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Differentiation and Integration of Vector-Valued Functions
A graph of the vector-valued function ( ) ( ) ( )
2 cos cos , 2 cos sin 2 2
r t t t t t
⎝ ⎝^ ⎠^ ⎠ ⎝ ⎝^ ⎠⎠
G
is
shown in Figure 1. Find the values of r t ( )
G
and r ′( t )
G
at t = 0, , 2
t
= t = π,and
t
Sketch each of the derivative vectors with their tails at the corresponding point on r t ( )
G
.
x
y
Figure 1: r t ( )
G
Table 1: r t ( )
G
and r ′( ) t
t r t ( )
G
r ′( t )
Three properties of r ′( t 0 )
G
when r ′ ( t 0 )≠ 0
G G
.
Figure 2 shows the function r ( ) t = sin ( ) t , cos( ) t
G
along with the
velocity vector ( v t ( ) = r ′( ) t
G G
) and acceleration vector (^) ( a t ( ) = r ′′( t ))
G G
at three different values of t.
Establish that the path described by r t ( )
G
is truly circular.
What appears to be the relationship between v t ( )
G
and a t ( )
G
at every value of t? Confirm this
relationship.
Figure 2
Figure 4 shows the function ( )
2 / 3 3/ 2 3 1 1 2 / 3 3 1 1 , 3 2
t (^) t r t
⎣ ⎦ +^ −
G
along with the velocity
vector and acceleration vector at three different values of t.
Upon what curve does this motion take place?
What appears to be the relationship between v t ( )
G
and a t ( )
G
at every value of t? Confirm this
relationship.
Figure 4
Show that the speed functions for both r 1 ( t ) = sin ( ) t , cos( ) t
G
and
2 / 3 3 / 2 2 / 3
2
t (^) t r t
⎣ ⎦ +^ −
G
are constant and that the speed function for
2 2 r 3 (^) t = sin t , cos t
G
is not constant.
Net displacement vs. distance travelled
In MTH 252 we talked about motion that takes place along a line. In that context, a position
function is a scalar function.
If s ( ) t is a scalar position function, then v t ( ) = s ′( t )is that function’s velocity function. Since
the motion is linear, the velocity value can describe the direction of motion with its sign; one
direction corresponding to positive velocity and the other to negative velocity.
In this context, we define the net displacement over the time interval (^) [ a b , (^) ]to be s (^) ( b (^) ) − s a ( ).
s ( b ) − s a ( ) is the distance between the starting and ending points of the motion and the sign on
s ( b ) − s a ( ) indicates the direction of the net movement (for example, a positive sign generally
means that the net movement was either rightward or upward).
From the Total Change Theorem (a weak statement of The Fundamental Theorem of Calculus), we
know that ( ) ( ) ( ) ( )
b b
a a
s b − s a = s ′ t dt = v t dt ∫ ∫ ; in other words, integrating the velocity function
over [ a b , ] gives us the net displacement over [ a b , ].
We can create a contrived situation to come up with the actual distance travelled over (^) [ a b , ] ; if all
of the motion is in the positive direction, then the distance traveled is equal to the net
displacement. We can force all of the motion into the positive direction by forcing the velocity to
always be positive, i.e., by taking the absolute value of the velocity. Thus, the distance traveled is
given by ( )
b
a
v t dt ∫
. Since v t ( ) is the speed function, we conclude that integrating the speed
function over (^) [ a b , ] results in the total distance traveled during the time interval (^) [ a b , ].
In Figure 5, the net displacement and total distance traveled
over (^) [ 0, 3] are, respectively:
3
0
s ′ t dt = 3 ∫
and ( )
3
0
s ′ t dt = 5 ∫ .
Net displacement over (^) [ 0,2]
Net displacement over (^) [ 0, 3] Net displacement over (^) [ 2, 3]
Figure 5: ( )
2 s t = 4 t − t
When working with a vector valued function that describes motion, we get similar results when
integrating the velocity and speed functions. The only difference is that the net displacement is a
vector rather than a scalar; geometrically, we can think of the net displacement as an arrow that
points from the point of origin to the point of termination.
The function r ( ) t = sin( 4 t ), cos( 3 t )
G
is shown in figures 5 and 6. If we think of that function as
describing a microbe moving along the curve, the microbe journeys between the points labeled A and
B over each of the time intervals
⎡ π^ π⎤ ⎢ ⎥ ⎣ ⎦
and
⎡ π^ π⎤ ⎢ ⎥ ⎣ ⎦
. Let’s verify on our calculators that
the net displacement vector over each time interval is indeed given by ( ) ( )
b b
a a
v t dt = r ′ t dt ∫ ∫
G G
and
let’s illustrate the net displacement as a vector subtraction on Figure 5.
Let’s illustrate symbolically why there is nothing even remotely surprising about the fact that
b
a
v t dt ∫
G
gives us the net displacement vector over (^) [ a b , ].
B
A
Figure 5: r ( t ) = sin 4( t ) , cos 3( t )
G
