Motion and paths in space

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

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.

Let $\vec {v}(t)$ be a vector-valued function denoting the velocity of some object in $\R ^2$ or $\R ^3$ :

$\vec {v}'(t)$ tells us the acceleration (instantaneous change in velocity) of the object, and
$\int _a^b \vec {v}(t) \d t$ tells us the displacement (position with respect to an origin) of the object.

Additionally, the speed of the object is the magnitude of velocity, $|\vec v(t)|$ .

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.

You whirl a ball, attached to a string, above your head in a counter-clockwise circle. The ball follows a circular path and makes $2$ revolutions per second. The string has length $2$ ft.

Find the position function $\vec {p}(t)$ that describes this situation.
Find the acceleration of the ball and give a physical interpretation.

The ball whirls in a circle. Since the string is $2$ ft long, the radius of the circle is $2$ . We start by writing down a position function that describes

a circle with radius $2$ ,
centered at the origin of the $(x,y)$ -plane,
that makes a full revolution every $2\pi$ seconds.

$\vector {\answer [given]{2\cos (t)}, \answer [given]{2\sin t}}$ However, we want two revolutions per second, not one revolution every $2\pi$ seconds. We modify the period by multiplying $t$ by $\answer [given]{4\pi }$ . The final position function is $\vec {p}(t) =\vector {\answer [given]{2\cos (4\pi t)}, \answer [given]{2\sin (4\pi t)}}.$ To find $\vec {a}(t)$ , we differentiate $\vec {p}(t)$ twice.

$\begin{align*} \vec{v}(t) = \vec{p}'(t) &= \vector{\answer[given]{-8\pi \sin (4\pi t)}, \answer[given]{8\pi \cos (4\pi t)}}\\ \vec{a}(t) =\vec{p}''(t) &= \vector{\answer[given]{-32\pi^2 \cos (4\pi t)}, \answer[given]{-32\pi^2 \sin (4\pi t) }} %&= -32\pi^2\vector{\cos (4\pi t), \sin (4\pi t)}. \end{align*}$

For the physical interpretation, recall the classic physics equation, “Force $=$ mass $\times$ 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 $\vec {a}(t)$ is parallel to $\vec {p}(t)$ , 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 .

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 $\vec {a}$ ? $\text {The angle equals } \answer {\pi /2}\unit {radians}$

Since the speed is constant, the velocity and the acceleration are perpendicular.

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.

Sydney shoots her Red Ryder BB gun across level ground from an elevation of $4\unit {ft}$ , where the barrel of the gun makes a $0.1$ radian angle with the horizontal. Find how far the BB travels before landing, assuming the BB is fired at the advertised rate of $350\unit {ft/s}$ and ignoring air resistance. Assume that acceleration due to gravity is $32\unit {ft/sec^2}$

Write with me $\begin{align*} \vec{v}(t) &= \int \vec{a}(t) \d t\\ \vec{v}(t) &= \int \vector{\answer[given]{0},\answer[given]{-32}} \d t\\ \vec{v}(t) &= \vector{\answer[given]{0},\answer[given]{-32t}} + \vec{v}(0). \end{align*}$

But from the statement of the problem, we see that $\vec {v}(0) = 350\vector {\cos (0.1),\sin (0.1)}$ . So $\vec {v}(t) = \vector {0,\answer [given]{-32t}}+ 350\vector {\cos (0.1),\sin (0.1)}.$ Please continuing writing with me

$\begin{align*} \vec{p}(t) &= \int \vec{v}(t)\d t \\ \vec{p}(t) &= \int \vector{0,-32t} + 350\vector{\cos(0.1),\sin(0.1)}\d t\\ \vec{p}(t) &= \vector{\answer[given]{0},\answer[given]{-16t^2}} +t\cdot 350\vector{\cos(0.1),\sin(0.1)}+ \vec{p}(0). \end{align*}$

and $\vec {p}(0) = \vector {\answer [given]{0},\answer [given]{4}}$ , so $\vec {p}(t) = \vector {0,-16t^2} +t\cdot 350\vector {\cos (0.1),\sin (0.1)}+ \vector {0,4}.$ Now we need to find when the BB lands, then we can find where it lands. We accomplish this by setting the $y$ -component equal to $0$ $-16t^2+350\sin (0.1)t+4 = 0$ and solving for $t$ : $t = \frac {-\answer [given]{350\sin (0.1)} \pm \sqrt {\left (\answer [given]{350\sin (0.1)}\right )^2-4\answer [given]{(-16)(4)}}}{\answer [given]{-32}}$ Discarding the negative solution that resulted from our quadratic equation, we have found that the BB lands approximately $2.3\unit {s}$ after firing; with $t\approx 2.3$ , we find the $x$ -component of our position function is $2.3\cdot 350\cos (0.1) = 800.98\unit {ft}$ .

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 $|\vec {v}(t)|$ .

Let $\vec {v}(t)$ be a velocity function for a moving object. The distance traveled by the object for $a\le t\le b$ is: $\text {distance traveled} = \int _a^b |\vec {v}(t)|\d t.$

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

Let $\vec {f}(t) = \vector {x(t),y(t),z(t)}$ be a vector-valued function where

$x'(t)$ , $y'(t)$ , and $z'(t)$ are continuous.
The curve defined by $\vec {f}(t)$ is traversed once for $a\le t\le b$ .

The arc length of the curve from $\vector {x(a),y(a),z(a)}\quad \text {to}\quad \vector {x(b),y(b),z(b)}$ is given by $\text {arc length} = \int _a^b |\vec {f}'(t)|\d t.$

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.

A particle moves in space with position function $\vec {p}(t) =\vector {t,t^2,\sin (\pi t)}$ on $[-2,2]$ , where $t$ is measured in seconds and distances are in meters.

Find:

An integral that computes the distance traveled by the particle on $[-2,2]$ .
The displacement of the particle on $[-2,2]$ .
The particle’s average speed.

First we’ll compute the distance traveled. $\begin{align*} \text{distance traveled} &= \int_{-2}^2 |\vec{v}(t)|\d t \\ &= \int_{-2}^2 \sqrt{\vec{v}(t)\dotp \vec{v}(t)}\d t.\\ &= \int_{-2}^2 \answer[given]{\sqrt{1+(2t)^2+ \pi^2\cos^2(t\pi)}}\d t. \end{align*}$

This cannot be solved in terms of elementary functions so we turn to numerical integration, finding the distance to be $12.88\unit {m}$ .

The displacement is the vector

$\begin{align*} \vec{p}(2)-\vec{p}(-2) &= \vector{\answer[given]{2},\answer[given]{4},\answer[given]{0}} - \vector{\answer[given]{-2},\answer[given]{4},\answer[given]{0}}\\ &= \vector{\answer[given]{4},\answer[given]{0},\answer[given]{0}}. \end{align*}$

That is, the particle ends with an $x$ -value increased by $4$ and with unchanged $(y,z)$ -values the same.

We found above that the particle traveled $12.88\unit {m}$ over $4$ seconds. We can compute average speed by dividing: $12.88/4 = \answer [given]{3.22}\unit {m/s}$ .

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

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

$\begin{align*} \frac{\text{displacement}}{\text{travel time}} &= \frac1{b-a}\int_a^b \vec{p}'(t)\d t\\ &= \frac{\vec{p}(b)-\vec{p}(a)}{b-a}; \end{align*}$

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.