On this page, we discuss what vector functions are and how to determine the domain of vector functions. These ideas will probably not be new to you but how to write vector functions and notation are different than you've probably seen before.
These three terms are easily confused and some books and instructors interchange them. In general, vector functions are parametric equations described as vectors. Vector fields usually define a vector to each point in the plane or in space to describe something like fluid flow, air flow and similar phenomenon. Vector-valued functions may refer to either vector functions or vector fields. Look carefully at the context and check with your instructor to make sure you understand what they are talking about.
In all three cases, you need to look at the context to see what is being discussed. To avoid confusion, we do not use the term vector-valued function on this site but some of the instructors in the videos we use refer to vector-valued functions.
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What Are Vector Functions?
Vector functions are a way of writing parametric equations of a set of points in the plane or space in vector form. So, for example, if we have a set of parametric equations with parameter \(t\)
\( x(t) = \cos(t), y(t) = \sin(t) \)
we can write this as the vector function \(\vec{v}(t)\)
\( \vec{v}(t) = x(t)\hat{i} + y(t)\hat{j} \) |
or more directly as |
\( \vec{v}(t) = \cos(t)\hat{i} + \sin(t)\hat{j} \) |
Key - - The key to using this notation is that the terminal point of the vector defines the \((x,y)\) value for each particular \(t\) value. So, if you think about it, this is just a different way to write parametric equations.
Note - - We use the word 'function' here rather loosely since, as you remember from parametrics, the resulting graphs probably will not pass the vertical line test and, therefore, cannot be correctly referred to 'functions'. However, we are following the standard terms used in most textbooks. So we will continue to call these 'vector functions' whether or not the graphs pass the vertical line test.
3-space - - The idea is the same in 3-space. We will just add a \(\hat{k}\) component so that the equation might look like
\(\vec{w}(t) = f(t)\hat{i} + g(t)\hat{j} + h(t)\hat{k} \)
Multiple parameters - - Note that we are not limited to just one parameter. When describing planes, we may have 2 parameters or even more. So we may have a vector function that looks like
\( \vec{A}(\lambda,\mu) = F(\lambda,\mu)\hat{i} + G(\lambda, \mu)\hat{j} + H(\lambda,\mu)\hat{k}\)
The important point to remember about vector functions is that the terminal point of the vector defines the points in the plane or in space and writing the equations as a vector is just convenient and compact notation that you already learned with parametric equations. You can do everything with vector functions that you can with parametric equations.
Okay, so vector functions are not that hard. They are just a matter of taking parametric equations and writing them in vector form. This first video explains this in more detail, showing how to graph vector functions and it contains some great examples. It's a bit long but well worth taking the time to watch to get this clear in your head.
video by Dr Chris Tisdell |
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The position of an object in the plane (2-dim) or in space (3-dim) can be described by vector functions using the same ideas as above. We go into more detail on the projectile motion page but here is a video to watch first to give you a better feel for vector functions and for what is coming up.
video by Khan Academy |
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Practice
Determine if any of the points \( (3,e^5, \ln(4))\), \((-1,1,0) \), \( (1/3, e^2, \ln(5)) \) lie on the curve \(\displaystyle{ \vec{r}(t) = \frac{1}{t^2-1}\hat{i} + e^t\hat{j} + [\ln(t+1)]\hat{k} }\)
Problem Statement |
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Determine if any of the points \( (3,e^5, \ln(4))\), \((-1,1,0) \), \( (1/3, e^2, \ln(5)) \) lie on the curve \(\displaystyle{ \vec{r}(t) = \frac{1}{t^2-1}\hat{i} + e^t\hat{j} + [\ln(t+1)]\hat{k} }\)
Solution |
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video by PatrickJMT |
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Domain of a Vector Function
The domain of a vector function is the intersection of domain of each term. To find the domain of a vector function, find the domain of the \(\hat{i}\), \(\hat{j}\) and, if it exists, \(\hat{k}\) terms and then take the intersection of those domains. Here are some good practice problems.
Practice
Find the domain of \(\displaystyle{ \vec{r}(t) = \frac{t-2}{t+2}\hat{i} + \sin(t)\hat{j} + \ln(9-t^2)\hat{k} }\).
Problem Statement |
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Find the domain of \(\displaystyle{ \vec{r}(t) = \frac{t-2}{t+2}\hat{i} + \sin(t)\hat{j} + \ln(9-t^2)\hat{k} }\).
Solution |
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video by PatrickJMT |
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Find the domain of \( \vec{r}(t) = \langle \sqrt{4-t^2}, e^{-3t}, \ln(t+1) \rangle \).
Problem Statement |
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Find the domain of \( \vec{r}(t) = \langle \sqrt{4-t^2}, e^{-3t}, \ln(t+1) \rangle \).
Final Answer |
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\( (-1,2] \)
Problem Statement |
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Find the domain of \( \vec{r}(t) = \langle \sqrt{4-t^2}, e^{-3t}, \ln(t+1) \rangle \).
Solution |
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video by Krista King Math |
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Final Answer |
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\( (-1,2] \)
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Calculus of Vector Functions
Okay, so now you know what vector functions are and how to graph them (from the video above). Let's discuss calculus on vector functions. The main topics we will discuss are limits, derivatives and integrals. These are all critical topics that you need to understand when you get to vector analysis. See these separate pages for discussion of limits, derivatives and integrals of vector functions.
Really UNDERSTAND Calculus
external links you may find helpful |
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The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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