Next I want to explore the effect on the image curve of transforming the parameter. Formally this means we are changing a parent function [math]\vec{c}\left(t\right)=\left(x\left(t\right),y\left(t\right)\right)[/math] into the function [math]\vec{p}\left(t\right)=\vec{c}\left(at+b\right)=\left(x\left(at+b\right),y\left(at+b\right)\right)[/math] for constants [math]a[/math] and [math]b[/math]. [br][br]We'll begin by analyzing the unit circle. Back in PreCal you learned to parameterize the unit circle via:[br][math]\vec{c}\left(t\right)=\left(\cos t,\sin t\right),t\in\left[0,2\pi\right][/math] [br]In this class I'll refer to this as the [color=#ff0000][b]standard parameterization[/b] [/color]of the unit circle.[br][br]In the applet below, experiment by changing the values of [math]a[/math] and [math]b[/math] to see the effect on the image curve. Jot down your observations.
Eliminate the parameter in the standard parameterization of the unit circle to obtain a rectangular equation for the image curve. How does transforming the parameter affect this rectangular equation?
Recall the pythagorean identity for sine and cosine tells us [math]\cos^2t+\sin^2t=1[/math]. Hence a rectangular equation for the image curve is [math]x^2+y^2=1[/math].[br][br]Transforming the parameter does not change the underlying rectangular equation, but it may change which points on the curve [math]x^2+y^2=1[/math] are actually in the image curve of the resulting path. For example the image curve of the path [math]\vec{p}\left(t\right)=\left(\cos\frac{t}{2},\sin\frac{t}{2}\right),t\in\left[0,2\pi\right][/math] is only the top half of the unit circle.[br][br]Changing the value of [math]b[/math] adjusts the location of the endpoints of the path but again does not affect the underlying rectangular equation. Adjusting [math]b[/math] is often referred to as [b][color=#ff0000]shifting the parameter[/color][/b].