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Motion in Space

- Calculus
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The other day I went to the park with my friends to do a scavenger hunt in which the organizer hid a series of clues to find a prize. However, due to an oversight, no one could find the prize in time! This was because, despite the fact that we found some clues, they were written like this:

"\(500\) feet from here, you can find the next clue".

Notice anything? The hint tells you how far you are from the next clue, but it does not tell you the direction! To make things worst, it turns out that the prize was on top of a tree! It looks like the organizer could have used a method to describe the location of things considering our three dimensions. Here you will learn about **motion in space**, another way of referring to **motion in three-dimensional space**.

Vectors are useful for describing the position of objects because, in addition to a magnitude (length), they also have a direction. This way there is no more ambiguity when you are told where to go!

In two dimensions you need to use vectors with two components. You can write vectors either in component form or in column vector form. For example, the two-dimensional vector \( \vec{A}\) written in component form

\[ \vec{A} = 2\hat{i}+3\hat{j},\]

can also be written in column vector form as

\[ \vec{A} = \begin{bmatrix} 2 \\ 3\end{bmatrix}.\]

Since each component of the vector consists of a real number, this vector is an element of \(\mathbb{R}^2\).

For vectors in space, you need a third component. You can describe this third component either with a third unit vector, \( \hat{k}\), or with a third entry in a column. For example, the three-dimensional vector \( \vec{V}\) written in component form

\[ \vec{V} = 2\hat{i}+3\hat{j}+5\hat{k},\]

can also be written in column vector form as

\[ \vec{V} = \begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix}.\]

In this case, we say that a vector in the three-dimensional space is an element of \( \mathbb{R}^3\).

So what about motion in space? In mathematics, especially in Calculus, when talking about space we generally do not refer to outer space (like rockets, stars, or galaxies). Space can be used in a wide variety of contexts, but within this context, we are talking about **three-dimensional space.** So motion in space refers to motion in three-dimensional space.

**Motion in space** refers to the change of position of an object described in three-dimensional space. The position of such an object is described by the vector-valued function \( \vec{r} (t)\), where \(t \) represents time.

This means that the vector-valued function

\[ \vec{r} (t) \]

gives you a vector in space for each value of \(t\) you give to it. You can picture this as if \( \vec{r} (t) \) was telling you how to move through the three-dimensional space!

The best way to picture motion in space would be through animation. Since you are dealing with **motion**, time plays a fundamental role in how the vectors will be pictured. Let's start with a simple example.

The position of an object in space is given by the vector-valued function

\[ \vec{r} (t) = (t+1)\,\hat{i} + t^2 \, \hat{j} + (t-1) \, \hat{k},\]

where \(t\) represents time, in seconds.

- What is the initial position of the object?
- Find the position of the object when \(t=3\).

**Solution:**

- Whenever you see the word
*initial*, it usually refers to the instant when \(t=0\), so you can find the initial position of the object by substituting \( t=0\) in the vector-valued function. This will give you\[ \begin{align} \vec{r} (0) &= (0+1)\,\hat{i} + (0)^2 \, \hat{j} + (0-1) \, \hat{k} \\ &= \hat{i} - \hat{k}.\end{align}\]This means that the initial position of the object is\[ \vec{r} (0) = \hat{i}-\hat{k}. \] - This time you just need to substitute \( t=3\) in the vector-valued function, so
\[ \begin{align} \vec{r} (3) &= (3+1)\,\hat{i} + (3)^2 \, \hat{j} + (3-1) \, \hat{k} \\ &= 4 \hat{i} +9\hat{j} + 2\hat{k}.\end{align}\]

This means that the position of the object \(3\) seconds after it starts its motion is

\[ \vec{r} (3) = 4\hat{i}+9\hat{j}+2\hat{k}.\]

Simple, isn't it? Unfortunately, to picture the whole motion you would need to use many more \(t\) values!

Luckily, there is software (like Computer Algebra Systems and plotters) that can do those animations. Software like this can plot thousands, even millions of vectors, and show them from time to time, doing a great description of the motion in space!

So what can you do to describe motion in space using paper and a pen? Since you are not a wizard, you do not have magic ink that can move at your will, so you will not be able to graph time itself. Instead, you need to plot what is known as a **trajectory**.

The **trajectory** of an object moving through space is a curve that describes all the points the object passes through.

A trajectory can also be seen as the **path **that an object follows through its movement.

Graphing trajectories in space is not a simple task. While in the two-dimensional space you can draw the curve \( y=f(x)\) to represent the trajectory, in the three-dimensional space the function \( z=f(x,y) \) actually represents a surface. To graph a trajectory in space you need to inspect the function and sketch some points, here is an example.

Sketch the trajectory of an object moving according to the vector-valued function

\[ \vec{r} (t) = \cos{t}\,\hat{i} + \sin{t}\,\hat{j} +t\,\hat{k},\]

where \(t\) is given in seconds.

