Vectors are used in many applications and describe a quantity that occurs in a certain direction. Within the context of vectors, we use the term scalars when referring to numbers like 3 or \(-\pi\). The only data we have on a scalar is a value (including its sign). However, if we have a vector, we have a value that occurs in a specific direction. This is extremely valuable when describing things like force, displacement or velocity, usually encountered in physics.
Recommended Books on Amazon (affiliate links) | ||
---|---|---|
Join Amazon Student - FREE Two-Day Shipping for College Students |
scalar |
a number, either positive or negative | |
vector |
a number, either positive or negative, and a direction |
If the idea of scalars versus vectors is new to you, we recommend that you watch this next video. It explains this difference very well. As a bonus, it also discusses the difference between speed and velocity, a key concept necessary in calculus and physics but sometimes not explained completely by instructors.
video by Khan Academy |
---|
Geometric-Based Introduction |
Before we jump into the details of vectors, here is a great video for you that gives you a gentle introduction to vectors from a geometric point of view. If vectors are completely new to you, this will get you started on the right foot. If you are already familiar with vectors, this video will give you some background that you may not have seen before. Either way, this is a great video to watch.
video by Dr Chris Tisdell |
---|
A gentle, video-only introduction to vectors by Dr Chris Tisdell can be found in this youtube playlist. |
Dr Chris Tisdell also has a free ebook called Introduction To Vectors, which can be downloaded from this page. |
Notation
Since vectors are special entities, we need special notation. There are several ways to describe vectors.
1. Most textbooks show vectors in bold-face type, i.e. v.
2. Since it's kind of difficult to write bold-face by hand, it is better to write vectors with an arrow above, i.e. \(\vec{v}\).
3. Some special vectors are written with a 'hat', i.e. \(\hat{i}\). This notation is saved for a special set of vectors called the primary unit vectors. You can read more about them on the unit vectors page.
4. When I am writing vectors, my personal favorite is to use a half-arrow (also called a harpoon) above the name of the vector. It's hard to show this here but the idea is to use something like this \( \rightharpoonup \) above the vector name.
5. A couple of less common ways to write vectors are with a straight line either above or below the vector name. Some of the videos show the instructors doing this but you will be able to tell from the context that they are talking about vectors.
On this site, we will use option 2, an arrow above the vector name and option 3, the 'hat' notation for unit vectors. It would be easier for us to make the vector name bold but we think it will help you to keep seeing the vectors like you need to write them. You need to ask your instructor to see what they expect. However, good instructors will require you to write vectors with the correct notation, i.e. \(\vec{v}\) is a vector, while \(v\) is a scalar.
In the next section, we discuss the basics of vectors and how to represent them. We will also show you the notation required to represent specific vectors.
Vector Representations
To help us visualize what is going on, vectors are usually drawn as arrows. It is easiest initially to show two dimensional vectors. There are applications that can draw vectors in three dimensions but it is difficult to accurately draw three dimensional vectors by hand. So for now, we will stay in two dimensions.
Vectors have two pieces of information that identify them as vectors, magnitude and direction. For visualization purposes, we will show vectors on graphs as follows.
Magnitude is shown by the length of the vector.
Direction is shown by the orientation of the vector.
The graph to the right shows five vectors labeled \( \vec{A}, \vec{B}, \vec{C}, \vec{D}, \) and \(\vec{E} \). Notice these things about the vectors:
1. Each vector has a starting point, called an initial point, and an ending point, called a terminal point. The arrow is on the terminal point end and indicates that the vector starts at the initial point and ends at the terminal point.
Important Point - - Unlike graphs of functions, the arrow does not indicate that the line or vector goes on forever. With vectors, the arrow indicates the direction of the vector and, by convention, is usually drawn at the terminal end of the vector.
2. The magnitude of the vector is indicated by it's length. For example, the magnitude of vector \(\vec{A}\) is greater than the magnitude of vector \(\vec{D}\).
