 In this video, we'll look at another example involving the speed of an object whose path is described parametrically. Consider a particle moving in the x-y plane whose path is given by the parametric equations x equals t cubed minus 3t squared and y equals 2t cubed minus 3t squared minus 12t. When is the particle at rest? Well, we know that the speed of the particle can be calculated by finding the square root of the sum of the squares of the derivatives of x and y respectively. So in our case, with x being t cubed minus 3t squared and y being 2t cubed minus 3t squared minus 12t, our expression for speed is then the derivative of x with respect to t, which is 3t squared minus 6t. We square that and add the derivative of y with respect to t squared as well. Now, since this is the expression for speed, and we want to know when the particle is at rest, well, that can be found by answering the question, when is this expression equal to zero? Because that means that the speed of the particle is zero. Well, we have two expressions here that are squared, which means both are positive and we're adding them. So the only way we can actually get this expression to be zero is to identify when the radicand is zero. Another way of looking at it is I could square both sides of this equation. That would ultimately give me 3t squared minus 6t squared plus 6t squared minus 6t minus 12 squared equals zero. And again, since this sum is zero and both terms are squared, neither one of these could be negative. The only way to get this sum to equal zero is if this expression equals zero and if this expression equals zero. In other words, we want to find a t value such that dx dt is equal to dy dt and they're both zero. So let's first take a look at dx dt and where it's zero. dx dt is 3t squared minus 6t. We can factor out 3t, leaving us with t minus 2. So this expression is zero when t is zero and t is two. dy dt, which is 6t squared minus 6t minus 12, that's equal to zero. I'm going to factor 6 out so I can take a look at this a little more clearly. This is t plus 1, t minus 2. So this expression is zero when t equals negative 1 and t equals 2. Now recall that we want to define when dx dt and dy dt were both zero. So we see that happens when t equals 2. So the particle must be at rest when t equals 2. Let's confirm this by looking at the graph. So here I've entered my equations into y equals. Let me take a look at our graph. There we see. Notice that the object stopped moving at that point and then started again.