Vector valued functions have values in ℝp where p is an integer at least as large as 1. Here are some
examples.
Example 8.1.1A rocket is launched from the rotating earth. You could define a function havingvalues in ℝ3as
(r (t),θ(t),ϕ(t))
where r
(t)
is the distance of the center of mass of the rocket fromthe center of the earth, θ
(t)
is the longitude, and ϕ
(t)
is the latitude of the rocket.
Example 8.1.2Let f
(x,y)
=
(sinxy,y3 + x,x4)
. Then f is a function defined on ℝ2which hasvalues in ℝ3. For example, f
(1,2)
=
(sin2,9,16)
.
As usual, D
(f)
denotes the domain of the function f which is written in bold face because it will
possibly have values in ℝp. When D
(f)
is not specified, it will be understood that the domain of fconsists
of those things for which f makes sense.
Example 8.1.3Let f
(x,y,z)
=
(x+y √ ----2- )
z , 1− x ,y
. Then D
(f)
would consist of the set of all
(x,y,z)
such that
|x|
≤ 1 and z≠0.
There are many ways to make new functions from old ones.
Definition 8.1.4Let f,g be functions withvalues in ℝp. Let a,b be points of ℝ (scalars). Then af + bg isthe name of a function whose domain is D
(f)
∩ D
(g )
which is defined as
(af + bg)(x) = af (x) +bg (x ).
f ⋅ g or
(f,g)
is the name of a function whose domain is D
(f)
∩ D
(g )
which is defined as
(f,g)(x) ≡ f ⋅g(x) ≡ f (x )⋅g(x).
If f and g have values in ℝ3, define a new functionf × gby
f × g(t) ≡ f (t) ×g (t).
If f : D
(f)
→ X and g : X → Y , then g ∘ f is the name of a function whose domain is
{x ∈ D (f) : f (x) ∈ D (g)}
which is defined as
g∘ f (x) ≡ g(f (x)).
This is called the composition of the two functions.
You should note that f
(x)
is not a function. It is the value of the function at the point x. The name of
the function is f. Nevertheless, people often write f
(x)
to denote a function and it does not cause too many
problems in beginning courses. When this is done, the variable, x should be considered as a generic
variable free to be anything in D
(f)
. I will use this slightly sloppy abuse of notation whenever
convenient.
Example 8.1.5Let f
(t)
≡
(t,1+ t,2)
and g
(t)
≡
(t2,t,t)
. Thenf ⋅ gis the name of the functionsatisfying
f ⋅g(t) = f (t) ⋅g (t) = t3 +t + t2 + 2t = t3 +t2 + 3t
Note that in this case is was assumed the domains of the functions consisted of all of ℝ because this was
the set on which the two both made sense. Also note that f and g map ℝ into ℝ3 but f ⋅ g maps ℝ into
ℝ.