Velocity and Acceleration
Part 1: Limits, Derivatives, and Antiderivatives
In R3 ; vector-valued functions are of the form
r (t) = hf (t) ; g (t) ; h (t)i ; t in [a; b]
If f (t) ; g (t) ; and h (t) are continuous over [a; b], then r (t) ; t in [a; b] ; parameterizes a curve C. If r (t1 ) 6= r (t2 ) for all t1 6= t2 in (a; b) ; then C is an oriented
curve with the orientation going from r (a) to r (b).
Conversely, r (t) ; t in [a; b] ; is called a parametrization of C: Sometimes we
will refer to r (t) as a vector-valued dependent variable. The variable t is called
the parameter of r (t) :
EXAMPLE 1 Sketch the curve parametrized by r (t) = hcos (t) ; sin (t) ; ti
for t in [0; 2 ] :
Solution: The computer algebra system Maple yields the following
curve
(Note: Click and drag on the image above to see it from di¤erent
views).
1
Because we have little experience with 3-dimensional curves, eliminating t from
x = f (t) ; y = g (t) ; and z = h (t) tells us little about the curve. Instead,
we must develop the calculus of vector-valued functions in R3 as a means of
studying curves and their parameterizations.
To begin with, we de…ne the limit of a vector-valued function r (t) = hf (t) ; g (t) ; h (t)i
to be
lim r (t) =
lim f (t) ; lim g (t) ; lim h (t)
t!p
t!p
t!p
(1)
t!p
when each of the limits exist. As a result, we de…ne the derivative of r (t) =
hf (t) ; g (t) ; h (t)i to be
r (t +
dr
= lim
t!0
dt
t)
t
r (t)
which via the de…nition of the limit (1) yields
dr
=
dt
lim
f (t +
t!0
t)
t
f (t)
; lim
g (t +
t!0
t)
t
g (t)
; lim
h (t +
t!0
t)
t
h (t)
We often denote dr=dt by v and call it the velocity of r (t) : Likewise, the derivative of v (t) is often called the acceleration of r (t) : The computation above
shows us that we have the following:
Theorem 6.1: The velocity v (t) of the function r (t) = hf (t) ; g (t) ; h (t)i is
v (t) =
dr
= hf 0 (t) ; g 0 (t) ; h0 (t)i
dt
Likewise, the acceleration of r (t) is given by
a (t) =
dv
= hf 00 (t) ; g 00 (t) ; h00 (t)i
dt
If v (t) exists for each t in (a; b) ; then we say that r (t) is di¤ erentiable over
(a; b).
EXAMPLE 2
Compute the velocity and acceleration of
r (t) = t2 ; e2t ; t3
Solution: To begin with, the velocity of r (t) is
v (t) =
d 2 d 2t d 3
t ; e ; t
dt dt
dt
2
= 2t; 2e2t ; 3t2
As a result, the acceleration of r (t) is
a (t) =
dv
=
dt
d
d
d
2t; 2e2t ; 3t2
dt
dt
dt
= 2; 4e2t ; 6t
Check your Reading: What is v (t) if r (t) = t2 ; t2 ; t2 ?
Velocity as a Tangent Vector
If r (t) is the position vector of a moving object at time t; then the vector
r (t + t) r (t) is a secant vector whose initial and terminal points are on the
curve traced by the terminal end of r (t) :
Thus, as t approaches 0, the resulting secant vectors more and more resemble
the curve itself in the vicinity of r (t) ;
so that as
t approaches 0, the di¤erence quotient
r (t +
t)
t
r (t)
approaches the tangent vector shown in red above. Since the di¤erence quotient
also approaches v (t) as t approaches 0, we obtain the following theorem:
3
Theorem 6.2: The velocity vector v = dr=dt is tangent to the
parametrized curve at the point with position r (t).
A curve which has a non-zero tangent vector at each point is said to be smooth,
and a parameterization r (t) ; t in [a; b], is said to be smooth if it parameterizes
a smooth curve.
EXAMPLE 3
t3 ; t 2 :
Sketch v (1) at the point of tangency to r (t) =
Solution: To do so, we …rst di¤erentiate to obtain
v (t) =
dr
= 3t2 ; 2t
dt
We then let t = 1 to obtain
v (1) = h3; 2i
which when graphed versus r (t) yields the following:
4
Notice in example 3 that v (0) = h0; 0i, so r (t) = t3 ; t2 is not smooth at r (0) :
As the …gure shows, the curve has a cusp at r (0) and there is no tangent line
(or a tangent vector) at a cusp.
