Introduction to calculus
In calculus, we often work with two different types of quantities: scalars and vectors.
Scalar quantities can be described by their magnitude (or size) alone. This means they can be represented by a single number. For example, if we say it is \(30^{\circ} C\) outside, we have a clear idea of how warm the weather is. Other examples of scalar quantities are \(40\textrm{km/hr}\), \(5\textrm{kg}\) and \(15\textrm{minutes.}\) This is because speed, mass and time are all scalar quantities. In other words, a scalar tells us “how much” but does not include any direction.
Vectors are quantities that have both a magnitude and a direction. For example, we might say that a car is travelling at \(60\textrm{km/hr}\) to the north, or that someone has moved \(5\textrm{m}\) to the right. These are vector quantities because each one gives us both a magnitude and a direction.
This page explains how to do different calculations that involve vectors.
In brief, it covers the following:
- How to represent and label vectors in a vector diagram
- How to write a vector in terms of its components
- How to add and subtract vectors
- How to calculate the magnitude of a vector
- How to multiply a vector by a scalar
- How to represent a vector using polar coordinates, and how to convert between these and Cartesian coordinates
- How to multiply two vectors using the dot product method
Vector diagrams
A vector can be shown in a diagram as an arrow between two points. The length of the arrow shows the magnitude of the vector, and the direction the arrow points shows which way the vector is going.
We can label a vector in two ways. The first way is to use a single lower case letter, which is usually written in bold. The second way is to use the names of the two points at each end of the arrow, with a small arrow written above them.
For example, the vector shown below in green goes from A to B. Point A is called the initial point, and point B is called the terminal point. This vector could be labelled as \(\mathbf{a}\) or as \(\overrightarrow{AB}.\)
If two vectors have the same magnitude and direction they are called equivalent or equal vectors.
We can add vectors together on a diagram using the nose to tail method (also known as the triangle law).
The nose to tail method says that if two vectors are positioned so that the initial point of one is at the terminal point of another, then we can add the two vectors together to get the displacement vector.
Example: Suppose a person has walked from point A to point B, then changed direction and walked from point B to another point, C:
If we label these vectors \(\mathbf{u}\) and \(\mathbf{v}\), where \(\mathbf{u} = \overrightarrow{AB}\) and \(\mathbf{v} = \overrightarrow{BC},\) what is the displacement vector indicating the distance and direction the person is from their starting point?
Solution: Since the initial point of \(\mathbf{v}\) is at the terminal point of \(\mathbf{u}\), our displacement vector will be \(\mathbf{u} + \mathbf{v}\)
Subtracting vectors on a diagram is done in a similar way. All we need to do is reverse the direction of the vector we want to subtract, and then add as normal. This however means that the addition becomes a subtraction since we would have \(\mathbf{u} + (-\mathbf{v})\) which is the same as \(\mathbf{u} - \mathbf{v}\).
Components
It can be difficult to show vectors accurately in diagrams. Because of this, vectors are usually written using coordinates when we need to perform calculations.
One way to do this is to split a vector into parts. These parts are the vector’s \(x\) and \(y\) components, which are also known as Cartesian coordinates. (For more information about Cartesian coordinates, and about using polar coordinates instead, see converting between coordinates.)
A vector’s \(x\) component tells us how many units ‘long’ the vector is in relation to the \(x\)-axis. A vector’s \(y\) component tells us how many units ‘high’ the vector is in relation to the \(y\)-axis.
If we have vector \(\mathbf{v}\), its components can be written as \(\mathbf{v_x}\) and \(\mathbf{v_y}\). These components are then expressed in the form \(\mathbf{v}=(\mathbf{v}_x, \mathbf{v}_y),\) similar to the coordinates of a graph, or as \(\begin{pmatrix} \mathbf{v}_x \\ \mathbf{v}_y \end{pmatrix}\)
Example:
What are the components of the vectors shown below?
