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Sometimes we are given the set of parametric equations and we are asked to write the equation without the parameter by eliminating the parameter. This not always possible but with some equations there are ways to do it. Here are some of the ways.

The easiest technique to try is to solve one of the equations for the parameter and then substitute the result in the other equation. Here is an example.
\( x = 2t \) and \( y = t^2 \)
Solve the first equation for \(t\) giving \( t=x/2\) and substitute into the second equation.
\( y = (x/2)^2 = x^2/4 \)

WARNING - This technique can sometimes lead to errors if you are not careful. One example is shown below in practice problem 4423. You need to be careful to watch that you do not lose information about the domain and range when you perform algebra.

For example, if you have \(x=\sqrt{t}\) and you square both sides to get \(x^2=t\). Now, even though you have performed a valid operation, these equations are NOT the same. In the first one, \(x\) and \(t\) are both greater than or equal to zero. However, in the second one, \(x\) can be negative. So certain operations require you to keep the domain (when working with \(x\)) and the range (when working with \(y\)) in mind.

Here are some situations you need to watch out for.
1. when you taking the even power or even root of both sides of an equation,
2. when you are working with trig functions and taking inverses
3. when you are working with logarithms
4. when you dividing by a term, make sure it is never zero

A second technique involves the use of trig identities. For example, given the parametric equations \( x=\cos(t)\) and \(y=\sin(t)\), we know that \( \cos^2(x) + \sin^2(x) = 1 \). So we can write \( x^2 + y^2 = 1 \) which eliminates the parameter.

A third way is by inspection. Sometimes it is obvious what a substitution might be. For example, if we have \(x=e^t\) and \(y=e^{3t}+1\) then we can rewrite \(y=(e^t)^3+1\). Then we can replace \(e^t\) with \(x\) in the last equation (because our first parametric equation was \(x=e^t\)) to get \(y=x^3+1\). Of course, the first technique would have worked too by solving \(x=e^t\) for \(t\) and then substituting and simplifying but by standing back and looking at the equations more carefully, the solution was much easier.
Okay, time for some practice problems.