and Directrix of a Parabola The focus of the parabola y = ax 2 is the point (0, c) where c = 1/(4a). The directrix is the line y = −c. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Focus Directrix Even though they may not look it, the two red line segments are the same length

Example 7 (a) Find the focus and the directrix of the parabola y = 3x 2. (b) Show that the point (1, 3) on the parabola is equidistant from the focus and the directrix. Solution We have a = 3 so the focus is at the point (0, 1/12) and the directrix is the line y = −1/12. We have the distance from (1, 3) to the focus (0, 1/12): d 1 = The distance from (1, 3) to the directrix y = −1/12 is the difference in the y-values, since the directrix is a horizontal line: d 2 = 3 + 1/12 = 37/12. Thus the distance from the point (1, 3) to the focus equals the distance from the point (1, 3) to the directrix. This property holds for all points (x, y) on the parabola. Locating the Focus and Directrix of a Parabola Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Reflective property of parabolas All beams parallel to the axis of symmetry reflect off the parabola to the focus. Reflective Properties of Parabolas Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Giant parabolic mirrors are used in telescopes to focus light and radio waves from outer space. Car headlights have parabolic reflectors with the lightbulb at the focus. Point of reflection Focus

Example 8 The parabolic reflector behind a car head light is 16 cm across and 3 cm deep. Where should the bulb be positioned? Solution The shape of the parabolic reflector is made by rotating a parabola y = ax 2 about the y-axis. Since the rim of the reflector lies on the parabola, the point (8, 3) satisfies the equation y = ax 2 so 3 = a(82) and a = 3/64. The focus is located at:c = 1/(4a) = 1/(4(3/64)) = 16/3. The bulb should be placed 16/3 cm from the vertex of the parabola along the axis of symmetry. Parabolic Mirrors in Car Headlights Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

14.6 HYPERBOLIC FUNCTIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Hyperbolic Sine and Hyperbolic Cosine From these definitions, we can see that cosh 0 = 1, sinh 0 = 0 cosh(−x) = cosh x, sinh(−x) = −sinh x Hyperbolic Sine and Hyperbolic Cosine From these definitions, we can see that cosh 0 = 1, sinh 0 = 0 cosh(−x) = cosh x, sinh(−x) = −sinh x Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

The figures show graphs of cosh x and sinh x together with dashed-line graphs of ½ e x and ½ e −x or -½ e −x. The graphs suggest that cosh x is even and sinh x is odd and that cosh x has a y-intercept of 1,whereas sinh x has a y-intercept of 0: Plots of the Hyperbolic Sine and Cosine y = -½ e −x y = ½ e x y = cosh x y = sinh x Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Example 4 Describe and explain the behavior of cosh x as x → ∞ and then as x → − ∞. Solution From the figure, we see that as x → ∞, the graph of cosh x resembles the graph of ½ e x. Similarly, as x → − ∞, the graph of cosh x resembles the graph of ½ e -x Since e −x → 0 as x → ∞ and e x → 0 as x → − ∞, we can predict these results algebraically: Long-Term Behavior of the Hyperbolic Cosine Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally y = -½ e −x y = ½ e x y = cosh x

Identities Involving cosh x and sinh x cosh 2 x − sinh 2 x = 1 Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally This can be shown using the fact that e x · e −x = 1:

Defining the Hyperbolic Tangent Function Hyperbolic Tangent tanh x = sinh x/cosh x. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Graph of the hyperbolic tangent and its horizontal asymptotes

Consider the curve parameterized by the equations x = cosh t, y = sinh t, − ∞ < t < ∞. We can eliminate the parameter t by using the identity just derived: x 2 − y 2 = cosh 2 x − sinh 2 x = 1 This is the implicit equation for the unit hyperbola. Notice that since cosh t > 0, we have x > 0, so the parameterization gives only the right branch of the hyperbola. Parameterizing the Hyperbola Using Hyperbolic Functions Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally As t → ∞, both cosh t and sinh t approach ½ e t, so for large values of t we see that cosh t ≈ sinh t. Thus, as t increases the curve draws close to the diagonal line y = x. Similarly, as t → − ∞, we know that x → ½ e −t and y → − ½ e −t, which means that y →−x. Thus, as t → − ∞, the curve draws close to the asymptote y = −x. y=x y=-x t=2 t=-2 t=-1 t=1 t=0 x = cosh t, y = sinh t