Solution:
Vector \(\mathbf{a}\) begins at \((0,0)\) and ends at \((2,4)\) therefore we have \(\mathbf{a} = (2,4)\)
Vector \(\mathbf{b}\) begins at \((1,3)\) and ends at \((4,8)\). To find the components of \(\mathbf{b}\), we first have to find the difference between the coordinates of these points:
\[\begin{aligned} \mathbf{b}_x &= 4 - 1 \\\ &= 3 \\\ \mathbf{b}_y &= 8 - 3 \\\ &= 5 \\\ \end {aligned}\]Therefore, the components for \(\mathbf{b}\) will be \(\mathbf{b} = (3,5)\)
Adding and subtracting vectors
Once vectors are written using components, we can add or subtract them quite easily. To do this, we add or subtract the matching \(x\) components, and then we add or subtract the matching \(y\) components. The results give us the \(x\) and \(y\) components of the new vector.
Example:
Add the two vectors below to find the displacement vector:
Solution:
The displacement vector’s components will be:
\[\begin{aligned} \mathbf{u} + \mathbf{v} &= (8+6,\,5+(-2)) \\\ &=(14,\,3) \\\ \end{aligned}\]Check your understanding of adding and subtracting vectors by working through the questions below:
\(\begin{aligned} \mathbf{a} + \mathbf{b} &= (3 + 4,\,5+(-7))\\\ &= (7, -2) \\\ \end{aligned}\)
\(\begin{aligned} \mathbf{c} - \mathbf{d} &= (12 - 7, -3 - 2)\\\ &= (5, -5) \\\ \end{aligned}\)
\(\begin{aligned} \mathbf{y} - \mathbf{x} + \mathbf{z} &= (8 - 2 + 7, -3 - 3 + 4)\\\ &= (13, -2) \\\ \end{aligned}\)
Magnitude of a vector
Vectors have both a magnitude and a direction. If we have a vector \(\mathbf{u}\), the magnitude of that vector is written as \(\vert \mathbf{u} \vert,\) or sometimes as \(\Vert \mathbf{u} \Vert\) so that it isn’t confused with the symbol for absolute value.
The magnitude of a vector is different to the components of a vector, as the magnitude is the numeric value of the vector. However, we can find the magnitude of a vector given its components using Pythagoras’ theorem.
The magnitude of a vector \(\mathbf{u} = (\mathbf{u}_x , \mathbf{u}_y)\) will be:
\[\vert \mathbf{u} \vert = \sqrt{\mathbf{u_x}^2 + \mathbf{u_y}^2}\]Example:
Find the magnitude of \(\mathbf{v} = (6,\,8)\)
Solution:
\[\begin{aligned} \vert \mathbf{v} \vert &= \sqrt{\mathbf{v_x}^2 + \mathbf{v_y}^2} \\\ &= \sqrt{6^2 + 8^2} \\\ &= \sqrt{100} \\\ &= 10 \\\ \end{aligned}\]Unit vectors
If a vector has a magnitude of \(1\) it is called a unit vector. There are two commonly known unit vectors that are often used in vector notation. The unit vectors are \(\mathbf{i}\) and \(\mathbf{j}\) and they sit along the \(x\) and \(y\) axes, respectively:
Vectors can also be represented using these unit vectors, for example as \(\mathbf{v} = x\mathbf{i} + y\mathbf{j}\)
Check your understanding of the magnitude of vectors and unit vector notation by working through the questions below:
\(\begin{aligned} \vert \mathbf{a} \vert &= \sqrt{\mathbf{a_x}^2 + \mathbf{a_y}^2} \\\ &= \sqrt{4^2 + (-3)^2} \\\ &= \sqrt{25} \\\ &= 5 \\\ \end{aligned}\)
\(\begin{aligned} \vert \mathbf{b} \vert &= \sqrt{\mathbf{b_x}^2 + \mathbf{b_y}^2} \\\ &= \sqrt{(-7)^2 + 5^2} \\\ &= \sqrt{74} \\\ &= 8.602 \\\ \end{aligned}\)
\(\begin{aligned} \vert \mathbf{c} \vert &= \sqrt{\mathbf{c_x}^2 + \mathbf{c_y}^2} \\\ &= \sqrt{4^2 + (-2)^2} \\\ &= \sqrt{20} \\\ &= 4.47 \\\ \end{aligned}\)
Scalar multiples
Vectors can be multiplied by scalar quantities in order to scale them up or down. To multiply a vector by a scalar, we multiply each of the vector’s components by the scalar.
If we have a vector \(\mathbf{u}\), the scaled vector will be \(k\mathbf{u}\) where \(k\) is the scalar and \(\mathbf{u}\) is the vector.
