Motion in Space

We have now seen how to describe curves in the plane and in space, and how to determine their properties, such as arc length and curvature. All of this leads to the main goal of this chapter, which is the description of motion along plane curves and space curves. We now have all the tools we need; in this section, we put these ideas together and look at how to use them.

Motion Vectors in the Plane and in Space

Our starting point is using vector-valued functions to represent the position of an object as a function of time. All of the following material can be applied either to curves in the plane or to space curves. For example, when we look at the orbit of the planets, the curves defining these orbits all lie in a plane because they are elliptical. However, a particle traveling along a helix moves on a curve in three dimensions.

Since $\text{r}\left(t\right)$

can be in either two or three dimensions, these vector-valued functions can have either two or three components. In two dimensions, we define $\text{r}\left(t\right)=x\left(t\right)\text{i}+y\left(t\right)\text{j}$

and in three dimensions $\text{r}\left(t\right)=x\left(t\right)\text{i}+y\left(t\right)\text{j}+z\left(t\right)\text{k}.$

Then the velocity, acceleration, and speed can be written as shown in the following table.

To gain a better understanding of the velocity and acceleration vectors, imagine you are driving along a curvy road. If you do not turn the steering wheel, you would continue in a straight line and run off the road. The speed at which you are traveling when you run off the road, coupled with the direction, gives a vector representing your velocity, as illustrated in the following figure.

However, the fact that you must turn the steering wheel to stay on the road indicates that your velocity is always changing (even if your speed is not) because your direction is constantly changing to keep you on the road. As you turn to the right, your acceleration vector also points to the right. As you turn to the left, your acceleration vector points to the left. This indicates that your velocity and acceleration vectors are constantly changing, regardless of whether your actual speed varies ([link]).

Components of the Acceleration Vector

We can combine some of the concepts discussed in Arc Length and Curvature with the acceleration vector to gain a deeper understanding of how this vector relates to motion in the plane and in space. Recall that the unit tangent vector T and the unit normal vector N form an osculating plane at any point P on the curve defined by a vector-valued function $\text{r}\left(t\right).$

The following theorem shows that the acceleration vector $\text{a}\left(t\right)$

lies in the osculating plane and can be written as a linear combination of the unit tangent and the unit normal vectors.

Proof

Because $\text{v}\left(t\right)=r^{\prime }\left(t\right)$

and $\text{T}\left(t\right)=\frac{r^{\prime }\left(t\right)}{\Vert r^{\prime }\left(t\right)\Vert },$

we have $\text{v}\left(t\right)=\Vert r^{\prime }\left(t\right)\Vert \text{T}\left(t\right)=v\left(t\right)\text{T}\left(t\right).$

Now we differentiate this equation:

Since $\text{N}\left(t\right)=\frac{T^{\prime }\left(t\right)}{\Vert T^{\prime }\left(t\right)\Vert },$

we know $T^{\prime }\left(t\right)=\Vert T^{\prime }\left(t\right)\Vert \text{N}\left(t\right),$

so

A formula for curvature is $\kappa =\frac{\Vert T^{\prime }\left(t\right)\Vert }{\Vert r^{\prime }\left(t\right)\Vert },$

so $\Vert T^{\prime }\left(t\right)\Vert =\kappa \Vert r^{\prime }\left(t\right)\Vert =\kappa v\left(t\right).$

This gives $\text{a}\left(t\right)=v^{\prime }\left(t\right)\text{T}\left(t\right)+\kappa {\left(v\left(t\right)\right)}^2\text{N}\left(t\right).$

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The coefficients of $\text{T}\left(t\right)$

and $\text{N}\left(t\right)$

are referred to as the tangential component of acceleration and the normal component of acceleration, respectively. We write $a_{\text{T}}$

to denote the tangential component and $a_{\text{N}}$

to denote the normal component.

