11.3 Motion in Space

We are finally at a point where we have sufficient mathematical machinery to describe the motion of an object in a three-dimensional setting. Problems such as this were among the earliest and most important applications of the calculus and continue to be of great importance today. For instance, if you launch a rocket (e.g., a space shuttle), you probably want to know where it will go. Problems such as this, dealing with motion, were one of the primary focuses of Newton and many of his contemporaries. While Newton certainly wasn't thinking about launching space shuttles into orbit, he used his newly invented calculus to explain all kinds of motion, from the motion of a projectile (such as a ball) hurled through the air, to the motion of the planets. His stunning achievements in this field unlocked mysteries that had eluded the greatest minds for centuries and form the basis of our understanding of mechanics today. In this section, we use vector-valued functions to describe objects in motion.

Suppose that an object moves along a curve described parametrically by

C: x = f (t ), y = g(t ), z = h(t ),
where t  represents time and where t [ a, b] . Recall that we can think of the curve as being traced out by the endpoint of the vector-valued function
We observed in section 11.2 that the value of r'(t ) for any given value of t  is a tangent vector pointing in the direction of the orientation of the curve. We can now give another interpretation of this. From (2.3), we have
and the magnitude of this vector-valued function is
(Where have you seen this expression before?) Notice that from (1.4), given any number t 0 [ a, b], the arc length of the portion of the curve from u = t 0 up to u = t  is given by
(3.1)
Part II of the Fundamental Theorem of Calculus says that if we differentiate both sides of (3.1), we get
Think about the physical interpretation of s' (t ). Since s(t ) represents arc length, s' (t ) gives the instantaneous rate of change of arc length with respect to time, that is, the speed of the object as it moves along the curve. So, for any given value of t, r' (t ) is a tangent vector pointing in the direction of the orientation of C (i.e., the direction followed by the object) and whose magnitude gives the speed of the object. What would you call r ' (t )? We call it the velocity vector, usually denoted v(t ). As in the case of one-dimensional motion, we refer to the derivative of the velocity vector v' (t ) = r '' (t ) as the acceleration vector, denoted a (t ) . When drawing the velocity and acceleration vectors, we locate both of their initial points at the terminal point of r(t ) (i.e., at the point on the curve), as shown in Figure 11.11.



Figure 11.11
Position, velocity and acceleration vectors.

3.1   
Finding Velocity and Acceleration Vectors
 
Find the velocity and acceleration vectors if the position of an object moving in the xy - plane is given by
 
 
We have
In particular, this says that at t = 1, we have , and . We plot the curve and these vectors in Figure 11.12.



Figure 11.12
Position, velocity and acceleration vectors.
 

Just as in the case of one-dimensional motion, the relationship between position, velocity and acceleration vectors goes both ways. That is, given the acceleration vector, we can determine the velocity and position vectors, provided we have some additional information.

3.2   
Finding Velocity and Position from Acceleration
 
Find the velocity and position of an object at any time t, given that its acceleration is , its initial velocity is and its initial position is .
 
 
Since a(t ) = v '(t ), we can integrate once to obtain
  = 3t 2 i+(6t 2+2t ) j+et  k+c1,
where c1 is an arbitrary constant vector. To determine the value of c1 , we use the initial velocity:
so that . This gives us the velocity
v(t ) = (3t 2+2)i+(6t 2+2t )j+et k.
Since v(t ) = r'(t ), we integrate once again, to obtain
  = (t 3+2t )i+(2t 3+t 2)j+et k+c2,
where c2 is an arbitrary constant vector. We can use the initial position to determine the value of c2, as follows:
so that . This gives us the position vector
r(t ) = (t 3+2t )i+(2t 3+t 2+3)j+(et +4)k.
We show the curve and indicate sample vectors for r(t ), v (t ) and a(t ) in Figure 11.13.



Figure 11.13
Position, velocity and acceleration vectors.
 

We have already seen Newton¹s second law of motion several times now. In the case of one-dimensional motion, we had that the net force acting on an object equals the product of the mass and the acceleration ( F = ma ). In the case of motion in two or more dimensions, we have the vector form of Newton's second law:
F = ma.
Here, m is the mass, a is the acceleration vector and F is the vector representing the net force acting on the object.

3.3   
Finding the Force Acting on an Object
 
Find the force acting on an object moving along a circular path of radius b centered at the origin, with constant angular speed.
 