Example 1:
Multiply the vector \(\mathbf{f} = (9,3)\) by \(4\)
Solution:
\[\begin{aligned} 4 \mathbf{f} &= (4 \times 9 ,\,4 \times 3) \\\ &= (36, 12) \\\ \end{aligned}\]Example 2:
Calculate \(2.5 \mathbf{u}\) when \(\mathbf{u} = (5, 8)\)
Solution:
\[\begin{aligned} 2.5 \mathbf{u} &= (2.5 \times 5,\,2.5 \times 8) \\\ &= (12.5, 20) \\\ \end{aligned}\]Check your understanding of calculating scalar multiples by working through the questions below:
\(\begin{aligned} 3 \mathbf{a} &= (3 \times 7,\,3 \times -6) \\\ &= (21, -18) \\\ \end{aligned}\)
\(\begin{aligned} 7 \mathbf{b} &= (7 \times -3) \mathbf{i} + (7 \times 4) \mathbf{j} \\\ &= -21 \mathbf{i} + 28 \mathbf{j} \\\ \end{aligned}\)
\(\begin{aligned} 3 \mathbf{c} - 4 \mathbf{d} &= (3 \times 12 - 4 \times (-3)) \mathbf{i} + (3 \times 3 - 4 \times 5) \mathbf{j} \\\ &= (36 + 12) \mathbf{i} + (9 - 20) \mathbf{j} \\\ &= 48 \mathbf{i} - 11 \mathbf{j} \\\ \end{aligned}\)
Converting between coordinates
So far, we have used Cartesian coordinates to write vectors in terms of their \(x\) and \(y\) components. However, vectors can also be written using polar coordinates. When we use polar coordinates, we state the magnitude of the vector and its direction instead of its components.
The direction is given as an angle. This angle can be measured in degrees or in radians. We will use degrees on this page. For example, we could write a vector \(\mathbf{v}\) as \((12, 30^\circ)\) using polar coordinates.
To convert between Cartesian and polar coordinates, we need to use trigonometry. To convert from Cartesian coordinates to polar coordinates, we use two steps. First, we use Pythagoras’ theorem to calculate the magnitude. Second, we use the tangent trigonometric ratio to calculate the direction.
To write the vector \(\mathbf{a} = (\mathbf{a}_x, \mathbf{a}_y)\) using polar coordinates \((r, \theta)\) we calculate:
\[r = \sqrt{\mathbf{a_x}^2 + \mathbf{a_y}^2}\] \[\theta = \tan^{-1}\frac{a_y}{a_x}\]Example:
Write \(\mathbf{a} = (10, 4)\) using polar coordinates rounded to two decimal places.
Solution:
\[\begin{aligned} r &= \sqrt{\mathbf{a_x}^2 + \mathbf{a_y}^2} \\\ &= \sqrt{10^2 + 4^2} \\\ &= \sqrt{116} \\\ &= 10.77 \\\ \end{aligned}\] \[\begin{aligned} \theta &= \tan^{-1}\frac{a_y}{a_x} \\\ &= \tan^{-1}\frac{4}{10} \\\ &= 21.80^\circ \end{aligned}\]Therefore \(\mathbf{a} = (10.77,\,21.80^\circ)\)
To convert from polar coordinates to Cartesian coordinates we use the sine and cosine trigonometric ratios.
To write the vector \(\mathbf{b} = (r, \theta)\) using Cartesian coordinates \((\mathbf{b}_x, \mathbf{b}_y)\) we calculate:
\[\mathbf{b}_x = r\,cos\,\theta\] \[\mathbf{b}_y = r\,sin\,\theta\]Example:
Write \(\mathbf{b} = (25, 30^\circ)\) using Cartesian coordinates rounded to two decimal places.
Solution:
\[\begin{aligned} \mathbf{b}_x &= r\,cos\,\theta \\\ &= 25\,cos\,30^\circ \\\ &= 21.65 \\\ \end{aligned}\] \[\begin{aligned} \mathbf{b}_y &= r\,sin\,\theta \\\ &= 25\,sin\,30^\circ \\\ &= 12.5 \\\ \end{aligned}\]Therefore \(\mathbf{b} = (21.65, 12.5)\)
Check your understanding of how to convert between coordinates by working through the questions below:
Therefore \(\mathbf{u} = (12.37,\,14.04^\circ)\)
Therefore \(\mathbf{v} = (6.36,\,12.47)\)
Therefore the vector can be expressed as \((-2.46\textrm{m},\,1.72\textrm{m})\)
Multiplying vectors
If we want to multiply two vectors together, instead of just multiplying a vector by a scalar, there are two different methods we can use. The first method is called the dot product, and it gives an answer that is a scalar. The second method is called the cross product, and it gives an answer that is a vector.