The normal component of acceleration is also called the centripetal component of acceleration or sometimes the radial component of acceleration. To understand centripetal acceleration, suppose you are traveling in a car on a circular track at a constant speed. Then, as we saw earlier, the acceleration vector points toward the center of the track at all times. As a rider in the car, you feel a pull toward the outside of the track because you are constantly turning. This sensation acts in the opposite direction of centripetal acceleration. The same holds true for noncircular paths. The reason is that your body tends to travel in a straight line and resists the force resulting from acceleration that push it toward the side. Note that at point B in [link] the acceleration vector is pointing backward. This is because the car is decelerating as it goes into the curve.

The tangential and normal unit vectors at any given point on the curve provide a frame of reference at that point. The tangential and normal components of acceleration are the projections of the acceleration vector onto T and N, respectively.

Projectile Motion

Now let’s look at an application of vector functions. In particular, let’s consider the effect of gravity on the motion of an object as it travels through the air, and how it determines the resulting trajectory of that object. In the following, we ignore the effect of air resistance. This situation, with an object moving with an initial velocity but with no forces acting on it other than gravity, is known as projectile motion. It describes the motion of objects from golf balls to baseballs, and from arrows to cannonballs.

First we need to choose a coordinate system. If we are standing at the origin of this coordinate system, then we choose the positive y-axis to be up, the negative y-axis to be down, and the positive x-axis to be forward (i.e., away from the thrower of the object). The effect of gravity is in a downward direction, so Newton’s second law tells us that the force on the object resulting from gravity is equal to the mass of the object times the acceleration resulting from to gravity, or $F_g=mg,$

where $F_g$

represents the force from gravity and g represents the acceleration resulting from gravity at Earth’s surface. The value of g in the English system of measurement is approximately 32 ft/sec² and it is approximately 9.8 m/sec² in the metric system. This is the only force acting on the object. Since gravity acts in a downward direction, we can write the force resulting from gravity in the form $F_g=\text{−}mg\text{j},$

as shown in the following figure.

Newton’s second law also tells us that $F=m\text{a},$

where a represents the acceleration vector of the object. This force must be equal to the force of gravity at all times, so we therefore know that

Now we use the fact that the acceleration vector is the first derivative of the velocity vector. Therefore, we can rewrite the last equation in the form

By taking the antiderivative of each side of this equation we obtain

for some constant vector ${\text{C}}_1.$

To determine the value of this vector, we can use the velocity of the object at a fixed time, say at time $t=0.$

We call this velocity the initial velocity: $\text{v}\left(0\right)={\text{v}}_0.$

Therefore, $\text{v}\left(0\right)=\text{−}g\left(0\right)\text{j}+{\text{C}}_1={\text{v}}_0$

and ${\text{C}}_1={\text{v}}_0.$

This gives the velocity vector as $\text{v}\left(t\right)=\text{−}gt\text{j}+{\text{v}}_0.$

Next we use the fact that velocity $\text{v}\left(t\right)$

is the derivative of position $\text{s}\left(t\right).$

This gives the equation

Taking the antiderivative of both sides of this equation leads to

with another unknown constant vector ${\text{C}}_2.$

To determine the value of ${\text{C}}_2,$

we can use the position of the object at a given time, say at time $t=0.$

We call this position the initial position: $\text{s}\left(0\right)={\text{s}}_0.$

Therefore, $\text{s}\left(0\right)=\text{−}\left(1\text{/}2\right)g{\left(0\right)}^2\text{j}+{\text{v}}_0\left(0\right)+{\text{C}}_2={\text{s}}_0$

and ${\text{C}}_2={\text{s}}_0.$

This gives the position of the object at any time as

Let’s take a closer look at the initial velocity and initial position. In particular, suppose the object is thrown upward from the origin at an angle $\theta$

to the horizontal, with initial speed $v_0.$

How can we modify the previous result to reflect this scenario? First, we can assume it is thrown from the origin. If not, then we can move the origin to the point from where it is thrown. Therefore, ${\text{s}}_0=0,$

as shown in the following figure.