3. Vector \(\vec{A}\) can be described as going from the point \( (-4, -5) \) to the point \((-1,4)\). This is one way to describe a vector (but is not the preferred method; this will be explained below).
4. Vectors are independent of their position, i.e. it doesn't matter where they are in the plane.
5. Two vectors are said to be equal if they have the same magnitude and direction.
Before we jump into a lot of detail, take a few minutes to watch this video. It explains very well what we are about to discuss below.
video by PatrickJMT |
---|
Let's discuss the last two points above in more detail.
Magnitude
To find the magnitude of a vector we use the Pythagorean Theorem. For example, let's look at vector \(\vec{E}\). Vector \(\vec{E}\) starts at the point \((4,1)\) and ends at \((2, 4)\). If you draw a triangle where the legs are parallel to the coordinate axes and the hypotenuse is the vector itself, using the Pythagorean Theorem, you have
\( \| \vec{E} \| = \sqrt{ 2^2 + 3^2 } = \sqrt{ 13 } \)
The magnitude of vector \( \vec{E} \) is \( \| \vec{E} \| = \sqrt{13} \). We usually do not say 'magnitude of a vector'. We use the term norm to indicate the magnitude of a vector. Also, we use double vertical bars on each side of the name of the vector (with the proper arrow notation) to indicate the norm of a vector. So when discussing this equation, we say the norm of vector \( \vec{E} \) is \(\sqrt{13}\). [For practice calculating the norm of vectors, see the practice problems below.]
Okay, so as you would expect, here is a video clip explaining how to find the magnitude of a vector.
video by PatrickJMT |
---|
Direction
The second part of describing a vector is to know it's direction. Above we mentioned that we could give the initial point and the terminal point of the vector. This would certainly work but there is an easier, shorter and more standard way to indicate it's direction. For starters, let's look at vectors \(\vec{B}\) and \(\vec{C}\) in the above plot.
Notice that vector \( \vec{B} \) starts at the point \((-1,-1)\) and ends at the point \((1,0)\). So we can say that vector \( \vec{B} \) goes right 2 units and up 1 unit. An easier way to write this vector is \( \vec{B} = \langle 2, 1 \rangle \). The angle brackets indicate that this is a vector.
Now, if you think about this you may notice that we have lost the information about where the vector starts and where it ends. But we are okay with that because of the second to the last item in the above list, i.e. vectors are independent of their position. This means that it doesn't matter where in the plane the vector is. We don't need to know where it is. All we need to know is its magnitude and its direction.
There are two ways to get these vector numbers. We can graph the vector and pull the information right off the graph like we did above. This really only works if the initial and terminal points can be clearly read off a graph. This will NOT usually be the case. A much better way is to calculate the vector information from the given initial and terminal points.
Let's use vector \( \vec{B} \) as an example. Let's label the initial point \( I = (-1,-1) \) and the terminal point \( T = (1,0) \). From your coordinate system background, you should know that the x-distance from \( I \) to \( T \) is \( T_x - I_x = 1 - (-1) = 2 \) and the y-distance from \( I \) to \( T \) is \( T_y - I_y = 0 - (-1) = 1 \). So we can write \( \vec{B} = \langle 2,1 \rangle \).
For another example, let's calculate the same information about vector \( \vec{D} \). This vector starts at \( (2,-4) \) and ends at \( (4,-3) \). So we can write \( \vec{D} = \langle 4-2 , -3-(-4) \rangle = \langle 2,1 \rangle \).
Now let's look at the last item in the list above, i.e. two vectors are said to be equal if they have the same magnitude and direction. Since vector \( \vec{B} = \langle 2,1 \rangle \) and vector \( \vec{D} = \langle 2,1 \rangle \), \( \vec{B} = \vec{D} \) since the vectors are in the same direction and have the same magnitude. It doesn't matter that they are in different positions. ( NOTE: This concept applies to vectors in general. Once we start doing application problems, it will matter where the vector is located, for example, in the case of force or acceleration. )
This video explains in more detail this concept of when two vectors are considered equivalent.
video by PatrickJMT |
---|
There is another way to write vectors using the standard unit vectors. You can find an explanation on the unit vectors page. See the practice problems below to practice finding the vector representation.
