The main area where you need to be able to describe a curve using parametric equations is when evaluating line integrals. The skills required to do this involve only graphs (precalculus) and parametrics.
This technique is called parameterization of a curve or parameterizing a curve. 

The idea here is that you are given the graph (and sometimes the equation) of a line segment (usually of a finite length but not necessarily a straight line) and you are told to describe the curve using parametric equations. There are an infinite number of ways to set up the parametric equations. We will use these guidelines in the following discussion.
1. On this page, we use the term line to describe a line segment of finite length, which is not necessarily a straight line. It could be a curved line. We could also use the term arc or curve. If the line is straight, then we will specificially say straight line.
2. We will use t as the parameter.
3. We will start t at zero and increase the value, i.e. \(t \geq 0\). The value of t will never be negative.
4. We assume that you have the equation of the curve. If you are given a graph only, you need to determine the associated equation with that curve in order to apply the techniques described here.
5. We will discuss only smooth curves, which are curves whose first derivative is both continuous and not equal to zero.
6. We will limit discussion with curves in the xyplane. It is easy to extend the ideas to other planes.
An Individual Line Segment in the xyplane
The easiest way to describe a curve in the plane is to let the dependent variable be equal to t and then plug in t where that variable occurs in the equation. For example, if you have an equation \(y=f(x)\) where x is the dependent variable, your parametric equations of the curve are
\( x(t) = t \) 
\( y(t) = f(t) \) 
Pretty easy, eh? Okay, since we are talking about a finite length curve we need a range on t. We suggest starting on one end and move to the other end, starting with the smallest value of the dependent variable (initial value \(t_i\)) up to the largest value (final value \(t_f\)). If t starts at zero or your instructor/textbook does not require that t start at zero, you are done. Here are two examples.
given \( y = f(x) \) 

set \(x=t\) 
\( y=f(t)\) 
\( t_i \leq t \leq t_f\) 
given \( x = g(y) \) 

set \(y=t\) 
\( x=g(t)\) 
\( t_i \leq t \leq t_f\) 
The best description for the range on t is for t to start at zero. So what do you do if you follow the instructions above and you get \( t_i \neq 0\) and you need \( t_i = 0 \)? There is a very easy solution. This works whether \(t_i > 0\) or \(t_i < 0\).
Here is the rule: Everywhere there is a t, replace t with \(t+t_i\). Let's see what this looks like. We will work with the case \(y=f(x)\) from the above table.
original 
adjusted to start t at zero  

\(x(t)=t\) 
\(x(t)=t+t_i\)  
\(y(t)=f(t)\) 
\(y(t)=f(t+t_i)\)  
\( t_i \leq t \leq t_f\) 
\( 0 \leq t \leq t_ft_i\) 
In the range for t in the adjusted column, we performed the following operations.
\(
\begin{array}{rcccl}
t_i & \leq & t+t_i & \leq & t_f \\
t_i t_i & \leq & t+t_it_i & \leq & t_ft_i \\
0 & \leq & t & \leq & t_ft_i
\end{array}
\)
Okay, it is time for an example. If you think you are ready, try this yourself before looking at the hidden solution.
Parameterize the curve shown in this plot. The equation of the curve is \(y=x^31\). Start t at \(t=0\).
\(x(t)=t1;\) \( y(t)=(t1)^31;\) \( 0 \leq t \leq 3\)
Parameterize the curve shown in this plot. The equation of the curve is \(y=x^31\). Start t at \(t=0\).
Solution 

Parameterize the curve shown in this plot. The equation of the curve is \(y=x^31\). Start t at \(t=0\).
Solution 
We are given most of the information we need. We know that the equation of the curve is \(y=x^31\) and from the graph we can see that x ranges from \(1\) to \(2\). Since x is the independent variable, we let \(x(t)=t\). The other equation gives us \(y(t)=t^31\). Now we let t be from the smallest value of x to the largest, giving us \( 1 \leq t \leq 2 \).
To get t starting at zero, we replace every \(t\) with \(t1\). This gives us our final answer.
\(x(t)=t1;\) 
\( y(t)=(t1)^31;\) 
\( 0 \leq t \leq 3\) 
One of the main reasons we use parametrics to describe curves is that we can specify that the curve be swept out in a specific direction. In this example, the curve is swept out left to right as shown in this plot. The next section describes how to switch directions of the curve, if the parameterization you come up with is going in the wrong direction.
Final Answer 