 
Here, by constant angular speed, we mean that if is the angle made by the position vector and the positive x - axis and t  is time (see Figure 11.14a, where the indicated orientation is for the case where > 0 ), then we have that



Figure 11.14a
Motion along a circle.

Notice that this says that = t+c , for some constant c. Further, we can think of the circular path as the curve traced out by the endpoint of the vector-valued function
Notice that the path is the same for every value of c. (Think about what the value of c affects.) For simplicity, we take = 0 when t = 0, so that = t  and
Now that we know the position at any time t, we can differentiate to find the velocity and acceleration. We have
and
 
From Newton's second law of motion, we now have
F(t ) = ma(t ) = -m 2r(t ).
Notice that since m 2 > 0, this says that the force acting on the object points in the direction opposite the position vector. That is, at any point on the path, it points in toward the origin (see Figure 11.14b). We call such a force a centripetal (center-seeking) force. Finally, observe that on this circular path, so that at every point on the path, the force vector has constant magnitude:



Figure 11.14b
Centripetal force.
 



 Notice that one consequence of the result F(t ) = -m 2 r(t ) from example 3.3 is that the magnitude of the force increases as the rotation rate increases. You have experienced this if you have been on a roller coaster with tight turns or loops. The faster you are going, the stronger the force that your seat exerts on you.

Just as we did in the one-dimensional case, we can use Newton's second law of motion to do more than simply identify the force acting on an object with a given position function. It's much more important to be able to determine the position of an object given only a knowledge of the forces acting on it. For instance, one of the most significant problems faced by the military is how to aim a projectile (e.g., a missile) so that it will end up hitting its intended target. This problem is harder than it sounds, even when the target is standing still. When the target is an aircraft moving faster than the speed of sound, the problem presents significant challenges. In any case, the calculus can be used to arrive at a meaningful solution. We present the simplest possible case (where neither the target nor the source of the projectile are moving) in the following example.

3.4   
Analyzing the Motion of a Projectile
 
A projectile is launched with an initial speed of 140 feet per second from ground level at an angle of to the horizontal. Assuming that the only force acting on the object is gravity (i.e., there is no air resistance, etc.), find the maximum altitude, the horizontal range and the speed at impact of the projectile.
 
 
From Newton's second law of motion, we have
F(t ) = ma(t ).
Notice that here, the motion is in a single plane (so that we need only consider two dimensions) and the only force acting on the object is the force of gravity, which acts straight down. In this case, we have that a(t ) is simply the acceleration due to gravity. Although this is not constant, it is nearly so at altitudes reasonably close to sea level. We will assume that
a(t ) = -gj,
where g is the constant acceleration due to gravity, g 32 feet/second 2. So, we have
v' (t ) = a(t ) = -32j.
Integrating this once gives us
(3.2)
where c1 is an arbitrary constant vector. If we knew the initial velocity vector v(0) , we could use this to solve for c1, but we only know the initial speed (i.e., the magnitude of the velocity vector). Referring to Figure 11.15a, notice that you can read off the components of v(0). From the definitions of the sine and cosine functions, we have



Figure 11.15a
Initial velocity vector.

From (3.2), we now have
Substituting this back into (3.2), we have
(3.3)
Integrating (3.3) will give us the position vector
where c2 is an arbitrary constant vector. Since the initial location was not specified, we choose it to be the origin. (This is usually most convenient.) This gives us
0 = r(0) = c2,
so that
(3.4)
We show a graph of the path of the projectile in Figure 11.15b.



Figure 11.15b
Path of a projectile.

Now that we have found expressions for the position and velocity vectors for any time, we can answer the physical questions. Notice that the maximum altitude occurs at the instant when the object stops moving up (just before it starts to fall). This says that the vertical ( j ) component of velocity must be zero. From (3.3), we get
0 = 70-32t,
so that the time at the maximum altitude is
The maximum altitude is then found from the vertical component of the position vector at this time:
To determine the horizontal range, we first need to determine the instant at which the object strikes the ground. Notice that this occurs when the vertical component of the position vector is zero (i.e., when the height above the ground is zero). From (3.4), we have that this occurs when
There are two solutions of this equation: t = 0 (the time at which the projectile is launched) and (the time of impact). The horizontal range is then the horizontal ( i ) component of position at this time:
Finally, the speed at impact is the magnitude of the velocity vector at the time of impact:
 

You might have noticed in example 3.4 that the speed at impact was the same as the initial speed. Don't expect this to always be the case. Generally, this will only be true for a projectile of constant mass that is fired from ground level and returns to ground level and that is not subject to air resistance or other forces.