We can rewrite the initial velocity vector in the form ${\text{v}}_0=v_0\text{cos}\theta \text{i}+v_0\text{sin}\theta \text{j}.$

Then the equation for the position function $\text{s}\left(t\right)$

becomes

The coefficient of i represents the horizontal component of $\text{s}\left(t\right)$

and is the horizontal distance of the object from the origin at time t. The maximum value of the horizontal distance (measured at the same initial and final altitude) is called the range R. The coefficient of j represents the vertical component of $\text{s}\left(t\right)$

and is the altitude of the object at time t. The maximum value of the vertical distance is the height H.

One final question remains: In general, what is the maximum distance a projectile can travel, given its initial speed? To determine this distance, we assume the projectile is fired from ground level and we wish it to return to ground level. In other words, we want to determine an equation for the range. In this case, the equation of projectile motion is

Setting the second component equal to zero and solving for t yields

Therefore, either $t=0$

or $t=\frac{2v_0\text{sin}\theta }{g}.$

We are interested in the second value of t, so we substitute this into $\text{s}\left(t\right),$

which gives

Thus, the expression for the range of a projectile fired at an angle $\theta$

is

The only variable in this expression is $\theta .$

To maximize the distance traveled, take the derivative of the coefficient of i with respect to $\theta$

and set it equal to zero:

This value of $\theta$

is the smallest positive value that makes the derivative equal to zero. Therefore, in the absence of air resistance, the best angle to fire a projectile (to maximize the range) is at a $45\text{°}$

angle. The distance it travels is given by

Therefore, the range for an angle of $45\text{°}$

is $v_0^2\text{/}g.$

Kepler’s Laws

During the early 1600s, Johannes Kepler was able to use the amazingly accurate data from his mentor Tycho Brahe to formulate his three laws of planetary motion, now known as Kepler’s laws of planetary motion. These laws also apply to other objects in the solar system in orbit around the Sun, such as comets (e.g., Halley’s comet) and asteroids. Variations of these laws apply to satellites in orbit around Earth.

Kepler’s third law is especially useful when using appropriate units. In particular, 1 astronomical unit is defined to be the average distance from Earth to the Sun, and is now recognized to be 149,597,870,700 m or, approximately 93,000,000 mi. We therefore write 1 A.U. = 93,000,000 mi. Since the time it takes for Earth to orbit the Sun is 1 year, we use Earth years for units of time. Then, substituting 1 year for the period of Earth and 1 A.U. for the average distance to the Sun, Kepler’s third law can be written as

for any planet in the solar system, where $T_P$

is the period of that planet measured in Earth years and $D_P$

is the average distance from that planet to the Sun measured in astronomical units. Therefore, if we know the average distance from a planet to the Sun (in astronomical units), we can then calculate the length of its year (in Earth years), and vice versa.

Kepler’s laws were formulated based on observations from Brahe; however, they were not proved formally until Sir Isaac Newton was able to apply calculus. Furthermore, Newton was able to generalize Kepler’s third law to other orbital systems, such as a moon orbiting around a planet. Kepler’s original third law only applies to objects orbiting the Sun.

Proof

Let’s now prove Kepler’s first law using the calculus of vector-valued functions. First we need a coordinate system. Let’s place the Sun at the origin of the coordinate system and let the vector-valued function $\text{r}\left(t\right)$

represent the location of a planet as a function of time. Newton proved Kepler’s law using his second law of motion and his law of universal gravitation. Newton’s second law of motion can be written as $\text{F}=m\text{a},$

where F represents the net force acting on the planet. His law of universal gravitation can be written in the form $\text{F}=-\frac{GmM}{{\Vert \text{r}\Vert }^2}·\frac{\text{r}}{\Vert \text{r}\Vert },$

which indicates that the force resulting from the gravitational attraction of the Sun points back toward the Sun, and has magnitude $\frac{GmM}{{\Vert \text{r}\Vert }^2}$

([link]).

