Part 1: Vector-Valued Functions

Now that we have introduced and developed the concept of a vector, we are ready to use vectors to define functions. To begin with, a vector-valued function is a function whose inputs are a parameter t and whose outputs are vectors r( t) .

In 2 dimensions, a vector-valued function is of the form

r( t) = á f( t) ,g( t) ñ

If f and g are continuous functions over [a,b], then the set of position vectors of the form r( t) = á f( t) ,g( t) ñ for t in [ a,b] forms a curve whose orientation is in the direction in which the parameter is increasing. The point r(a) = (f(a),g(a)) is called the initial point of the curve, and the point r(b) = (f(b),g(b))  is the curve's terminal point.

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The function r( t) = á f(t) ,g( t) ñ , t in [ a,b] , is called a parametrization of the curve.  To sketch the curve, we often let

x = f( t) ,        y = g( t) ,        t  in [ a,b]

and then eliminate the parameter t in hopes of obtaining a familiar equation in x and y.      

EXAMPLE 1    Sketch the curve parametrized by r(t) = á 1-t²,2t²-t⁴ ñ for t in [ 0,1] .       

Solution: To begin with, x = 1-t², which means that t² = 1-x. Since y = 2t²-t⁴, that means that

y = 2( 1-x) -( 1-x) 2

(1)

which simplifies to y = 1-x². Moreover, the initial and terminal points are

initial: ( t = 0) x = 1-02 = 1, y = 2·0-04 = 0 terminal: ( t = 1) x = 1-12 = 0, y = 2·1-14 = 1

That is, the graph of r( t) = á1-t²,2t²-t⁴ ñ for t in [ 0,1] is the section of the graph of y = 1-x² between the initial point (1,0) and the terminal point ( 0,1) .

       

EXAMPLE 2    Sketch the curve parameterized by r(t) = á cos( t) ,sin( t) ñ for t in [ 0,2p] .       

Solution: Since x = cos( t) and y = sin(t) , we begin with

cos2( t) +sin2( t) = 1

Substituting x = cos( t) and y = sin( t) into the identity yields

x2+y2 = 1

which is the unit circle. Moreover, the initial and terminal points are

initial: ( t = 0) x = cos( 0) = 1, y = sin( 0) = 0 terminal: ( t = 2p) x = cos( 2p) = 1, y = sin( 2p) = 0

Since the initial and terminal points are both ( 1,0) , we must determine the orientation by choosing a value of t near an endpoint of [0,2p] :

t = p 6 :    x = cos æ è p 6 ö ø = 3 2 ,    y = sin æ è p 6 ö ø = 1 2

Thus, the orientation of the curve is counterclockwise: