 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Parametric Equations and Motion - Problem 1

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Let's take a look at another parametric equations problem. The position of a particle is given by the parametric equations; x equals -1 plus 4t, y equals 15 minus 3t for t between 0 and 4. Eliminate the parameter to obtain a rectangular equation for the particle's path.

Again, eliminating the parameter means, I want to end up with an equation only in x and y. So let me start by taking the x equation, and solving it for t. So I get x equals -1, plus 4t. I add 1 to both sides, and I get x plus 1 equals 14. Then I divide both sides by 4. So x plus 1 over 4 equals t.

Now I can substitute this for t in this equation. Y equals 15 minus 3t becomes y equals 15 minus 3, times x plus 1 over 4. So I get y equals 15, minus 3/4 x, and minus 3/4 of 1. Now 15 minus 3/4 is 14 and a quarter, but it would be nicer to get it in terms of fraction. So 15 is 60 over 4 minus 3 is 57 over 4. So y equals -3/4 x, plus 57 over 4.

Now this equation represents a line, and so the shape of the path is going to be a line. Let's not forget that there is a restriction on the t values. So it's not going to be the entire line. When I graph it, it's important to know that it's a line, because all I have to do is plot two points. That's what the next part of the problem asks me to do. Plot points and graph the particle's path.

Let me plug in the two end points; 0 and 4. That will tell where my line segment is going to be, so just a little piece of a line. When t equals 0, x is -1 plus 4 times 0, -1. Y is 15 minus 3 times 0, 15. When t equals 0, I get -1/15. When t equals 4, I get -1 plus 4 times 4, plus 16 that's 15. I get 15 minus 3 times 4, 15 minus 12, 3. Let me plug -1, 15 and 15, 3. So -1 is here, 15 is up here. So that's -1, 15. Then 15, 3 is right here. Since I know it's a line, I can just draw the line segment connecting. I don't need anymore points.

I also know that the particle is here when t equals 0, and down here when t equals 4. Sometimes your teacher may actually want you to draw a little arrow indicating the direction of motion. This is a graph of the particle's path. The shape of the graph is the line that we found in part a. This is the shape of the graph.

When you're graphing a set of parametric equations, it's always good to indicate some motion, and give some time values so you have an idea of where the particle started, and if necessary where it ends.