Calculus help - parametric equations/graphs?

Question

I am just wondering how you are supposed to match parametric equations with their graphs? Like if you are given a bunch of funny looking curves, and a bunch of equations, where do you start in determining which equation belongs to which graph? For example, one question gives
x=sin(t + sint), y=cos(t+cost)
So I don't know how you can tell which of the given graphs this equation belongs to. Do I actually have to plot some points on my own and see if i get one of the graphs? It says not to use a graphing device. Please help and explain how you can see this from the equations, thanks

Answer 1

it depends what other graphs are given. if they are "obviously" not the equation you cross them off. for instance its a straight line but your equations is not.

next yes you should plot some points, with the t's not too far apart.

Answer 2

That is a particularly hard question.

First of all, if I saw an equation that looked like it could be easily converted to polar coordinates (identical terms in the trig functions), I would do so.

Normally, I just graph x with respect to (wrt) t and y wrt t and compare the graphs, if that's something easy to do. Usually, that gives me a good idea of what the graph will look like.

If it's something like x = sin(at), y = cos(bt), I would sometimes find the least common multiple (LCM) of a and b and plot (connecting while plotting) all of the points in order at each interval where sin(at) = 0 or 1 and where cos(bt) = 0 or 1 until both at and bt have reached the LCM at least once.
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The next section is a lengthy way to solve this particular problem, and I realized later that the part afterward might be easier anyway while taking a test. So feel free to skip it, but I think it may help.
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Looking at the equations themselves and figuring out what they tell me is where I start. It almost always uses the least amount of time and energy once you're used to doing it.

I would look at x = sin(t+sin(t)) and think "ok, what's going into the outer function?" Well, think about the graph of t+sin(t). It is a diagonal sine wave. This tells you that what goes into the outer sine function increases, but not steadily. In fact, its derivative is cos(t), but that (itself) won't help you much.

If you plotted x = sin(t+sin(t)) on a graph with x on the veritcal axis, what you'd probably get is a sine wave that progresses quickly and then slowly. I bet you'd get something that looks very similar to a sine wave, except it compresses and rarifies with the same period as the wave itself (at t = 0 or n*pi, the graphs of sin(t) and sin(t+sin(t) coinside, but at t= (1+2n)pi/2, the graph will be advanced through the curve more, and at t = (1-2n)pi/2 the graph will lag behind.)

Now that we know what x =sin(t+sin(t)) would look like, we can take a look at y = cos(t+cos(t)). It is the same thing, except the graphs of cos(t) and cos(t+cos(t)) coinside at the points where cos(t) = 0 and t + cos(t) = 1 when t = 0, so your function for y will start at cos(1) from the beginning and hit the equilibrium point at the same time cos(t) would.

Knowing all this, you can graph what x and y would look like wrt t. Once you've done that, you should be able to figure out what the graph would look like. My guess is that it would be some kind of ameoba-looking circle, closely resembling the graph of x = sin(t) and y = cos(t).
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In hindsight, you could probably get the answer to this problem by a purely qualatative analysis. You know x will be a wierd sine wave that intercepts sin(t) whenever t = 0 or n*pi, and you know y will be a wierd cosine wave that intercepts cos(t) whenever t = (1+2n)pi. This information alone would lead me to believe it's going to be an amoeba-looking circle.
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The bottom line is, this problem is probably going to be among the harder ones from your test. You should look for the easiest graphs to identify and solve them first.

Answer 3

I would plot a couple points. Such the point (t=0) is not hard to solve for x and y. By just picking arbitrary points for t, and finding the x and y coords, you should be able to begin determining the aproximate shape of the graph. Especially if you are given some graphs, then you should be able to quickly eliminate a few. Its not the most elegant method, but I hope it helps.
Another trick I just remembered that sometimes helps is transforming the equation back into a y = x type deal (if possible), as those are a bit easier to visualize.

Answer 4

You have the right idea--you should pick a value of the parameter (in your example, t) and compute the x and y values. Hopefully, that point will fall on only one of the curves. If it happen to be on more than one, pick another parameter value. Sometimes you can judge just from the nature of the functions; trig functions will result in some form of conic-section figure, linear functions in straight lines, etc.