EXAMPLE 4 Sketch v (5 =2) at the point of tangency to the helix
r (t) = h4 cos (t) ; 4 sin (t) ; ti.
Solution: To do so, we notice that
v (t) = h 4 sin (t) ; 4 cos (t) ; 1i
so that the velocity vector at t = 5 =2 is
v
5
2
=
4 sin
5
2
; 4 cos
5
2
;1
= h 4; 0; 1i
which is graphed along with the curve below:
Check your Reading: Which plane is v ( ) parallel to if r (t) = hcos (t) ; sin (t) ; ti?
5
Antiderivatives of Vector-Valued Functions
In addition, we de…ne antiderivatives of q (t) = hf (t) ; g (t) ; h (t)i by
Z
Z
Z
Z
q (t) dt =
f (t) dt; g (t) dt; h (t) dt + hC1 ; C2 ; C3 i
where C1 ; C2 ; C3 are arbitrary constants. The fundamental theorem of calculus
then implies that
*Z
+
Z b
Z b
Z b
b
q (t) dt =
f (t) dt;
g (t) dt;
h (t) dt
a
a
a
a
or equivalently, if Q0 (t) = q (t) ; then
Z
b
q (t) dt = Q (b)
Q (a)
(2)
a
If a (t) is the acceleration of an object with an initial position r0 = r (0) and
an initial velocity v0 = v (0) ; then (2) impies that
Z t
a ( ) d = v (t) v (0)
0
where is the lowercase Greek letter tau and is used here because t is being
used as the parameter of v (t) : Similarly for r (t) ; so that
Z t
Z t
v (t) = v0 +
a( )d
and r (t) = r0 +
v( )d
0
0
are the velocity and position of the objects at time t:
EXAMPLE 5 Find the velocity and position of an object subject
to an acceleration of
a (t) = h 8 sin (2t) ; 8 cos (2t) ; 32i
given that v0 = h0; 0; 0i and r0 = h0; 0; 10i :
Solution: The velocity is given by
Z t
v (t) = h0; 4; 0i +
h 8 sin (2 ) ; 8 cos (2 ) ; 32i d
0
Z t
Z t
Z t
=
8
sin (2 ) d ; 8
cos (2 ) d ;
32d
0
0
0
D
E
t
t
t
=
4 cos (2 )j0 ;
4 sin (2 )j0 ;
32 j0
6
Thus, v (t) = h4 cos (2t) 4;
4 sin (2t) ; 32ti, so that
Z t
r (t) = h0; 0; 10i +
v( )d
0
Z t
Z t
(4 cos (2 ) 4) d ; 4
sin (2 ) d ; 10
=
0
E
D 0
t
t
=
2 sin (2 ) 4 j0 ; 2 cos (2 )j0 ; 10 16t2
=
2 sin (2t)
4t; 2 cos (2t)
2; 10
Z
t
32 d
0
16t2
A projectile is an object whose motion is due solely to the force of gravity. The
scientist Galileo showed that if a projectile moves only a short distance, then
its acceleration is practically constant. That is, its acceleration is given by
a (t) = g
(3)
where g is a constant vector called the acceleration due to gravity. The velocity
v (t) of the projectile follows from integration of (3) and a given initial velocity
v0 :
The position r (t) at time t follows from integration of v (t) and a given initial
position r0 :
EXAMPLE 6 Near the surface of Mars, g = 0; 0; 12:2 f t= sec2 :
Find the position of a projectile with an initial position of r0 =
h0; 0; 0i and an initial velocity of v0 = h10; 20; 64i.
Solution: To begin with, (3) says that a (t) = h0; 0; 12:2i : The
velocity is
Z
Z
v (t) = a (t) dt = h0; 0; 12:2i dt = h0; 0; 12:2ti + hC1 ; C2 ; C3 i
7
and since v0 = h10; 20; 64i ; we must have
h10; 20; 64i = v (0) = h0; 0; 12:2 0i + hC1 ; C2 ; C3 i = hC1 ; C2 ; C3 i
Thus, the velocity of the object is
v (t) = h0; 0; 12:2ti + h10; 20; 64i = h10; 20; 12:2t + 64i
To …nd r (t) ; we antidi¤erentiate again to obtain
Z
Z
r (t) = v (t) dt = h10; 20; 12:2t + 64i dt = 10t; 20t; 6:1t2 + 64t +hC1 ; C2 ; C3 i
Since r0 = h0; 0; 0i ; we must have hC1 ; C2 ; C3 i = h0; 0; 0i ; so that
the position of the object at time t is
r (t) = 10t; 20t; 64t
6:1t2
Check your Reading: Would you weigh more or less if you were
on Mars?