Here is a good video clip explaining components and magnitude of vectors.
video by PatrickJMT |
---|
In this section, we've been working with two dimensional vectors, also called vectors in the xy-plane. We can add a third component to describe vectors in space. Such a vector would look like \( \vec{v} = \langle v_x, v_y, v_z \rangle \), where
\( v_x \) is the magnitude of the vector in the x-direction;
\( v_y \) is the magnitude of the vector in the y-direction;
\( v_z \) is the magnitude of the vector in the z-direction;
Okay, so up until now, we've discussed vectors that have obvious initial and final points. But what if you are just given two points and told to determine the vector between those two points. The wording is important.
If you are given two points, you can determine a vector between those points. You have two possible directions of the vector, depending on how you perform the operation. Here is how it works.
Given the points \( P = (p_x, p_y, p_z) \) and \( Q = (q_x, q_y, q_z) \), you form a vector by subtracting components. The direction of the vector is determined by the order of the subtraction.
The vector from \(P\) to \(Q\) is \( \overrightarrow{PQ} = Q - P = \langle q_x - p_x, q_y - p_y, q_z - p_z \rangle \). Notice that the direction is \(P \to Q \) and the idea can be written as \( \overrightarrow{PQ} = \) terminal - initial. We write the name of the resulting vector in the expected order, i.e. \( P \to Q = \overrightarrow{PQ} \).
Here is a good video demonstrating this process.
video by Krista King Math |
---|
Basic Vector Operations
As you probably expect, since vectors are special entities, there are only very specific operations that can be done with vectors. This next panel explains some of them and ends with a rather thorough list of operations as well as some properties that will help you work with vectors.
We have already discussed one vector operation above under vector representations, finding the norm (magnitude) of a vector. It's pretty boring to work with just one vector. We need ways to combine vectors. We can perform other operations like adding and multiplying vectors.
Adding vectors is pretty easy. Basically we just add each corresponding components, i.e. we add the x-component of one vector to the x-component of the second component and similarly for the y and z components. Let's do an example.
Using the plot on the right, let's add vectors \( \vec{A} \) and \( \vec{B} \).
\(\begin{array}{rcl}
\vec{A} + \vec{B} & = & \langle 3,9 \rangle + \langle 2,1 \rangle \\
& = & \langle 3+2, 9+1 \rangle \\
& = & \langle 5,10 \rangle
\end{array}\)
A way to visualize this addition process is to use what is called the 'tip to tail' method. This really only works for vectors that have integer components and can be easily graphed and is meant to just give you a feel for vectors. It will not be used most of the time. If you would like an idea of how this works, you can find an explanation on this wiki page.
To subtract vectors, we need another concept first. We can adjust the length of a vector by multiplying the vector by a scalar. The resulting vector will be a new vector in the same direction (when the scalar is positive) but have a different length. For example, we can take vector \( \vec{C} \) in the graph on the right and multiply it by 2 to double the length as follows.
\( 2\vec{C} = 2\langle 2,1 \rangle = \langle 4,2 \rangle \)
From this example, you can see that we multiply the scalar by each component of the vector. The resulting vector is a vector in the same direction as \(\vec{C}\) but with twice the magnitude.
If you multiply a vector by a negative number, you essentially reverse the direction. For example, if we multiply vector \( \vec{C} \) by \( -1 \), the result is \( (-1)\vec{C} = (-1)\langle 2,1 \rangle = \langle -2,-1 \rangle \).
Now for subtraction, let's start with subtraction of scalars. If you subtract \( a - b \), you can write this as \( a+(-b) \), i.e. you add a to the negative of b. The same applies to vectors. To subtract vectors, we multiply the second one by \(-1\) and then add it to first vector. Let's do an example.
