We interpret vector-valued functions as paths of objects in space.
Position, velocity, and acceleration
From single-variable calculus, we know that if is a function that represents the velocity (signed speed) of an object at time , then
- tells us the acceleration (instantaneous change in velocity) of the object, and
- tells us the displacement (position with respect to an origin) of the object.
There is a similar story to be told with vector-valued functions.
- tells us the acceleration (instantaneous change in velocity) of the object, and
- tells us the displacement (position with respect to an origin) of the object.
Additionally, the speed of the object is the magnitude of velocity, .
Note, from the definition above, we also see that if the position of an object with respect to some origin is given by a vector-valued function , then
- gives the instantaneous velocity of the object at time , and
- give the instantaneous acceleration of the object at time .
Now let’s see an example.
- Find the position function that describes this situation.
- Find the acceleration of the ball and give a physical interpretation.
- a circle with radius ,
- centered at the origin of the -plane,
- that makes a full revolution every seconds.
However, we want two revolutions per second, not one revolution every seconds. We modify the period by multiplying by . The final position function is To find , we differentiate twice.
For the physical interpretation, recall the classic physics equation, “Force mass acceleration.” A force acting on a mass induces acceleration (i.e., the mass moves); acceleration acting on a mass induces a force (gravity gives our mass a weight). Thus force and acceleration are closely related. A moving ball “wants” to travel in a straight line. Why does the ball in our example move in a circle? It is attached to your hand by a string. The string applies a force to the ball, affecting it’s motion: the string accelerates the ball. This is not acceleration in the sense of “it travels faster;” rather, this acceleration is changing the velocity of the ball. In what direction is this force/acceleration being applied? In the direction of the string, towards your hand.
Hence it makes sense is parallel to , but has a different magnitude and points in the opposite direction. The magnitude of the acceleration is related to the speed at which the ball is traveling. A ball whirling quickly is rapidly changing direction/velocity. When velocity is changing rapidly, the acceleration must be smalllarge .
Projectile motion
An important application of vector-valued position functions is projectile motion: the motion of objects under the influence of gravity. We will measure time in seconds, and distances will either be in meters or feet. We will show that we can completely describe the path of such an object knowing its initial position and initial velocity (where it is and where it is going.)
Suppose an object has initial position and initial velocity of Here, is often called the angle of elevation. Since the acceleration of the object is known, namely where is the gravitational constant, we can find knowing our two initial conditions. We first find :
We integrate once more to find :You can adjust the initial position, , angle, magnitude of the velocity, and magnitude of the acceleration below:
We demonstrate how to solve for a position function in the context of projectile motion in the next example.
From distance traveled to arc length
Consider a driver who sets her cruise-control to , and travels at this speed for an hour. We can ask:
- How far did the driver travel?
- How far is the driver from her starting position?
The first is easy to answer: she traveled miles. The second is impossible to answer with the given information. We do not know if she traveled in a straight line, on an oval racetrack, or along a slowly-winding highway.
This highlights an important fact: to compute distance traveled, we need only to know the speed, given by .
This theorem is a specific instance of the more general theorem for arc length:
- , , and are continuous.
- The curve defined by is traversed once for .
The arc length of the curve from is given by
One more example, again interpreting our vector-valued function as giving the position of an object in space.
Find:
- An integral that computes the distance traveled by the particle on .
- The displacement of the particle on .
- The particle’s average speed.
The displacement is the vector
That is, the particle ends with an -value increased by and with unchanged -values the same.We found above that the particle traveled over seconds. We can compute average speed by dividing: .
Looking back
Finally, let’s think about what we learned in our previous calculus courses, in terms of what we know now.
Arc length
When you first learned how to compute arc length in calculus, you probably use a formula like next you learned that for parametric functions, the arc length was Note, the first formula, is really just the second one, where so that and . Finally in this class, we view arc length as Since the magnitude of a vector can be given via the dot product, this is a very general formula.
Average value
In your first calculus course, you defined the average value of a function to be Above we computed the average speed as that is, we just found the average value of on .
Likewise, given position function , the average velocity on is
that is, it is the average value of , or , on . As we learn new material, we must constantly reconcile, and reintegrate what have learned before.