**Solution:**

In order to gain more insight, you can start by looking for any particular relationship between the components of this vector-valued function. In particular, note that the \(x\) and the \(y\) components are related by the Pythagorean identity

\[ \sin^2{t}+\cos^2{t}=1.\]

The unit vector \(\hat{i}\) represents directions in the \(x\)-axis, and the unit vector \(\hat{j}\) represents directions in the \(y\)-axis, which means that

\[\begin{align} x^2+y^2 &= \cos^2{t}+\sin^2{t} \\ &=1. \end{align}\]

Since

\[ x^2+y^2=1\]

is the equation of a circle with radius \(1\) and center at the origin, you can conclude that the object will be moving between the \(x\) and \(y\) axes circularly.

However, since this is motion in space, you also have to take the movement in the \(z\) direction as well. This will be clearer when you do a table of values to graph some points of the trajectory. Try using multiples of \( \pi\) because of the trigonometric functions!

\[t\] | \[x\] | \[y\] | \[z\] |

\[0\] | \[1\] | \[0\] | \[0\] |

\[\frac{\pi}{2}\] | \[0\] | \[1\] | \[\frac{\pi}{2}\] |

\[\pi\] | \[-1\] | \[0\] | \[\pi\] |

\[\frac{3\pi}{2}\] | \[0\] | \[-1\] | \[\frac{3\pi}{2}\] |

\[2\pi\] | \[1\] | \[0\] | \[2\pi\] |

Table 1. values for \(\pi\), motion in space.

The curve of the trajectory will pass through these points. Finally, you can connect them, but keep in mind that you found a circular shape between the \(x\) and \(y\) axes!

It is worth noting that the downside of looking at a trajectory is that you cannot tell whether the object is slow or fast, since there is no trace of time in a trajectory. To address this issue, you need to see how Calculus is used when describing motion in space.

We usually associate the word motion with change. Given different values of \(t\), the vector-valued function \( \vec{r} (t)\) describes the position of an object, which means that position is a quantity that **changes** over time. This is where Calculus comes in.

Let \(\vec{r} (t) \) be a vector-valued function describing the position of an object as a function of time \(t\). The **velocity** of the object denoted as \( \vec{v} (t) \), is defined as the derivative of the position with respect to time, that is

\[ \vec{v} (t) = \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}.\]

Since it is a derivative, it is possible that you find velocity written with prime notation, that is

\[ \vec{v} (t) = \vec{r}'(t),\]

or even with dot notation, which is more common in Physics, that is

\[ \vec{v} (t) = \dot{\vec{r}}(t),\]

where the dot above a quantity is used to represent a derivative with respect to time.

The magnitude of \( \vec{v} (t) \) receives a special name, which will probably sound familiar to you.

Let \( \vec{v}\) be the velocity of an object. The magnitude of \( \vec{v} \) is called **speed,** and it is usually denoted by \( v\).

Please note that speed and velocity are **not** synonyms. Velocity is a vector quantity, while speed is a scalar quantity. You can remember this by noting that speed does not have a vector arrow above it!

To find the speed of an object you just need to find the magnitude of the vector \( \vec{v}\), which you can do by adding the squares of each component and then taking its square root.

The position of an object is described by the vector-valued function

\[ \vec{r} (t) = t \, \hat{i} + (2t-1) \, \hat{j} + t^2\, \hat{k}.\]

- Find its velocity.
- Find its speed.

**Solution:**

- To find the velocity of the object, you just need to differentiate the position\[ \vec{v}(t) = \vec{r}'(t).\]Since the position is a vector-valued function, you can do it from component to component using the Power Rule, which is\[ \frac{\mathrm{d}}{\mathrm{d}t} t = 1,\]\[\frac{\mathrm{d}}{\mathrm{d}t}(2t-1) = 2, \] and\[\frac{\mathrm{d}}{\mathrm{d}t}(t^2)=2t. \] This means that\[ \vec{v} (t) = \hat{i}+2\hat{j}+2t\,\hat{k}.\]
- To find the speed you just need to find the magnitude of the velocity you just found. You can achieve this by adding the squares of each component and then taking the square root, that is\[ \begin{align} v (t) &= \sqrt{(1)^2+(2)^2+(2t)^2} \\ &= \sqrt{1+4+4t^2} \\ &= \sqrt{4t^2+5}. \end{align}\]

You have seen how velocity is the derivative of position with respect to time, which you can picture as how fast or how slow an object is moving. But what about the change of velocity itself?