\(\begin{array}{rcl}
\vec{A} - \vec{B} & = & \langle 3,9 \rangle - \langle 2,1 \rangle \\
& = & \langle 3,9 \rangle + (-1)\langle 2,1 \rangle \\
& = & \langle 3,9 \rangle + \langle -2,-1 \rangle \\
& = & \langle 3-2,9-1 \rangle \\
& = & \langle 1,8 \rangle
\end{array}\)
Now it's time for a video. This video explains these ideas again and shows how to get vector components using trig, an important skill.
video by Khan Academy |
---|
This video explains addition of vectors and multiplication of vectors by a scalar. It also contains plenty of good examples.
video by PatrickJMT |
---|
Okay, so far we know how to find the norm of a vector, add two vectors and subtract two vectors. Those are the basic operations you need for now. But wait! You may be thinking, what about multiplying two vectors? Multiplication is much more involved and there are actually two ways to multiply vectors, dot product and cross product. You can find more information on those two pages.
This theorem contains a rather thorough list of properties of vectors that you need to be familiar with.
Theorem - Properties of Vector Operations | |||
---|---|---|---|
Let \(\vec{u}\), \(\vec{v}\) and \(\vec{w}\) be vectors in space, and let a and b be scalars. | |||
1. | \(\vec{u} + \vec{v} = \vec{v} + \vec{u}\) |
commutative property | |
2. | \((\vec{u} + \vec{v}) + \vec{w} = \vec{u} + (\vec{v}+\vec{w})\) |
||
2. Associative Property, Vector Addition ProofLet \(\vec{u}\), \(\vec{v}\) and \(\vec{w}\) be vectors in space as follows Dr Chris Tisdell - Associative Law for Vector Addition [6mins-42secs]
| |||
3. | \(\vec{u} + \vec{0} = \vec{u}\) |
additive identity property | |
4. | \(\vec{u} + (-\vec{u}) = \vec{0}\) |
additive inverse property | |
5. | \((a+b)\vec{u} = a\vec{u}+b\vec{u}\) |
||
5. Scalar Distributive Property ProofLet \(\vec{u} = \langle u_x, u_y, u_z \rangle \) be a vector in space and \(a\) and \(b\) be scalars.
| |||
6. | \(a(\vec{u}+\vec{v}) = a\vec{u} + a\vec{v}\) |
||
6. Vector Distributive Property ProofHere is a video proving this property. Dr Chris Tisdell - Distributive Law for Vectors [4mins-8secs]
| |||
7. | \(a(b\vec{u}) = (ab)\vec{u}\) |
||
7. Associative Property, Scalar Multiplication ProofHere is a video proving this property. Dr Chris Tisdell - Scalar Multiplication of Vectors: Associative law [4mins-10secs]
| |||
8. | \(0(\vec{u}) = \vec{0}\) |
zero property |
Important Note - - In the above list of properties, \(0\) is the zero scalar while \(\vec{0}\) is the zero vector, i.e. a vector with zero length. This idea may seem strange but you will get used to it over time. However, you need to keep them straight when performing vector operations.
Here is a list of some important vector properties and definitions.
Definitions | |
---|---|
Parallel Vectors |
Two nonzero vectors \( \vec{u} \) and \(\vec{v}\) are parallel if there exists a scalar \(c\) such that \(\vec{u} = c\vec{v}\). |
Okay, time for some practice problems.
Once you are done with those, we suggest unit vectors as the next topic.
Practice
Unless otherwise instructed, give your answers in exact, simplified form.
Basic |
---|
a) For each of the vectors in the graph, find the vector representation and calculate the norm.
b) Which vectors are equal? Explain.
Problem Statement |
---|
a) For each of the vectors in the graph, find the vector representation and calculate the norm.
b) Which vectors are equal? Explain.
Solution |
---|
a) Vector \( \vec{A} \): initial point \((-4,-5)\) and terminal point \((-1,4)\). It goes right \( -1 - (-4) = 3 \) units and it goes up \( 4 - (-5) = 9 \) units. So its vector representation is \( \langle 3,9 \rangle \). This can also be written \( 3\hat{i}+9\hat{j} \).
To calculate its norm we use the Pythagorean Theorem.
