Lecture Note
University
California State UniversityCourse
MATH 150B | Calculus IIPages
3
Academic year
2023
yung dump
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11.1 Parametrization of Plane Curves 84 We may describe the movement of a particle in the xy plane at position t by (x(t), y(+)) = (f(+), g(+)) Def If X and y are given as functions position of the particle at time t not a function x = (tz) and y = g(+) , TEI, then the set of points (x,y) = (F(+1, 9(+1) is a parameteric curve. Note that X= F(+) and y= g(t) are called parametric equations. 12 the variable t is called the parameter of the curve. 3 the interval I is called the parameter interval. If I = [a, b] closed interval, then the point (f(a), g(a)) is the initial point and the point (f(b), g(b)) is the terminal point. 4 We say that we have parametrized the curve, if we find III and [3] . That is and (3) give a parametrization of the curve. Exp Given the parametric equation and parameter interval: & X = t 2 y=t+1 / -oo
y 85 The curve that represents the t=2 3 particle movement. 2 t=o 1 t -3 -2 3 4 X 9410149 t=-2 y 1-2-101234 Exp Graph the parametric curve of X= cost, y=sint ost S 2TT We can eliminate the parameter t by: cartesian equation = cost + sin't =1 (0,1) Initial point is ( coso, sino). = (1,0) t=o Terminal point is (COS TIT, sinzii) ( 1,0) (1,0) (-1,0) t= II the position is (-1,0) (0,-1) Direction: counterclockwix Exp Graph the particle's movement and direction if its parametric equation and parameter interval is 1 X=VE / y=t , tz o y y=x We can eliminate the parameter t x>,0 y t=x2 = y=x Cartesian equation t=1 to x Initial point is (vo, o) = (0,0) ( No terminal point t=1 the position is (1,1) 2 2 kt / y=t we can eliminate the parameter t y ystr y=t2 = Cartesian y=x ((-2,4) t=-2 for equation no initial point t>o tco no terminal point x t20
Exp Find a parametrization for the line passes throw the 86 points (a,b) and (c,d) A cartesian cyvation is y-b= m(x-a) where the slope m= d-b , cfa set the parameter t = x-a c-a Hence, X = a+t, / y= b+mt, parameterizes the line. the line segment with endpoints (-1,3) and (3,-2) m= -2-3--5 3+1 toessy (I( X = -1 +t, / y = 3-st, / ostsy 3 1 - X { X=-1 +4t, Y=3-5t , osts)3 -2 t=4 Both parametrization give the same segment Exp sketch and identify the path by the point P(x,y) if X= t+1/2 , y=t- , t>o We can eliminate the parameter t by: 2 2 i-y-4 x = 4+y x +y = 2t (x+y)(x-y)=4 x-y=? y X= + 4+y2 t at t= 0.5 the positite is (0.5+2,0.5-2) t=2 (2.5,3 (2.5,-3) t=1 at t=2 the posithho is (2.5,2) X (25,-3) t=0.5 Note that 1=t+1 ,y=t- , +>0} X >0 since +>0 y=t, - wet
Parametrizations of Plane Curves
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