Vector valued functions have values in ℝ^{p} 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 ℝ^{3}as

(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 ℝ^{2}which hasvalues in ℝ^{3}. 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 ℝ^{p}. 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 ℝ^{p}. 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 ℝ^{3}, 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 ℝ^{3} but f ⋅ g maps ℝ into
ℝ.