Vectors and Parametric Equations
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Sometimes when graphing a shape or equation we want to add a parameter, something like time, which requires us to use parametric equations. One example of the use of parametric equations is to describe the motion along a line. Parametric equations can also be used to describe line segments or circles. Parametric equations are represented by two functions of x and y dependent on t.
One of the neat applications of parametric equations is using the de-model motion. Suppose t represents time in seconds and the position xy at time t is given by the equations x equals t squared plus 1, y=4t for t greater than or equals 0. Now let's start by plotting out a few points, x and y for each of these values of time t just to see where the particle is. Now first when t=0 I get x equals 0 squared plus, x=1 and y equals 4 times 0, 0. t=1, I get 1 squared plus 1 for x, 2 and 4 times 1 for y, 4. Now let me go through and just do the x's it'll be a little bit quicker. For x's I square the time value and add 1. So 2 squared plus 1 is 4+1 5, 9+1 10, 16+1 17 and 25+1 26 and for the y values I multiply the time by 4, 2 times 4, 8, 3 times 4, 12, 16 and 20. So now I have a list of points and I can plot these points and get some idea of what the path looks like and that's the next thing I have to do.
Now here are the actual points already plotted let me just connect these points with a curve because as you can imagine the particle is somewhere in between t=0, t=1, t=1 and t=2 and so forth, so let's fill that in all the way up to t=5 and now that I've got this path drawn there's something missing, I have no information on this graph about when the particle is at each of these points. So sometimes it's nice to actually fill that in, so let me put t=0 here for this point t=1 for this one, t=2, t=3, t=4 and t=5. You get a better sense for motion when you see the t values, the particle starts here and move up into the right as time progresses. So part c asks me to find a rectangular equation of the path, so I have my parametric equations a rectangular equation is going to be an equation that's only in x any and doesn't have the t, the parameter.
And so what I'm going to do start with this equation and solve it for t, subtract 1 I get x-1 equals t squared and observing that t is greater than or equal to 0 I can just take the square root t equals root x-1, now y equals 4 times t and so I can substitute t in here, t is root x-1 and so my rectangular equation is y equals 4 times root x-1. This equation doesn't have any t values in it but what it does represent is the actual shape of the path. y equals 4 root x-1 is the actual path of the particle but it doesn't tell you anything about times. So something is lost when find the rectangular equation of the path, this process by the way is called eliminating the parameter and doing this often makes it much easier to graph parametric curve. So we'll be doing this a lot in future episodes.