We interpret vector-valued functions as paths of objects in space.

1 Position, velocity, and acceleration

From single-variable calculus, we know that if \(v(t)\) is a function that represents the velocity (signed speed) of an object at time \(t\), then

  • \(v'(t)\) tells us the acceleration (instantaneous change in velocity) of the object, and
  • \(\int _a^b v(t) \d t\) 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.

An object is moving with velocity \(\vec {v}(t) = \vector {t^2-4t,4-t^2}\) for \(0\le t\le 2\). Let speed be measured in the units of “feet-per-second.” Give a vector valued formula for the acceleration of our object at time \(t\). The acceleration is given by \(\vector {\answer {2t-4},\answer {-2t}}\) feet per second per second.
Supposing further that our object starts at the point \((2,3)\) relative to the origin, give a vector valued formula for the position of of our object at time \(t\). The position is given by \(\vector {\answer {t^3/3-2t^2+2},\answer {3+4t-t^3/3}}\).
What is speed of our object at time \(t\)? The speed of our object is \(\answer {\sqrt {2t^4-8t^3+8t^2+16}}\) feet per second.
When is the object’s speed maximized?
It might be easiest to maximize the square of the formula for speed.
The speed is maximized when \(t=\answer {1}\).

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 \(\vec {p}(t)\), then

  • \(\vec {p}'(t)\) gives the instantaneous velocity of the object at time \(t\), and
  • \(\vec {p}''(t)\) give the instantaneous acceleration of the object at time \(t\).

Now let’s see an example.

An object moves in a spiral with position function
\[ \vec {p}(t) = \vector {\cos t, \sin t, t}, \]
where distances are measured in meters and time is in minutes. Describe the object’s velocity and acceleration at time \(t\).
\begin{align*} \vec {v}(t) &= \vector {\answer {-\sin t}, \answer {\cos t},\answer {1}} \\ \vec {a}(t) &= \vector {\answer {-\cos t}, \answer {-\sin t},\answer {0}}. \end{align*}
What is the speed of this object?
\[ |\vec {v}(t)| = \answer {\sqrt {2}}\unit {m}/\unit {min} \]
What is the angle between \(\vec {v}\) and \(\vec {a}\)?
\[ \text {The angle equals } \answer {\pi /2}\unit {radians} \]

2 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

\[ \vec {p}(0) = \vector {x(0),y(0)} \]

and initial velocity of

\[ \vec {v}(0) = v_0\vector {\cos (\theta ),\sin (\theta )}. \]

Here, \(\theta \) is often called the angle of elevation. Since the acceleration of the object is known, namely

\[ \vec {a}(t) =\vector {0,-g}, \]

where \(g\) is the gravitational constant, we can find \(\vec {p}(t)\) knowing our two initial conditions. We first find \(\vec {v}(t)\):

\begin{align*} \vec {v}(t) &= \int \vec {a}(t) \d t\\ \vec {v}(t) &= \int \vector {0,-g} \d t\\ \vec {v}(t) &= \vector {0,-gt} + \vec {v}(0). \end{align*}

We integrate once more to find \(\vec {p}(t)\):

\begin{align*} \vec {p}(t) &= \int \vec {v}(t)\d t \\ \vec {p}(t) &= \int \vector {0,-gt} + \vec C_1 \d t\\ \vec {p}(t) &= \vector {0, \frac {-gt^2}{2}} +t\cdot \vec {v}(0)+ \vec {p}(0). \end{align*}

You can adjust the initial position, \(P_0\), 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.

3 From distance traveled to arc length

Consider a driver who sets her cruise-control to \(60\unit {mph}\), 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 \(60\) 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 \(|\vec {v}(t)|\).

This theorem is a specific instance of the more general theorem for arc length:

Knowing that
\[ \vector {R\cdot \cos (\theta ), R \cdot \sin (\theta )} \]
is a vector valued function that plots a circle of radius \(R\) for \(0\le \theta < 2\pi \), can you use an arc length integral to confirm its circumference is \(2\pi R\)? First note that
\[ |\vector {R\cdot \cos (\theta ), R \cdot \sin (\theta )}| = \answer {R}, \]
and now compute
\[ \int _0^{2\pi } R \d \theta = \answer {2\pi R}. \]

One more example, again interpreting our vector-valued function as giving the position of an object in space.

4 Looking back

Finally, let’s think about what we learned in our previous calculus courses, in terms of what we know now.

4.1 Arc length

When you first learned how to compute arc length in calculus, you probably use a formula like

\[ \text {arc length} = \int _a^b \sqrt {1+ f'(x)^2} \d x \]

next you learned that for parametric functions, the arc length was

\[ \text {arc length} = \int _a^b \sqrt {x'(t)^2+ y'(t)^2} \d t \]

Note, the first formula, is really just the second one, where \(x(t) = t\) so that \(\d x = \d t\) and \(y(t) = f(t)\). Finally in this class, we view arc length as

\[ \text {arc length} = \int _a^b |\vec {f}'(t)|\d t. \]

Since the magnitude of a vector can be given via the dot product, this is a very general formula.

4.2 Average value

In your first calculus course, you defined the average value of a function to be

\[ \frac {1}{b-a}\int _a^bf(x)\d x. \]

Above we computed the average speed as

\[ \frac {\text {distance traveled}}{\text {travel time}} = \frac 1{2-(-2)}\int _{-2}^2|\vec {v}(t)|\d t; \]

that is, we just found the average value of \(|\vec {v}(t)|\) on \([-2,2]\).

Likewise, given position function \(\vec {p}(t)\), the average velocity on \([a,b]\) is

\begin{align*} \frac {\text {displacement}}{\text {travel time}} &= \frac 1{b-a}\int _a^b \vec {p}'(t)\d t\\ &= \frac {\vec {p}(b)-\vec {p}(a)}{b-a}; \end{align*}

that is, it is the average value of \(\vec {p}'(t)\), or \(\vec {v}(t)\), on \([a,b]\). As we learn new material, we must constantly reconcile, and reintegrate what have learned before.