740 CHAPTER 10 Analytic Geometry Example 4 illustrates the versatility of parametric equations for replacing complicated rectangular equations, while providing additional information about orientation. These characteristics make parametric equations very useful in applications, such as projectile motion. 4 Use Time as a Parameter in Parametric Equations If we think of the parameter t as time, then the parametric equations x x t( ) = and y y t( ) = specify how the x - and y -coordinates of a moving point vary with time. For example, we can use parametric equations to model the motion of an object, sometimes referred to as curvilinear motion . Using parametric equations, we can specify not only where the object travels—that is, its location x y , ( ) —but also when it gets there—that is, the time t. When an object is propelled upward at an inclination θ to the horizontal with initial speed v ,0 the resulting motion is called projectile motion . Using vectors and calculus, we can prove the following result. Figure 68 Projectile motion x u h (x(t), y(t)) v0 y The parametric equations of the path of a projectile fired at an inclination θ to the horizontal, with an initial speed v ,0 from a height h above the horizontal, are x t v t y t gt v t h cos 1 2 sin 0 2 0 θ θ ( ) ( ) ( ) ( ) = = − + + (2) where t is time and g is the constant acceleration due to gravity (approximately 32 ft sec ,2 or 9.8 m sec2 ). See Figure 68. Projectile Motion Suppose that Jim hit a golf ball with an initial speed of 150 feet per second at an angle of 30° to the horizontal. See Figure 69. (a) Find parametric equations that describe the position of the ball as a function of time. (b) How long was the golf ball in the air? (c) When was the ball at its maximum height? Find the maximum height of the ball. (d) Find the distance that the ball traveled. (e) Using a graphing utility, simulate the motion of the golf ball by simultaneously graphing the equations found in part (a). EXAMPLE 5 Figure 69 308 Solution (a) We have v h 150 ftsec, 30 , 0 ft 0 θ = = ° = (the ball is on the ground), and g 32 ft sec2 = (since the units are in feet and seconds). Substitute these values into equations (2) to get x t v t t t y t gt v t h t t t t cos 150 cos30 75 3 1 2 sin 1 2 32 150 sin30 0 16 75 0 2 0 2 2 θ θ ( ) ( ) ( ) ( ) ( ) ( ) = = ° = = − + + = − ⋅ ⋅ + ° + = − + Figure 67 π ( ) ( ) =− =− ≤ ≤ > x t a t y t a t t a sin cos 0 0 y x (0, a) (2a, 0) (0, 2a) The plane curve defined by the parametric equations lies on a circle with radius a and center at 0, 0 . ( ) The curve begins at the point a t 0, , when 0; ( ) − = passes through the point a t , 0 , when 2 ; π ( ) − = and ends at the point a t 0, , when .π ( ) = The parametric equations define the left semicircle of a circle of radius a with a clockwise orientation. See Figure 67. The rectangular equation is x a y a y a 2 2 = − − − ≤ ≤
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