Plane of Motion and Maximum Altitude
Let’s suppose that r (t) is the position of a projectile at time t. Since v and a
span the same plane at each time t, the motion of the projectile is in the plane
through the point r0 with normal
n = v0
a
The maximum altitude occurs when v is horizontal, which is when the 3rd
component of v is identically zero. The time to maximum altitude is denoted
tmax and is obtained by setting the third component of v equal to 0 and solving
for t:
8
Finally, the point at which the projectile “strikes”the ground (i.e., the xy-plane)
is the point at which the third component of r (t) is equal to 0. Setting the third
component of r (t) equal to 0 and solving for t thus yields the time of impact,
denoted timp : The point of impact is r (timp ) :
EXAMPLE 7
Suppose a projectile has a position at time t of
16t2
r (t) = 10t; 20t; 64t
Find the plane of motion, the time to maximum height, the maximum height, the time of impact, and the point of impact of the
projectile.
Solution: To begin with, the velocity of the object is
v (t) = h10; 20; 64
32ti
and the acceleration is as expected, a (t) = h0; 0; 32i : Since
n = v0
a = h10; 20; 64i
h0; 0; 32i = h 640; 320; 0i
the plane of motion is the plane through r0 = (0; 0; 0) with normal
n = h 640; 320; 0i : Thus, the equation of the plane of motion is
640 (x
0) + 320 (y
0) + 0 (z
0) = 0
which is the vertical plane passing through the line y = 2x:
To …nd tmax ; we set the third component of v (t) equal to 0 and solve
for tmax :
64
32t
32t
tmax
=
=
=
0
64
2
Since tmax = 2; the maximum altitude itself is simply the z-coordinate
of r (tmax ) : Since
r (tmax ) = r (2) = h20; 40; 128
64i = h20; 40; 64i
the maximum altitude of the projectile is 64 feet. Finally, the time
of impact occurs when the third component of r (t) is equal to 0:
64t 16t2 = 0
16t (t 4) = 0
t = 0 or t = 4
9
Initially, the projectile is on the ground (i.e., r0 = h0; 0; 0i ). Thus,
the time of impact must be timp = 4; so that the point of impact is
r (timp ) = r (4) = (40; 80; 0)
Exercises
Find the velocity and acceleration of the following vector-valued functions:
1.
3.
5.
7.
9.
11.
13.
r (t) = t2 ; t3 ; t4
1
2
2.
3=2
r (t) = t ; t ; t
r (t) = h3 cos (t) ; 5 sin (t) ; 4 cos (t)i
r (t) = htan (t) ; cot (t) ; csc (t)i
r (t) = h1; 1; ln [sec (t)]i
r (t) = t2 i + tan 1 (t) j
r (t) = e t hsin (t) ; cos (t) ; ti
4.
6.
8.
10.
12.
14.
r (t) = Dt2 ; t5 ; t4
E
1=2
1
r (t) = t2 1
; t2 + 1
;1
r (t) = h3 cos (t) ; 5 sin (t) ; 4 cos (t)i
r (t) = et ; e t ; ln t2 + 1
r (t) = h2; 3; 1i et h1; 4; 7i e t
r (t) = i + t j + t2 k
r (t) = tan (t) hsin (t) ; cos (t) ; sec (t)i
Find the velocity at the given time and sketch it along with the curve r (t) :
15.
17.
19.
21.
r (t) = t2 ; t4 ; t = 1
r (t) = hcos (t) ; sin (t)i ; t = 6
r (t) = het ; e t i ; t = 0
r (t) = h3t + 1; 2t + 3i ; t = 2
16.
18.
20.
22.
10
r (t) = 2t; 64 16t2 ; t = 1
r (t) = hcos (t) ; sin (t)i ; t = 6
r (t) = htan (t) ; sec (t)i ; t = 4
r (t) = bt2 ; 2bt ; t = 1
Find the velocity v (t) and the position r (t) for the given acceleration and initial
conditions, and determine the plane of motion (if it exists). In exercises 23 - 26,
also …nd the time to maximum height, the maximum height, the time of impact,
and the point of impact of the projectile (note: 27 and 28 are projectile motions
with a small amount of atmospheric drag).
