Can you help me with the circled problems 62 and 68? They are circled on the 9th photo.

AT&T S 11:45 PM 01 19% 11 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS We can study the interaction between two animal species with populations q(f) and This chapter introduces two important new tools. First, we consider parametric equa- port. where each population is a function of tions, which describe curves in a form that is particularly useful for analyzing mo- and, to investigate how the two populations tion and is indispensable in fields such as computer graphics and computer-aided design. change. Combining the functions in the form (milk pill yields a parametric representation We then study polar coordinates, an alternative to rectangular coordinates that simplifies of a curve in the qp-plane. Tracing this curve computations in many applications. The chapter closes with a discussion of the conic as I changes creates a story about this sections (ellipses, hyperbolas, and parabolas). interaction and its impact on population size. 11.1 Parametric Equations We use the term "particle" when we treat Imagine a particle moving along a curve C in the plane as in Figure 1. We would like to an object as a moving point, ignoring its be able to describe the particle's motion along the curve. To express this motion mathe- internal structure. matically, we consider how its coordinates x and y are changing in time, that is, how they depend on a time variable f. Thus. x and y are both functions of time, r, and the location Position at time of the particle at f is given by c(1) = (x(1). >(1)) Curve C This representation of the curve C is called a parametrization with parameter , and C 1-0 is called a parametric curve. In a parametrization, we often use / for the parameter, thinking of the dependent variables as changing in time, but we are free to use any other variable (such as s or (). In plots of parametric curves, the direction of motion is often indicated by an arrow as in FIGURE | Particle moving along a curve C Figure I. in the plane. Specific equations defining a parametrization, such as x = 2t - 4 and y = 3 + = in the next example, are called parametric equations. EXAMPLE 1 Sketch the curve with parametric equations x = 2 -4. y'=3+1 1 Solution First compute the x- and y-coordinates for several values of : as in Table 1. and plot the corresponding points (.x. y) as in Figure 2. Then join the points by a smooth curve, indicating the direction of motion (direction of increasing () with an arrow. /1-4 TABLE 1 (4, 19) 1=2-4 y=3+1- -2 (-8, 7) (0. 7) Graphing calculators or CAS software can be used to sketch and examine parametric OF FIGURE 2 The parametric curve curves *=2-4, y=3+1-. CONCEPTUAL INSIGHT The graph of y = x can be parametrizationay. We take x = t and y = r'. Then, since y = ro and r = x, it follows that y == x-. There- fore, the parabola y z x? is parametrizatione generally, we can 619 1/9AT&T S 11:45 PM 01 19% 620 CHAPTER 11 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS parametrizationh of y = f(x) by taking x = t and y = f(1). Therefore, cina (f. f()) parametrization of y = f(x). For another example, the graph of ) = is parametrization advantage of parametric equations is that they enable us to describe curves that are not graphs of functions. For example, the curve is Figure 3 is not of the form y = f(x) but it can be expressed parametrically. As we have just noted, a parametric curve c(f) need not be the graph of a function If it is, however, it may be possible to find the function f by "eliminating the parameter" as in the next example. EXAMPLE 2 Eliminating the Parameter Describe the parametric curve FIGURE 3 The parametric curve * = 5 cost31) cos( { sin(5)). c(1) = (21 -4.3+12 ) v = 4 sin(3/) cos({ sin(5r)). of the previous example in the form y = f(x). Solution We eliminate the parameter by solving for y as a function of x. First, express in terms of x: Since x = 21 - 4, we have r = fx + 2. Then substitute 1=3+1 =3+ + +2)"=7+2x+4x2 Thus. c(f) traces out the graph of f(x) = 7+ 2x + Jx- shown in Figure 2. >(m) EXAMPLE 3 A model rocket follows the trajectory -20.4 c(t) = (80r. 200r - 4.912) 2000 until it hits the ground, with / in seconds and distance in meters (Figure 4). Find: (a) The rocket's height at t = 5 s. (b) Its maximum height. 1000- Solution The height of the rocket at time ? is y(1) = 2001 - 4.912. ( = 40.8 -x(m) (a) The height at / = 5 s is 1,000 2.000 3,000 "(5) = 200(5) - 4.9(5-) = 877.5 m OF FIGURE 4 Trajectory of rocket. (b) The maximum height occurs at the critical point of y() found as follows: CAUTION The graph of height versus time y'(1) = (200r -4.913) = 200 - 9.81 for an object tossed in the air is a parabola (by Gahtieo's formula). But keep in mind Thus, y' = 0 when 200 - 9.8: = 0, that is, for that Figure 4 is not a graph of height 200 versus time. It shows the actual path of the 9.8 20.4 s rocket (which has both a vertical and a horizontal displacement). The rocket's maximum height is y(20.4) = 200(20.4) - 4.9(20.4)2 2 2041 m. We now discuss parametrizations of lines and circles. They will appear frequently is later chapters. To begin, note that the parametric equations y = mi - 00 0) 69. Find the points where the tangent has slope }. Show that the projectile is launched at an angle d = tan" 70. Find the points where the tangent is horizontal or vertical. at a distance from the origin. 9/9