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Example 1 Sketch the parametric curve for the following set of parametric equations. 𝑥 =𝑡2+𝑡𝑦 =2⁢𝑡−1 At this point our only option for sketching a parametric curve is to pick values of 𝑡, plug them into the parametric equations and then plot the points. So, let’s plug in some 𝑡’s. We’ll discuss an alternate graphing method in later examples that will help to explain how these values of 𝑡 were chosen. This may seem like an unimportant point, but as we’ll see in the next example it’s more important than we might think. Before addressing a much easier way to sketch this graph let’s first address the issue of limits on the parameter. In the previous example we didn’t have any limits on the parameter. Without limits on the parameter the graph will continue in both directions as shown in the sketch above. We will often have limits on the parameter however and this will affect the sketch of the parametric equations. To see this effect let’s look a slight variation of the previous example. Example 2 Sketch the parametric curve for the following set of parametric equations. 𝑥 =𝑡2+𝑡𝑦 =2⁢𝑡−1⁢−1≤𝑡≤1 Notice that with this sketch we started and stopped the sketch right on the points originating from the end points of the range of 𝑡’s. Contrast this with the sketch in the previous example where we had a portion of the sketch to the right of the “start” and “end” points that we computed. Just how we eliminate the parameter will depend upon the parametric equations that we’ve got. Let’s see how to eliminate the parameter for the set of parametric equations that we’ve been working with to this point. Example 3 Eliminate the parameter from the following set of parametric equations. 𝑥 =𝑡2+𝑡𝑦 =2⁢𝑡−1 One of the easiest ways to eliminate the parameter is to simply solve one of the equations for the parameter (𝑡, in this case) and substitute that into the other equation. Note that while this may be the easiest to eliminate the parameter, it’s usually not the best way as we’ll see soon enough. Sure enough from our Algebra knowledge we can see that this is a parabola that opens to the right and will have a vertex at (−14,−2). We won’t bother with a sketch for this one as we’ve already sketched this once and the point here was more to eliminate the parameter anyway. The reality is that when writing this material up we actually did this problem first then went back and did the first problem. Plotting points is generally the way most people first learn how to construct graphs and it does illustrate some important concepts, such as direction, so it made sense to do that first in the notes. In practice however, this example is often done first. So, how did we get those values of 𝑡? Well let’s start off with the vertex as that is probably the most important point on the graph. We have the 𝑥 and 𝑦 coordinates of the vertex and we also have 𝑥 and 𝑦 parametric equations for those coordinates. So, plug in the coordinates for the vertex into the parametric equations and solve for 𝑡. Doing this gives, Getting a sketch of the parametric curve once we’ve eliminated the parameter seems fairly simple. All we need to do is graph the equation that we found by eliminating the parameter. As noted already however, there are two small problems with this method. The first is direction of motion. The equation involving only 𝑥 and 𝑦 will NOT give the direction of motion of the parametric curve. This is generally an easy problem to fix however. Let’s take a quick look at the derivatives of the parametric equations from the last example. They are, Note that the 𝑥 derivative isn’t as useful for this analysis as it will be both positive and negative and hence 𝑥 will be both increasing and decreasing depending on the value of 𝑡. That doesn’t help with direction much as following the curve in either direction will exhibit both increasing and decreasing 𝑥. The second problem with eliminating the parameter is best illustrated in an exa