Equations of Motion

We now derive the equations of motion for a projectile in a slightly more general setting than that described in example 3.4. Consider a projectile fired from an altitude h above the ground at an angle to the horizontal and with initial speed v0. We can use Newton's second law of motion to determine the position of the projectile at any time t  and once we have this, we can answer any questions about the motion.

We again start with Newton's second law and assume that the only force acting on the object is gravity. We have

F(t ) = ma(t ),
where F(t ) = -mg j. This gives us (as in example 3.4)
v' (t ) = a(t ) = -gj.
(3.5)
Integrating (3.5) gives us
(3.6)
where c1 is an arbitrary constant vector. In order to solve for c1, we need the value of v(t ) for some t, but we are given only the initial speed v0 and the angle at which the projectile is fired. Notice that from the definitions of sine and cosine, we can read off the components of v(0) from Figure 11.16a. From this and (3.6), we have



Figure 11.16a
Initial velocity.

This gives us the velocity vector
(3.7)
Since r' (t ) = v(t ), we integrate (3.7) to get the position:
To solve for c2, we want to use the initial position r (0), but we're not given it. We're only told that the projectile starts from an altitude of h feet above the ground. If we select the origin to be the point on the ground directly below the launching point, we have
so that
 
(3.8)
Notice that the path traced out by r(t ) (from t = 0 until impact) is a portion of a parabola (see Figure 11.16b).



Figure 11.16b
Path of the projectile.

Now that we have derived (3.7) and (3.8), we have all we need to answer any further questions about the motion. For instance, if we need to know the maximum altitude, this occurs at the time at which the vertical ( j ) component of velocity is zero (i.e., at the time when the projectile stops rising). From (3.7), we solve

0 = v0sin -gt,
so that the time at which the maximum altitude is reached is given by
 Time to reach maximum altitude
The maximum altitude itself is the vertical component of the position vector at this time. From (3.8), we have
Maximum
altitude
 
  Maximum altitude
To find the horizontal range or the speed at impact, we must first find the time of impact. To get this, we set the vertical component of position to zero. From (3.8), we have
Notice that this is simply a quadratic equation for t. Given v0, and h, we can solve for the time t  using the quadratic formula.

In all of the foregoing analysis, we left the constant acceleration due to gravity as g. You will usually use one of the two approximations:

g 32 ft/sec2 or g 9.8 m/sec2.
When using any other units, simply adjust the units to feet or meters and the time scale to seconds or make the corresponding adjustments to the value of g.

In the following example, we examine a fully three-dimensional projectile motion problem for the first time.

3.5   
Analyzing the Motion of a Projectile in Three Dimensions
 
A projectile of mass 1kg is launched from ground level toward the east at 200 meters/ second, at an angle of to the horizontal. If a gusting northerly wind applies a steady force of 2 newtons to the projectile, find the landing location of the projectile and its speed at impact.
 
 
Notice that because of the cross wind, the motion is fully three-dimensional. We orient the x -, y -, and z - axes so that the positive y - axis points north, the positive x - axis points east and the positive z - axis points up, as in Figure 11.17a, where we also show the initial velocity vector and vectors indicating the force due to the wind. The two forces acting on the projectile are gravity (in the negative z - direction with magnitude 9.8m = 9.8 newtons) and wind (in the y - direction with magnitude 2 newtons). Newton's second law is F = m a = a . We have



Figure 11.17a
The initial velocity and wind velocity vectors.

Integrating gives us the velocity function
(3.9)
where c1 is an arbitrary constant vector. Note that the initial velocity is
From (3.9), we now have
This gives us
We integrate this to get the position vector:
for a constant vector c2. Taking the initial position to be the origin, we get
0 = r(0) = c2,
so that
(3.10)
Note that the projectile strikes the ground when the k component of position is zero. From (3.10), we have that this occurs when
0 = 100t-4.9t 2 = t (100-4.9t ).
So, the projectile is on the ground when t = 0 (time of launch) and when 20.4 seconds (the time of impact). The location of impact is then the endpoint of the vector and the speed of impact is
 
We show a computer-generated graph of the path of the projectile in Figure 11.17b. In this figure, we also indicate the shadow made by the path of the projectile on the ground. In Figure 11.17c, we show the projection of the projectile's path onto the xz plane. Observe that this parabola is analogous to the parabola shown in Figure 11.15b.



Figure 11.17b
Path of the projectile.



Figure 11.17c
Projection of path onto the xz plane.
 

© 2002 McGraw-Hill Companies, Inc.