{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

parametric equations

# parametric equations - Sec 8.6 Parametric Equations...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Sec. 8.6: Parametric Equations Parametric equations provide a way to examine a graph where the x and y coordinates of points on the graph are deﬁned separately as functions of a parameter t, frequently representing time. If a projectile' 1s launched at initial elevation angle 9 and initial speed Va feet per second from initial height of Se feet, the parametric equations that describe the trajectory are developed by using right triangle trig ratios to "decompose" the initial velocity vector V0 into horizontal and vertical components X0 and y0 The formula for y(t) in step 2' IS derived using calculus. xo_ 1) V h-=s sine—)yomvosine x(t) = xﬂt ' 2) y(t) = —16t1 +y0t+Sﬂ Se .- 3) x(t) =(Vﬂ cosG)t ‘ y(t) = —16t2 +(va sin9)t +3., This allows for the height and the horizontal distance traveled by the projectile to be examined separately. The time in the air can be calculated by determining when the y parametric equation equals 0, and this value can be substituted into the x parametric equation to determine how far the projectile traveled during this time period. A rectangular equation can be determined by solving the x equation for t and substituting this into the y equation. This equation describes height in terms of distance, and can be used to determine the height of the projectile when it has traveled a given horizontal distance. =cose —) x0: V6 cosB g: Suppose the initial velocity is 100 feet per second, the initial angle is 50°, and the projectile is launched from x = (100 cosSO°)t an 64.3t ground level. The parametric equations would then be 2 . 2 . The projectile y = ~16t + (1003111 50°)t at —16t + 76.6t would .be on the ground when y = 0, which occurs at t as 4.8 seconds. ”This is represented by the graph on the left. . 1 I t I 1 1 . 0.5 l 1.5 2 2.5 S 3.5 4 4.5 \5 The projectile would travel about 309 feet. lfwe substitute t = a)“; into the y parametric equation, we get 2 y = —16[E}-3—] + 76.4%] e. 4.9038732 + 1.191: . The graph in the middle represents this equation, which is the same as the graph on the left, except that the horizontal axis now represents horizontal distance traveled instead of time in the air. The graph on the right is the linear graph of horizontal distance traveled as a function of time. ...
View Full Document

{[ snackBarMessage ]}