23.
25.
27.
29.
a (t) = h0; 0; 32i
r0 = h0; 0; 0i ; v0 = h1; 2; 64i
a (t) = h0; 0; 9:8i
r0 = h1; 3; 0i ; v0 = h72; 38; 65i
a (t) = 0; 0; 32e t=10
r0 = h0; 0; 64i ; v0 = h32; 32; 0i
a (t) = h 3 sin (t) ; 5 cos (t) ; 4 sin (t)i
r0 = h0; 0; 0i ; v0 = h0; 0; 0i
24.
26.
28.
30.
a (t) = h0; 0; 32i
r0 = h0; 0; 0i ; v0 = h1; 1; 96i
a (t) = h0; 0; 9:8i
r0 = h2; 1; 3i ; v0 = h1; 1; 0i
a (t) = 0; 0; 32e t=20
r0 = h0; 0; 4i ; v0 = h0; 0; 0i
a (t) = h 3 sin (t) ; 3 cos (t) ; 12:2i
r0 = h0; 0; 0i ; v0 = h0; 0; 0i
31. In example 6 a projectile is moving near the surface of Mars, and we found
that its position at time t is given by
r (t) = 10t; 20t; 64t
6:1t2
Find the plane of motion, the time to maximum height, the maximum height,
the time of impact, and the point of impact of the projectile, and then compare
the result to example 7, in which a projectile with the same initial conditions is
traveling near the earth.
32. When r (0) is in the xy-plane, then the range of the projectile— i.e., the
distance the projectile travels between leaving the earth and striking the earth—
is the length of the vector r (timp ) r (0) : That is,
range = kr (timp )
r (0)k
Find the range of the projectile in example 7, and then …nd the range of the
projectile in problem 31. How much farther would the projectile travel on Mars
11
than it would on the earth?
Exercises 33 - 38 explore Major League Baseball’s "GameDay" pitch tracking
system, in which the trajectory of a pitch is calculated using an imaging system,
the coordinate system below,
and projectile motion. Speci…cally, the Game Day pitch tracking system estimates the trajectory of a baseball in terms of a constant acceleration a (shown
in black in the image above), a position r0 when tracking begins (the beginning
of the blue curve), and an initial velocity v0 when tracking begins (40, 50, or
55 feet from the back of home plate).
33. On September 9, 2009 in a game between Anaheim and Seattle, Mark
Jipsen through Ken Gri¤ey, Jr., a sweeping curveball with the following gameday
vectors:
r0 = h 2:924; 50; 5:895i ; v0 = h2:264; 121:647; 2:486i
a = h10:529; 30:971; 41:439i
Find the parameterization of the ball’s trajectory r (t) by integrating a twice
and using the given initial data. What is the plane of motion of the curveball?
What is its position as it crosses the front of the plate ( i.e., when y = 1:417
feet)? How much did it move from left to right
34. Repeat Exercise 33 for Mark Jipsen’s last pitch of that game, which is
r0 = h 2:901; 50; 5:773i ; v0 = h8:42; 141:714; 4:76i
a = h 1:669; 36:404; 13:16i
Is this also a curveball (in your opinion)?
35. On July 3, 2009 in a game between Minnesota and Detroit, Joel Zumaya
threw a pitch clocked at well over 100 mph (and possibly the fastest pitch ever
recorded – a possible 105 mph when it left his hand at approximately y = 55
feet). The data for that pitch is
12
r0 = h 1:901; 50; 5:928i ; v0 = h9:588; 147:939; 5:642i
a = h 15:819; 39:136; 6:834i
How fast was the pitch traveling initially (i.e., when y = 50 feet)? How long
did it take to travel from y = 55 (approximate point of release) to y = 1:417
feet (the front of the plate)? How fast was it traveling when it crossed the front
of the plate? (hint: 1 mile = 5280 feet).
36. Try it out! The data and information above can be obtained by beginning
at Alan M. Nathan’s website http://webusers.npl.illinois.edu/~a-nathan/pob/tracking.htm.
Try …nding the information above for an outing of your favorite pitcher and then
repeating exercise 33.
