 In this video, we'll calculate the equation of a tangent line to a curve defined parametrically. Consider the curve defined by the parametric equations x of t equals t squared minus 4t plus 1 and y being t cubed. An object traveling along the curve decides to go off the path and follow the tangent line to the graph after three seconds. Let's find the equation of this tangent line. We'll first calculate the slope of the line which is the slope of the curve at t equals 3 and dy dx we find by differentiating y with respect to t and dividing that by the derivative of x with respect to t and given that x was t squared minus 4t plus 1 and y was t cubed. This slope is 3t squared over 2t minus 4. Since we're interested in the slope when t is 3, we'll evaluate this when t is 3. That gives me 27 over 2. So here's our slope. Let's find the coordinate when t is 3 x of 3 is equal to 9 minus 12 plus 1. That gives us negative 2. And y of 3 is 27. So our tangent line is y minus 27 equals 27 over 2 times x plus 2. So our equation is 27 over 2 times x plus 2 plus 27. Let's take a look at the graph to see if we get a sense that this is reasonable. So we're going to graph the equations. We see our coordinate is negative 2 and 27 and we see we have a tangent line that has a positive slope. So this seems to be like a reasonable result.