Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 12 Section 1
1
Vector-Valued Function
Def.: A vector-valued function is a function of the form…

r (t ) =
(in the plane)

r (t ) =
(in space)
Note: The component functions, f, g, and h are
vector-valued function
input
of the
output (function value)



r (t )  f (t )i  g (t ) j




r (t )  f (t )i  g (t ) j  h(t )k
Exercise 1a: Sketch the plane curve represented by the vector-valued function



r (t )  (3  2t ) i  (2  3t ) j , and indicate its orientation.
Exercise 1b: Sketch the plane curve represented by the vector-valued function
  13 3t  

r (t )  t i     j , and indicate its orientation.
2 2
Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 12 Section 1
Note 1: The plane curve is not the output of the function, but it is associated with/to the
output of the function. The actual output of the function is a set of vectors, whereas
the curve is the graph of the connected “heads” (tips) of the vectors.
Note 2: Although the vector-valued functions in Exercises 1a and 1b have the same
plane curve, they are technically different functions because they have different
orientations. Moreover, we see by these exercises that a curve does not have a unique
parametric representation.
Domain of a Vector-Valued Function
The domain of a vector-valued function is the intersection of the domains of the
component functions.



1
Exercise 2: Determine the domain of r (t )  i  4  t j  ln( t  5) k .
t
Note 3: A parameterization of the line segment that connects the point P1  x1 , y1 and
the point P2  x2 , y2  and that is oriented from P1 to P2 is given by

r (t ) =

r ( 0) =
= the vector in standard position whose terminal point is

r (1) =
= the vector in standard position whose terminal point is
Note 4: A parameterization of the line that passes through the point P1  x1 , y1  and the
point P2  x2 , y2  and that is oriented from P1 to P2 is given by

r (t ) =
Exercise 3: Determine a parameterization of the line segment that connects P1 = (1,3,5)
and P2 = (4,0,-1) and that is oriented from P1 to P2.
2
Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 12 Section 1
Note 1: To parameterize a plane curve y = f(x), a “natural” choice is to let
3
and
Exercise 4: Determine vector-valued functions that form the boundary of the region
below.
Note 2: Later, when we study line integrals, we will have to consider the two curves as a single curve
(the boundary of a region) and parameterize the boundary with a single vector-valued function.
Exercise 5: Sketch the space curve represented by the vector-valued function




r (t )  2 cos(t ) i  t j  2 sin( t ) k , 0  t  4 .

t
r (t )
0

2
π
2π
3π
4π
your attempt (don’t look over there  )
z
y
x
Maple spacecurve([2*cos(t),t,2*sin(t)],t=0..4*Pi);
Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 12 Section 1
Limit of a Vector-Valued Function
Def.:
vector-valued function
4

lim r (t )
t a



r (t )  f (t )i  g (t ) j




r (t )  f (t )i  g (t ) j  h(t )k

Note 1: lim r (t ) exists provided that the limit as t  a of each of the component
t a
functions exists.
Continuity of a Vector-Valued Function

Def.: A vector-valued function r (t ) is continuous at t = a if
(i)
and
(ii)
Note 2: From this definition we see that a vector-valued function is continuous at t = a if
and only if each component function is continuous at t = a.

Def.: A vector-valued function r (t ) is continuous on an interval I if it is continuous at
every point in the interval.
Exercise 6 (Section 12.1 #67): Evaluate the limit
  sin( t ) 
lim  e t i 
j  e t
t 0
t


k .