\(\begin{array}{rcl} \norm{\vec{A}} & = & \sqrt{(3)^2 + (9)^2} \\ & = & \sqrt{9+81} \\ & = & \sqrt{90} \\ & = & \sqrt{9 \cdot 10} \\ & = & 3\sqrt{10} \end{array}\)
\(\vec{A} = \langle 3,9 \rangle = 3\hat{i}+9\hat{j}\) |
\(\norm{\vec{A}} = 3\sqrt{10}\) |
Vector \( \vec{B} \): initial point \((-1,-1)\) and terminal point \((1,0)\). It goes right \( 1 - (-1) = 2 \) units and it goes up \( 0 - (-1) = 1 \) unit. So its vector representation is \( \langle 2,1 \rangle \).
To calculate its norm we use the Pythagorean Theorem.
\( \norm{\vec{B}} = \sqrt{(2)^2 + (1)^2} = \sqrt{5} \)
\(\vec{B} = \langle 2,1 \rangle = 2\hat{i}+\hat{j}\) |
\(\norm{\vec{B}} = \sqrt{5}\) |
Vector \( \vec{C} \): initial point \((0,0)\) and terminal point \((2,1)\). It goes right \( 2-0 = 2 \) units and it goes up \( 1-0 = 1 \) unit. So its vector representation is \( \langle 2,1 \rangle \).
To calculate its norm we use the Pythagorean Theorem.
\( \norm{\vec{C}} = \sqrt{(2)^2 + (1)^2} = \sqrt{5} \)
\(\vec{C} = \langle 2,1 \rangle = 2\hat{i}+\hat{j}\) |
\(\norm{\vec{C}} = \sqrt{5}\) |
Vector \( \vec{D} \): initial point \((2,-4)\) and terminal point \((4,-3)\). It goes right \( 4-2 = 2 \) units and it goes up \( -4 - (-3) = 1 \) unit. So its vector representation is \( \langle 2,1 \rangle \).
To calculate its norm we use the Pythagorean Theorem.
\( \norm{\vec{D}} = \sqrt{(2)^2 + (1)^2} = \sqrt{5} \)
\(\vec{D} = \langle 2,1 \rangle = 2\hat{i}+\hat{j}\) |
\(\norm{\vec{D}} = \sqrt{5}\) |
Vector \( \vec{E} \): initial point \((4,1)\) and terminal point \((2,4)\). It goes right \( 2 - 4 = -2 \) units and it goes up \( 4-1=3 \) unit. So its vector representation is \( \langle -2,3 \rangle \).
To calculate its norm we use the Pythagorean Theorem.
\( \norm{\vec{E}} = \sqrt{(-2)^2 + (3)^2} = \sqrt{4+9} = \sqrt{13} \)
Notice that for this vector we said the vector goes right \( -2 \) units. It would have also been correct to say that the vector goes left \( 2 \) units but then we would have had to remember to put the minus sign in the vector representation. So it is probably better to say the vector always goes to the right and the sign will be correct.
\(\vec{E} = \langle -2,3 \rangle = -2\hat{i}+3\hat{j}\) |
\(\norm{\vec{E}} = \sqrt{13}\) |
b) Vectors \(\vec{B}\), \(\vec{C}\) and \(\vec{D}\) are all equivalent. We know this because they all have the same magnitude of \( \sqrt{5}\) and the same direction, indicated by the representation \( \langle 2,1 \rangle \).
Log in to rate this practice problem and to see it's current rating. |
---|
Find the sum of \( \langle 1, -2 \rangle \) and \( \langle 3, 4 \rangle \).
Problem Statement |
---|
Find the sum of the vectors \( \langle 1, -2 \rangle \) and \( \langle 3, 4 \rangle \).
Solution |
---|
video by Krista King Math |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
Find the component form and the magnitude of the vector from \((2,3)\) to \((7,6)\).
Problem Statement |
---|
Find the component form and the magnitude of the vector from \((2,3)\) to \((7,6)\).
