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This text explains the concept of parametrizing paths and velocity vectors in mathematics, using examples such as circles and lines. It also explores the calculation of speed along a path and finding tangent vectors.
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The circle of radius 1 centered at the origin in R2 can be represented by the Cartesian equation x2 + y2 = 1. y This circle can also be represented by the joining the following sets of points: x (t , 1 – t2 ) for –1 t 1 (t , –1 – t2 ) for –1 <t < 1 Another (and “cleaner”) representation of this circle is the following set of points: Note that we can parametrize a circle of radius R with (R cos t , R sint) for 0 t 2 (cos t , sint) for 0 t 2 When we map a subset of R onto a path C in Rn, we say that we have parametrized the path and the variable which takes on values in R is called the parameter.
Suppose we want to parametrize the line in R3 through the points (6 , –2 , –3) and (5 , –7 , 2). The direction of the line can be described by the vector (–1 , –5 , 5) . The line can be described by t(–1 , –5 , 5) + (6 , –2 , –3) . The line can be represented by the set of points (– t + 6 , –5t– 2 , 5t – 3) for – < t < . Suppose we want to parametrize the graph of y = x2 in R2. (t , t2 ) for – < t < y y = x2 x
Suppose a circle rolls along the x axis as a wheel rolls on a level surface. Let b(t) be the path of the point on the circle initially touching the x axis. If the radius of the circle (wheel) is R, then parametrize the path of the point in one revolution of the circle (wheel). (This path is called a cycloid.) b(t) (0,R) (2R,R) x If the center of the circle was at the origin and did not move, then we could describe the path of the point at the bottom of the circle in one clockwise revolution as ( Rcos(3/2 – t) , Rsin(3/2 – t) ) for 0 t 2 . If the center of this circle was shifted to the point (0, R), then the path would be changed to y ( Rcos(3/2 – t) , R + Rsin(3/2 – t) ) for 0 t 2 .
b(t) (0,R) (2R,R) c(t) = (Rt, R) x If the center of the circle was at the origin and did not move, then we could describe the path of the point at the bottom of the circle in one clockwise revolution as ( Rcos(3/2 – t) , Rsin(3/2 – t) ) for 0 t 2 . If the center of this circle was shifted to the point (0, R), then the path would be changed to y (Rcos(3/2 – t) , R + Rsin(3/2 – t) ) for 0 t 2 . If this circle is now allowed to roll along the x axis, observe that c(t) = (Rt, R) is the path of the center of the circle as it rolls, and the path of the point of interest would be changed to b(t) = ( Rt + Rcos(3/2 – t) , R + Rsin(3/2 – t) ) for 0 t 2 .
Definition of differentiable path, velocity of a path, and speed of a path (page 145) z c(t) c(t) y Consider a path in R3 defined by c(t) = [x(t) , y(t) , z(t)] (a position vector) The velocity vector c(t) = [x (t) , y (t) , z (t)] at each point is tangent to the path c(t) at that point. x The speed at each point on the path is ||c(t)|| . These definitions of course apply in R2 or Rn for any n in the natural way.
Find the velocity vector for the path of the circle c(t) = (6 cos t , 6 sint) for 0 t 2 y c(t) = (– 6 sin t , 6 cost) for 0 t 2 Note that the speed at each point along the circle is x ||c(t)|| = 6 . Find the velocity vector for the path of the line c(t) = (5t + 6 , –5t – 2 , – t – 3) for – < t < c(t) = (5 , –5 , –1) for – < t < ||c(t)|| = 51 . Note that the speed at each point along the line is Find a vector tangent to the path c(t) = (cos t , sin t , t2) for 0 t < c(t) = (– sin t , cost , 2t) for 0 t < ||c(t)|| = 1 + 4t2 . Note that the speed at each point along this path is
Consider the path c(t) = (cos t , sin t , t2) for 0 t < . (a) Find t when the path goes through the point (–1 , 0 , 92). (b) Find the line tangent to the path at the point (–1 , 0 , 92). t = 3 c(t) = (– sin t , cost , 2t) for 0 t < The direction of the line can be described by the vector (0 , –1 , 6). The line can be described by s(0 , –1 , 6) + (–1 , 0 , 92) . The line can be represented by the set of points (–1, – t , 6s + 92) for – < s < .
Consider the path c(t) = (6 cos t , 2 sin t) for 0 t 2 . y (a) Sketch the path of a particle traveling along this path. (0 , 2) (32 , 2) This is the ellipse x2/36 + y2/4 = 1. x (–6 , 0) (6 , 0) (b) Find the position and speed of the particle when t = /4 . (0 , –2) c(/4) = (32 , 2) . The position of the particle when t = /4 is The velocity of the particle for any tis c (t) = ( 6 sin t , 2 sin t) . The velocity of the particle when t = /4 is c (/4) =(–32, 2) . The speed of the particle when t = /4 is ||c (/4)|| = ||(–32, 2)|| = 20 (c) Suppose that at the point where t = /4, the particle were to fly off the elliptical path and travel in a straight line in the direction of the tangent. What would the particle’s position be when t = 4?
Consider the path c(t) = (6 cos t , 2 sin t) for 0 t 2 . y (32 , 2) . c(/4) = (0 , 2) c(t) = (–6 sin t , 2 cost) for 0 t 2 (32 , 2) c (/4) = (–32, 2) x (–6 , 0) (6 , 0) (0 , –2) (c) Suppose that at the point where t = /4, the particle were to fly off the elliptical path and travel in a straight line in the direction of the tangent. What would the particle’s position be when t = 4? The point at which the particle flies off the elliptical path is c(/4) = (32 , 2) . The direction in which the particle travels immediately after flying off the elliptical path is c (/4) =(–32, 2) . The line on which the particle travels immediately after flying off the elliptical path can be parametrized as s( , ) + ( , )
(c) Suppose that at the point where t = /4, the particle were to fly off the elliptical path and travel in a straight line in the direction of the tangent. What would the particle’s position be when t = 4? The point at which the particle flies off the elliptical path is c(/4) = (32 , 2) . The direction in which the particle travels immediately after flying off the elliptical path is c (/4) =(–32, 2) . The line on which the particle travels immediately after flying off the elliptical path can be parametrized as s(–32, 2) + (32 , 2) where s = t /4 When t = 4, the particle’s position is (4 – /4)(–32, 2) + (32 , 2) = 32 2 – 92 , 52 . 4 4
