Parametric Equations

Transcription

1 Parametric Equations Suppose a cricket jumps off of the round with an initial velocity v 0 at an anle θ. If we take his initial position as the oriin, his horizontal and vertical positions follow the equations: = t y = v 0 sinθ t t These are called parametric equations, with time bein the parameter upon which the positions depend. If we had numeric values for the initial velocity and anle, it would be easy to make a chart of the motion. Supposin v 0 = m/s and θ = 30º. = 1.73 t y = t 4.9 t Time (s) (m) y (m) The raph of which is a simple parabola y (m) (m)

2 We can also eliminate the parameter throuh alebraic substitution: = t so t = y = v 0 sinθ t t y = tanθ - so y = v 0 sinθ ( ) - ( ) which is eactly the parabola raphed above So the question arises, if we can etract the parameter and simply have y as a function of, what is the value of parametric equations in the first place? There are several answers: 1. It is not always possible to etract the parameter from simultaneous equations. Often, even when possible, the result is uninformatively comple.. In many physical situations, we would like to know how the positions depend upon the parameter. For the cricket, the parametric equations tell us where he is at any iven time. 3. The parametric equations provide a direction for the curve. In our eample, the cricket was clearly jumpin from left to riht, but etractin time from the equations and writin y as a function of removed this information. We can see those advantaes with another eample, a car drivin around in circle, counterclockwise. Suppose the speed is a constant v, the radius of the track is R, the oriin is at the center of the track, and the car has the position shown at time zero. The parametric equations for the car s horizontal and vertical positions are then: = R cos( v R t) y = R sin( v R t)

3 We can etract the parameter of time as follows: + y = R cos ( v t) + R sin ( v t) = R R R or just + y = R But, aain, this Cartesian version has the followin disadvantaes: 1. + y = R is not a function, so cannot be written in simple functional notation. + y = R does not tell us where the car is at any iven time 3. + y = R does not tell us that the car is movin counter-clockwise What calculus can be applied to parametric equations? Suppose we wanted the instantaneous trajectory of the cricket at any time or position, that is, we want to find the tanent to the spatial curve, or. One way is to simply differentiate the equation y = tanθ - = tanθ - v 0 cos θ and find Another is to use the parametric equations. Let s bein with the chain rule: = or = If = t If y = v 0 sinθ t t then then = = v 0 sinθ t = = v 0 sinθ t or = tanθ - t as a function of time

4 We can also replace t with from earlier to et = tanθ - v 0 cos θ as we found before Incidentally, it s a ood habit, once equations are derived, to play with various values. Do the equations make sense when t = 0? Do they make sense where = 0? Why is t in the numerator and cosθ in the denominator? Now suppose we want the arc lenth traveled by the cricket. Let s define ds to be an infinitely small piece of this arc lenth. Then ds = + By the chain rule, =, so ds = + ( ) = [1 + ( ) ] Takin the square root of both sides and interatin yields: Arc lenth = 1 + ( ) By symmetry, we could also use Arc lenth = 1 + ( ) If we use the first form for our cricket, we would have Arc lenth = 1 + (tanθ v 0 cos θ ) where the boundaries for the full fliht would be = 0 to = v 0 sin (θ) (from the rane equation) That is not a particularly pleasant interal, so let s try usin parametric equations: If we remember = then arc lenth = 1 + ( ) becomes:

5 Arc lenth = 1 + ( ) or arc lenth = ( ) + ( ) For our cricket, arc lenth = (v 0 cosθ) + (v 0 sinθ t) where the boundaries for the full fliht are t = 0 and t = v 0sinθ This is still a comple interal and it takes some work, but the solution is: Arc lenth = v 0 cos θ 1+sin (θ) [ sec(θ) tan(θ) + ln ] 1 sin (θ) Lastly, what is the area under the arc jumped by the cricket? We could simply interate y = tanθ - = v 0 sin(θ)cos (θ) between the boundaries of = 0 and This would be: Area = tanθ = tanθ v 0 cos θ And with the boundaries entered: Area = v 0 4 sin 3 θ cosθ 3 What would this look like usin parametric equations? f Area = y() = y(t) ( ) i t f t i by usin substitution for definite interals For our cricket, y(t) = v 0 sinθ t t With the boundaries of t = 0 and t = v 0sinθ and = v0sinθ Area = (v 0 sinθ t t ) (v 0 cosθ) 0 Area = v 0 4 sin 3 θ cosθ 3 as we found before