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Calculus with
vector functions
In this section we need to talk briefly about limits, derivatives and integrals of vector functions
...
We will be doing all of the work in
3
but we can naturally extend the formulas/work in this section to n (i
...
n-dimensional
space)
...
Here is the limit of a vector function
...
sin ( 3t − 3) 2t
Example 1 Compute lim r ( t ) where r ( t ) = t ,
,e
...
sin ( 3t − 3)
lim r ( t ) = lim t , lim
, lim e2t
t →1
t →1
t →1
t →1
t −1
3
3cos ( 3t − 3)
= lim t , lim
, lim e 2t
t →1
t →1
t →1
1
3
= 1,3, e2
Notice that we had to use L’Hospital’s Rule on the y component
...
r′ ( t ) = f ′ ( t ) , g′ ( t ) , h′ ( t ) = f ′ ( t ) i + g′ ( t ) j + h′ ( t ) k
Example 2 Compute r′ ( t ) for r ( t ) = t 6 i + sin ( 2t ) j − ln ( t +1) k
...
1
r ′ ( t ) = 6t i + 2 cos ( 2t ) j −
k
t +1
5
Most of the basic facts that we know about derivatives still hold however, just to make it clear
here are some facts about derivatives of vector functions
...
A smooth curve is any curve for which r′ ( t ) is continuous and r′ ( t ) ≠ 0 for any t except
possibly at the endpoints
...
Finally, we need to discuss integrals of vector functions
...
∫ r ( t ) dt = ∫ f ( t ) dt , ∫ g ( t ) dt , ∫ h ( t ) dt
b
b
b
b
a
a
a
a
∫ r ( t ) dt = ∫ f ( t ) dt i + ∫ g ( t ) dt
b
b
b
a
a
a
j + ∫ h ( t ) dt k
b
a
With the indefinite integrals we put in a constant of integration to make sure that it was clear that
the constant in this case needs to be a vector instead of a regular constant
...
Example 3 Compute ∫ r ( t ) dt for r ( t ) = sin ( t ) , 6, 4t
...
∫ r ( t ) dt =
Example 4 Compute
− cos ( t ) , 6t, 2t
∫ r ( t ) dt for r ( t ) =
1
0
+c
2
sin ( t ) , 6, 4t
...
∫ r ( t ) dt = (
1
0
− cos ( t ) , 6t , 2t
2
)
1
0
= − cos (1) , 6, 2 − −1, 0, 0
= 1 − cos (1) , 6, 2