37. Magnus Force: The pitches in exercise 33 to 36 do not fall straight down
because the spin of a baseball creates a force called the Magnus force which is
perpendicular both to the direction of motion and to the axis about which the
ball spins.
For example, an angular velocity of about 1500 revolutions per minute induces
a Magnus force which is about 1/3 the acceleration due to gravity –i.e., about
11 feet per sec per sec! Drag must also be considered when Magnus force is
considered, giving us an approximate acceleration of
a = h11; 32; 32i
13
Use this information to construct the trajectory of such a baseball given that
r0 = h0; 50; 6i and v0 = h0; 132; 0i
(i.e., ball 6 feet above the ground traveling 90 mph at 50 feet from the back of
home plate).
38. Magnus Force Theory: In reality, the drag, magnus, and gravitational
forces lead to an acceleration vector of the form
a=
KCD vv
KCL v (v
!)
g
where K is a constant equalt to about 5:44 10 3 per foot, v is the magnitude
of the velocity v vector, g is the acceleration due to gravity, ! is the unit vector
parallel to the axis of rotation of the ball, and CD and CL are the drag and
lift constants, respectively. Explain why a cannot be constant (i.e., the same at
all times), and use this to explain why the Gameday curves are only statistical
approximations and not actual physical trajectories. (more with Magnus force
in later sections!)
39. When r0 = h0; 0; 0i ; then projectile motion begins at the origin. Show
that if r0 = h0; 0; 0i ; then timp = 2tmax :
40. * A Basketball’s Initial Velocity: An NBA player shoots a 3-pointer
from 24 feet ( 3 point line is at 23’9”), and 1.4 seconds elapse before it passes
through the hoop. If we assume that the player released the ball from a height
of 8 feet, then what is the initial velocity of the ball? And what is the ball’s
maximum altitude?
41. Bezier Curves: The Bezier curve determined by the 4 points P0 (x0 ; y0 ) ;
P1 (x1 ; y1 ) ; P2 (x2 ; y2 ) ; and P3 (x3 ; y3 ) is the curve with endpoints P0 and P3
!
!
which is tangent to the vectors P0 P1 and P2 P3 at the points P0 and P3 ; respec-
14
tively.
In this exercise, we show that a Bezier curve is the graph of the vector valued
function
r (t) = hx0 ; y0 i t3 + 3 hx1 ; y1 i t2 (1
t) + 3 hx2 ; y2 i t (1
2
t) + hx3 ; y3 i (1
3
t)
for t in [0; 1] :
1. (a) Compute r (0) and r (1) :
(b) Compute v (t) ; and then compute v (0) and v (1) :
(c) Explain how (a) and (b) relate to the …gure shown above.
42. Bezier Curves: Construct the Bezier curve using the parameterization
in exercise 39 using the 4 points P0 (0; 0) ; P1 (0; 1) ; P2 (1; 1) ; and P3 (1; 0) ; and
!
!
then sketch the result along with the vectors P0 P1 and P2 P3 :
43. Write to Learn: Bezier curves are often used to connect a sequence of
points (position vectors) P0 ; P1 ; : : : ; Pn with a curve subject to a set of control
15
points (position vectors) Q0 ; : : : ; Qn
1 ; R0 ; : : : ; Rn 1 :
Write short essay in which you show that for each j = 1; : : : ; n, the curve
rj (t) = Pj
1t
3
+ 3Rj
1t
2
for t in [0; 1] satis…es rj (0) = Pj
vj (0) = 3 (Rj
1
(1
1
t) + 3Qj
1 t (1
2
t) + Pj (1
3
t)
and rj (1) = Pj , as well as
Pj ) ; and vj (1) = 3 (Pj
1
Qj
1)
What conditions on Pj ; Rj 1 and Qj , j = 0; : : : ; n 1; are necessary for the
collection of "pieces" r1 (t) ; : : : ; rn (t) to form a smooth curve connecting all the
points P0 ; P1 ; P2 ; : : : ; Pn .
44.Write to Learn: Suppose that r (t) = hf (t) ; g (t) ; h (t)i ; t in [a; b] ; parametrizes a curve C; and suppose that is a di¤erentiable, 1-1 function. Write
a short essay explaining why
r (t) = hf ( (u)) ; g ( (u)) ; h ( (u))i ; u in [c; d]
parameterizes the same curve C when c = 1 (a) and d = 1 (b) : What is the
velocity vector for the new parameterization? How is it related to the velocity
vector for the original parameterization?
16