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
Find the component form and the magnitude of the vector from \((3,1)\) to \((2,-3)\).
Problem Statement |
---|
Find the component form and the magnitude of the vector from \((3,1)\) to \((2,-3)\).
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
Calculate the magnitude of the vector \( \vec{a} = \langle 3,1,-2 \rangle \).
Problem Statement |
---|
Calculate the magnitude of the vector \( \vec{a} = \langle 3,1,-2 \rangle \).
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
Calculate the magnitude of the vector \( \vec{b} = \langle 5,-1,2 \rangle \).
Problem Statement |
---|
Calculate the magnitude of the vector \( \vec{b} = \langle 5,-1,2 \rangle \).
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
Determine the components of the vector with magnitude \(8\) at an angle of \(60^{\circ}\).
Problem Statement |
---|
Determine the components of the vector with magnitude \(8\) at an angle of \(60^{\circ}\).
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
Determine the magnitude and direction of the vector \( \langle -3,4 \rangle \).
Problem Statement |
---|
Determine the magnitude and direction of the vector \( \langle -3,4 \rangle \).
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
Determine the magnitude and direction of the vector \( \langle -2,-5 \rangle \).
Problem Statement |
---|
Determine the magnitude and direction of the vector \( \langle -2,-5 \rangle \).
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
Determine the magnitude and direction of the vector \(\langle 2,6 \rangle\).
Problem Statement |
---|
Determine the magnitude and direction of the vector \(\langle 2,6 \rangle\).
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
Determine the magnitude and direction of the vector \(\langle 3, -10 \rangle\).
Problem Statement |
---|
Determine the magnitude and direction of the vector \(\langle 3, -10 \rangle\).
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
For \(\vec{w}=\langle 5,-3 \rangle\) and \( \vec{u} = \langle -1,4 \rangle \), find \( \vec{w} - \vec{u} \).
Problem Statement |
---|
For \(\vec{w}=\langle 5,-3 \rangle\) and \( \vec{u} = \langle -1,4 \rangle \), find \( \vec{w} - \vec{u} \).
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
For \(\vec{w}=\langle 2,-5 \rangle\) \(\vec{v}=\langle 1,3 \rangle\), find \(\vec{w} +\vec{v}\) and \(2\vec{v}\) and sketch all vectors.
Problem Statement |
---|
For \(\vec{w}=\langle 2,-5 \rangle\) \(\vec{v}=\langle 1,3 \rangle\), find \(\vec{w} +\vec{v}\) and \(2\vec{v}\) and sketch all vectors.
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
Intermediate |
---|
Determine the components of the vector with magnitude \(10\sqrt{2}\) at an angle of \(135^{\circ}\).
Problem Statement |
---|
Determine the components of the vector with magnitude \(10\sqrt{2}\) at an angle of \(135^{\circ}\).
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
Determine the components of the vector with magnitude 14 at an angle of \(245^{\circ}\).
Problem Statement |
---|
Determine the components of the vector with magnitude 14 at an angle of \(245^{\circ}\).
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
For \( \vec{w}=\langle 5,-3 \rangle \), \(\vec{u}=\langle -1,4 \rangle \), \(\vec{v}=\langle 0,2 \rangle\), find \(3\vec{w}+2\vec{v}-4\vec{u}\).
Problem Statement |
---|
For \( \vec{w}=\langle 5,-3 \rangle \), \(\vec{u}=\langle -1,4 \rangle \), \(\vec{v}=\langle 0,2 \rangle\), find \(3\vec{w}+2\vec{v}-4\vec{u}\).
Solution |
---|
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
Really UNDERSTAND Calculus
external links you may find helpful |
---|
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
---|---|---|---|
\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
---|---|---|
\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
---|---|
\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
---|---|
\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
To bookmark this page and practice problems, log in to your account or set up a free account.
Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
| |
Save Up To 50% Off SwissGear Backpacks Plus Free Shipping Over $49 at eBags.com! |
---|
Practice Instructions
Unless otherwise instructed, give your answers in exact, simplified form.