The cross product is not covered on this page. However, how to calculate the dot product is explained below.
Dot product
The dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is written as:
\[\mathbf{a} . \mathbf{b}\]The dot product can be calculated in two different ways. The method we use depends on whether the vectors are written using Cartesian coordinates or polar coordinates.
If the vectors are written using Cartesian coordinates, we follow three steps. First, we multiply the two \(x\) components together. Second, we multiply the two \(y\) components together. Third, we add the two results.
The dot product of two vectors written using Cartesian coordinates is:
\[\mathbf{a} . \mathbf{b} = \mathbf{a}_x \times \mathbf{b}_x + \mathbf{a}_y \times \mathbf{b}_y\]Example:
Calculate \(\mathbf{a} . \mathbf{b}\) given that \(\mathbf{a} = (-2, 4)\) and \(\mathbf{a} = (3, 5)\)
Solution:
\[\begin{aligned} \mathbf{a} . \mathbf{b} &= \mathbf{a}_x \times \mathbf{b}_x + \mathbf{a}_y \times \mathbf{b}_y \\\ &= -2 \times 3 + 4 \times 5 \\\ &= -6 + 20 \\\ &= 14 \\\ \end{aligned}\]If the vectors are written using polar coordinates, we multiply the magnitude of each vector together and then multiply by the cosine of the angle between them.
The dot product of two vectors written using polar coordinates is:
\[\mathbf{a} . \mathbf{b} = \vert \mathbf{a} \vert \times \vert \mathbf{b} \vert \times cos\,\theta\]where \(\theta\) is the angle between the two vectors.
Example:
Calculate \(\mathbf{a} . \mathbf{b}\) given that \(\mathbf{a} = (4, 45^\circ)\) and \(\mathbf{b} = (7, 105^\circ)\)
Solution:
The angle between the two vectors is \(105^\circ - 45^\circ = 60^\circ,\) so:
\[\begin{aligned} \mathbf{a} . \mathbf{b} &= \vert \mathbf{a} \vert \times \vert \mathbf{b} \vert \times cos\,\theta \\\ &= 4 \times 7 \times cos\,60^\circ \\\ &= 28 \times 0.5 \\\ &= 14 \\\ \end{aligned}\]Check your understanding of using the dot product to multiply vectors by working through the questions below:
The angle between the two vectors is \(75^\circ - 22^\circ = 53^\circ,\) so:
\[\begin{aligned} \mathbf{u} . \mathbf{v} &= \vert \mathbf{u} \vert \times \vert \mathbf{v} \vert \times cos\,\theta \\\ &= 3 \times 8 \times cos\,53^\circ \\\ &= 24 \times 0.602 \\\ &= 14.44 \\\ \end{aligned}\]From question 2 above, we have: \(\mathbf{c} . \mathbf{d} = 33\)
Furthermore, calculating the magnitude of the vectors gives:
\[\begin{aligned} \vert \mathbf{c} \vert &= \sqrt{\mathbf{c_x}^2 + \mathbf{c_y}^2} \\\ &= \sqrt{6^2 + 7^2} \\\ &= \sqrt{85} \\\ &= 9.22 \\\ \end{aligned}\] \[\begin{aligned} \vert \mathbf{d} \vert &= \sqrt{\mathbf{d_x}^2 + \mathbf{d_y}^2} \\\ &= \sqrt{9^2 + (-3)^2} \\\ &= \sqrt{90} \\\ &= 9.49 \\\ \end{aligned}\]So we have:
\[\begin{aligned} \vert \mathbf{u} \vert \times \vert \mathbf{v} \vert \times cos\,\theta &= 33\\\ \therefore 9.22 \times 9.49 \times cos\,\theta &= 33 \\\ \therefore cos\,\theta &= 0.38 \\\ \therefore \theta &= cos^{-1}0.38 \\\ \therefore \theta &= 67.83^\circ \\\ \end{aligned}\]So the angle between the two vectors is \(67.83^\circ\)