Setting these two forces equal to each other, and using the fact that $\text{a}\left(t\right)=v^{\prime }\left(t\right),$

we obtain

which can be rewritten as

This equation shows that the vectors $d\text{v}\text{/}dt$

and r are parallel to each other, so $d\text{v}\text{/}dt\times \text{r}=0.$

Next, let’s differentiate $\text{r}\times \text{v}$

with respect to time:

This proves that $\text{r}\times \text{v}$

is a constant vector, which we call C. Since $\text{r}$

and v are both perpendicular to C for all values of t, they must lie in a plane perpendicular to C. Therefore, the motion of the planet lies in a plane.

Next we calculate the expression $d\text{v}\text{/}dt\times \text{C}\text{:}$

The last equality in [link] is from the triple cross product formula (Introduction to Vectors in Space). We need an expression for $\text{r}·\text{v}.$

To calculate this, we differentiate $\text{r}·\text{r}$

with respect to time:

Since $\text{r}·\text{r}={\Vert \text{r}\Vert }^2,$

we also have

Combining [link] and [link], we get

Substituting this into [link] gives us

However,

Therefore, [link] becomes

Since C is a constant vector, we can integrate both sides and obtain

where D is a constant vector. Our goal is to solve for $\Vert \text{r}\Vert .$

Let’s start by calculating $\text{r}·\left(\text{v}\times \text{C}\right)\text{:}$

However, $\text{r}·\left(\text{v}\times \text{C}\right)=\left(\text{r}\times \text{v}\right)·\text{C},$

so

Since $\text{r}\times \text{v}=\text{C},$

we have

Note that $\text{r}·\text{D}=\Vert \text{r}\Vert \Vert \text{D}\Vert \text{cos}\theta ,$

where $\theta$

is the angle between r and D. Therefore,

Solving for $\Vert \text{r}\Vert ,$

where $e=\Vert \text{D}\Vert \text{/}GM.$

This is the polar equation of a conic with a focus at the origin, which we set up to be the Sun. It is a hyperbola if $e>1,$

a parabola if $e=1,$

or an ellipse if $e<1.$

Since planets have closed orbits, the only possibility is an ellipse. However, at this point it should be mentioned that hyperbolic comets do exist. These are objects that are merely passing through the solar system at speeds too great to be trapped into orbit around the Sun. As they pass close enough to the Sun, the gravitational field of the Sun deflects the trajectory enough so the path becomes hyperbolic.

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Key Concepts

• If $\text{r}\left(t\right)$

  represents the position of an object at time t, then

  $\text{r}\prime \left(t\right)$

  represents the velocity and

  $\text{r″}\left(t\right)$

  represents the acceleration of the object at time t. The magnitude of the velocity vector is speed.

• The acceleration vector always points toward the concave side of the curve defined by $\text{r}\left(t\right).$

  The tangential and normal components of acceleration

  $a_{\text{T}}$

  and

  $a_{\text{N}}$

  are the projections of the acceleration vector onto the unit tangent and unit normal vectors to the curve.

• Kepler’s three laws of planetary motion describe the motion of objects in orbit around the Sun. His third law can be modified to describe motion of objects in orbit around other celestial objects as well.
• Newton was able to use his law of universal gravitation in conjunction with his second law of motion and calculus to prove Kepler’s three laws.

Key Equations

• Velocity

  ---

  $\text{v}\left(t\right)=r^{\prime }\left(t\right)$
• Acceleration

  ---

  $\text{a}\left(t\right)=v^{\prime }\left(t\right)=\text{r″}\left(t\right)$
• Speed

  ---

  $v\left(t\right)=\Vert \text{v}\left(t\right)\Vert =\Vert r^{\prime }\left(t\right)\Vert =\frac{ds}{dt}$
• Tangential component of acceleration

  ---

  $a_{\text{T}}=\text{a}·\text{T}=\frac{\text{v}·\text{a}}{\Vert \text{v}\Vert }$
• Normal component of acceleration

  ---

  $a_{\text{N}}=\text{a}·\text{N}=\frac{\Vert \text{v}\times \text{a}\Vert }{\Vert \text{v}\Vert }=\sqrt{{\Vert \text{a}\Vert }^2-a_{}^{\text{T}}}$