\(x(t)=t1;\) \( y(t)=(t1)^31;\) \( 0 \leq t \leq 3\) 
Curve With a Specific Direction
Notice in the last example and in all our discussion so far, we have not said anything about the direction of the curve. Remember from parametrics that describing a curve parametrically gives you added information about the direction of the curve. So, many times, your graph will have an indicated direction and you will be asked to describe the curve taking the direction into account. This is important when working with line integrals.
So what do you do if you follow the techniques described above but your curve is going the wrong way? There are two ways to handle this. First, you could just reverse the direction t but this is not considered good mathematics. So, what we need to do is change the equations.
Go back to the last example. If we graph the parametric equations that we found and include the direction of the curve, we get the graph above (inside the solution panel). But, let's say we want the same graph but we want the arrow to go the other direction.
To do that we leave the range on t alone and replace every t in the equations \(x(t)\) and \(y(t)\) with \(t_ft\). So our answer for the last example above but with the direction of the graph in the opposite direction is
\(x(t)=2t\) 
\( y(t)=(2t)^31\) 
\( 0 \leq t \leq 3\)  
The plot to the right is the result. 
Multiple Curves Described In Parallel
Okay, so now you know how to describe one line (or curve) in the plane. Sometimes we need to describe multiple line segments that are governed by different equations. In this case, we have multiple sets of equations. It is easiest to describe them in parallel, meaning that each segment is described by equations where t starts at zero for each segment. The resulting answer will be in multiple parts. Here is an example. If you are feeling particularly comfortable with this technique, try it yourself before looking at the solution.
Parameterize the curve shown in this plot. Start each segment with \(t_i=0\).
A 
\(x(t)=t\) 
\(y(t)=0\) 
\(0 \leq t \leq 3\)  
B 
\(x(t)=3\) 
\(y(t)=t\) 
\(0 \leq t \leq 3\)  
C 
\(x(t)=3t\) 
\(y(t)=3t\) 
\(0 \leq t \leq 3\) 
Parameterize the curve shown in this plot. Start each segment with \(t_i=0\).
Solution 

Parameterize the curve shown in this plot. Start each segment with \(t_i=0\).
Solution 
First, let's label each segment so that we can be on the same page in the following discussion.
A  blue line segment on the xaxis 
B  green line segment at \(x=3\) 
C  red line segment from \((0,0)\) to \((3,3)\) 
Line Segment A   This one is pretty easy. The equation is \( y(x) = 0 \) and x ranges between \(0\) and \(3\). Using the ideas above, the parametric equations are \(x(t) = t; y(t) = 0; 0 \leq t \leq 3 \).
Line Segment B   Again this one should be pretty easy to see. The equation is \(x(y) = 3\) and \(y\) ranges between \(0\) and \(3\). So the parametric equations are \( x(t) = 3; y(t) = t; 0 \leq t \leq 3 \)
Line Segment C   This one is not too bad but we have to be careful about the direction. The equation of the line is \( y(x) = x \) and x ranges between \(0\) and \(3\). This gives us the parametric equations (but with the wrong direction) \( x(t) = t; y(t) = t; 0 \leq t \leq 3 \). To reverse the direction we replace t with \(3t\) in the equations for \(x(t)\) and \(y(t)\) but leave the range alone, giving us \( x(t) = 3t; y(t) = 3t; 0 \leq t \leq 3 \).
Below is a summary of the final answer.
A 
\(x(t)=t\) 
\(y(t)=0\) 
\(0 \leq t \leq 3\)  
B 
\(x(t)=3\) 
\(y(t)=t\) 
\(0 \leq t \leq 3\)  
C 
\(x(t)=3t\) 
\(y(t)=3t\) 
\(0 \leq t \leq 3\) 
Notes
1. The ranges on t are all the same but this will not happen all the time. It is a result of the configuration of this particular graph.
2. To solve this problem, we actually picked the starting point at the origin. We could have started elsewhere but it made sense to start there and go counterclockwise. (This will be more important in the discussion below)
Final Answer  


Multiple Lines Described Sequentially
Okay, so let's say you are asked to describe the line segments sequentially, i.e. with ranges on t run in sequential order. In the last example, this would look something like this.
A  \(0 \leq t \leq 3\) 
B  \(3 \leq t \leq 6\) 
C  \(6 \leq t \leq 9\) 
You can probably deduce what to do here but let's go through it. Remember in the first example above, when we wanted to shift the initial time from \(t_i\) (which we assumed was nonzero) to zero, we replaced t with \(t+t_i\). Now we are doing the opposite. We have \(t_i=0\) and we want to start from a nonzero time. As you would expect, we do the opposite, i.e. we replace t with \(tt_i\). In the previous example, the answers are now
A 
\(x(t)=t\) 
\(y(t)=0\) 
\(0 \leq t \leq 3\)  
B 
\(x(t)=3\) 
\(y(t)=t3\) 
\(3 \leq t \leq 6\)  
C 
\(x(t)=9t\) 
\(y(t)=9t\) 
\(6 \leq t \leq 9\) 
Notes
1. \( 3(t6) = 3t+6 = 9t \)
2. Implied in this discussion is that we chose the starting point at the origin.
Special Case: Straight Line Between Two Points
Okay, so you probably know by now that we prefer not to highlight special cases or shortcuts in calculus. We want you to learn the underlying calculus and how and when it works. However, here is a special case that comes up quite often and is not fundamental to calculus. So we present it here as something that might help you.
If you need to parameterize a straight line between two points in either the plane or in 3space, you can use a special equation. Let's call the initial point \(P= (p_x, p_y, p_z)\) and the final point \(Q=(q_x, q_y, q_z)\). One possible parameterization is
\( \vec{c}(t) = (1t)\langle p_x, p_y, p_z \rangle +\) \( (t)\langle q_x, q_y, q_z \rangle \) 
\(0 \leq t \leq 1\) 
Note   This is a straight line between the points, which is the only time this special equation can be used.
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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  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle 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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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