Let \(\vec{v} (t) \) be a vector-valued function describing the velocity of an object as a function of time \(t\). The **acceleration **of the object, denoted as \( \vec{a} (t) \), is defined as the derivative of the velocity with respect to time, that is

\[ \vec{a} (t) = \frac{\mathrm{d}\vec{v}}{\mathrm{d}t}.\]

Just like velocity, acceleration can be written with prime notation\[ \vec{a} (t) = \vec{v}'(t),\]

or with dot notation

\[ \vec{a} (t) = \dot{\vec{v}}(t).\]

Note that, since velocity is the derivative of position with respect to time, then acceleration is the second derivative of position with respect to time, that is

\[ \vec{a} (t) = \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2}.\]

This can also be written using prime notation

\[ \vec{a}(t) = \vec{r}''(t),\]

or dot notation

\[ \vec{a} (t) = \ddot{\vec{r}}(t).\]

Find the acceleration of an object moving according to the vector-valued function

\[ \vec{r} (t) = t \, \hat{i} + (2t-1) \, \hat{j} + t^2\, \hat{k}.\]

**Solution:**

You previously found that the velocity of this object is

\[ \vec{v} (t) = \hat{i}+2\hat{j}+2t\,\hat{k},\]

so you can save some time. You can find the acceleration by differentiating the velocity, where once again you will use the Power Rule, which is

\[ \begin{align} \vec{a} (t) &= (0)\,\hat{i} + (0)\,\hat{j}+(2)\,\hat{k} \\& = 2\, \hat{k}. \end{align}\]

Here you can take a quick look at the formulas used for describing motion in space.

The velocity of an object moving through space can be found by differentiating its position, that is

\[ \vec{v} (t) = \vec{r}'(t).\]

The speed of an object is the magnitude of its velocity, so

\[ v = ||\vec{v}(t)||.\]

If you need a reminder on how to find the magnitude of a vector, please check our article about Vectors in Space!

The acceleration of an object can be found from its velocity by differentiating once,

\[ \vec{a} (t) = \vec{v}'(t),\]

or from the position by differentiating twice,

\[ \vec{a} (t) = \vec{r}''(t).\]

Here you can look at more examples involving motion in space.

Sketch the trajectory of an object moving according to the vector-valued function

\[ \vec{r} (t) = t\, \hat{i} + t\,\hat{j} + e^t\,\hat{k}.\]

**Solution:**

You should begin by inspecting if there is any relation between the components of the position. In this case, note that both the \(x\) and \(y\) components are both equal, so they will increase at the same rate. You can picture this as if the shadow of the curve will be the function \(y=x\) on the \(xy\)-plane. Be sure to correctly identify the axes! Here the \(x\)-axis is highlighted.

Next, note that the \(z\)-component is an exponential function. Since the \(x\) and \(y\) components are just \(t\), the height of \(\vec{r}\) with respect to the \(xy\)-plane will increase just like the exponential function \(e^t\). Knowing this, you can now sketch the trajectory of this object.

Practice some derivatives by finding velocities and accelerations!

Consider the trajectory of an object moving according to the vector-valued function

\[ \vec{r} (t) = t\, \hat{i} + t\,\hat{j} + e^t\,\hat{k}.\]

- Find the velocity of the object.
- Find the speed of the object.
- Find the acceleration of the object.

**Solution:**

- To find the velocity, you should differentiate the position. Note that you can use the Power Rule for the \(x\) and \(y\) components, but for the \(z\) component you need to know that\[ \frac{\mathrm{d}}{\mathrm{d}t} e^t = e^t.\]This means that\[ \begin{align} \vec{v} (t) &= \vec{r}'(t) \\ &= (1) \vec{i} + (1) \vec{j} + e^t \, \hat{k} \\ &= \hat{i}+\hat{j} + e^t\,\hat{k}. \end{align}\]
- The speed is given by the magnitude of \(\vec{v}\), so\[ \begin{align} v &= || \vec{v} || \\ &= \sqrt{(1)^2+(1)^2+(e^t)^2} \\ &=\sqrt{1+1+e^{2t}} \\ &= \sqrt{2+e^{2t}}. \end{align}\]
- Finally, differentiate the velocity to find the acceleration. You will need to differentiate the exponential function once again, but it does not matter! It stays the same!\[ \begin{align}\vec{a} (t) &= \vec{v}'(t) \\ &= (0)\hat{i} + (0)\hat{j} + e^t\,\hat{k} \\ &= e^t\,\hat{k}. \end{align}\]

**Motion in space****Velocity,**usually denoted as \( \vec{v}\), is defined as the derivative of position with respect to time, that is\[ \vec{v} (t) = \vec{r}'(t).\]- Its magnitude \(v\) is known as
**speed**.

- Its magnitude \(v\) is known as
**Acceleration**usually denoted as \(\vec{a}\), is defined as the derivative of velocity with respect to time, that is\[ \vec{a} (t) = \vec{v}'(t).\]- Acceleration can also be thought of as the second derivative of position with respect to time,\[ \vec{a} (t) = \vec{r} '' (t).\]

- The
**trajectory**of an object moving through space is a curve that describes all the points the object passes through.- To graph trajectories in three-dimensional space, you need to inspect the function to find if there is a relation between its three coordinates.

More about Motion in Space

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