Parametric Spirals

Recall x(t)=cos(t) & y(t) = sin(t) gives a circle. Also remember that sin(t) = cos(t-Pi/2).

>	plot([cos(t),cos(t-Pi/2),t=0..2*Pi],x=-2..2,y=-2..2,color=blue,thickness=2);

Next we plot x(t) = cos(t) & y(t) = 2 sin(t). An, ellipse! Show why this is.

>	plot([cos(t),2*cos(t-Pi/2),t=0..2*Pi],x=-2..2,y=-2..2,color=blue,thickness=2);

What happens if we shift the phase of the x(t)?

>	plot([cos(t-Pi/3),2*cos(t-Pi/2),t=0..2*Pi],x=-2..2,y=-2..2,color=blue,thickness=2);

If the difference in the two phase angles is Pi/2 we get a "nonrotated" ellipse. What happens if the phase angles are equal?  Here is another example.

>	plot([cos(t-Pi/3),2*cos(t+Pi/4),t=0..2*Pi],x=-2..2,y=-2..2,color=blue,thickness=2);

Now, use some trig to show this is the same as,

>	plot([(1/2)*cos(t)-sqrt(3)/2*sin(t),       sqrt(2)*cos(t)+sqrt(2)*sin(t) ,t=0..2*Pi],x=-2..2,y=-2..2,color=blue,thickness=2);

In your problems, you start with x(t) & y(t) in the form, Acos(ct)+Bsin(ct), that we just got. But, by converting to the form Rcos(ct-d), that I started with above, you should see where the ellipse comes from. Now, we just multiple by an exponential factor to get a spiral.

>	plot([exp(-t/5)*((1/2)*cos(t)-sqrt(3)/2*sin(t)),       exp(-t/5)*(sqrt(2)*cos(t)+sqrt(2)*sin(t)) ,t=0..6*Pi],x=-2..2,y=-2..2,color=blue,thickness=2);

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