In Linear Algebra I taught you that a line through the origin is a one dimensional subspace. One dimensional, meaning it could be represented as all linear combinations of a single (non-zero) vector. Let's think about that in terms of parameterized curves:[br][br]Suppose [math]\vec{m}=\left(\Delta x,\Delta y\right)[/math] is a non-zero vector in [math]\mathbb{R}^2[/math]. Then the line through the origin parallel to [math]\vec{m}[/math] is the image of the vector function:[br][br][math]\vec{l}\left(t\right)=t\vec{m}[/math][br]We can represent this parametrically as:[br][math]\vec{l}\left(t\right)=\left(t\Delta x,t\Delta y\right),t\in\mathbb{R}[/math][br][br]In Multi we will be concerned with lines that may not pass through the origin. We can accomplish this by simply shifting a line through the origin by adding a fixed vector. [br]Hence, the general form for a line in [math]\mathbb{R}^2[/math] is:[br][br][math]\vec{l}\left(t\right)=t\vec{m}+\vec{b}=\left(t\Delta x+b_x,t\Delta y+b_y\right),t\in\mathbb{R}[/math][br]where the vector [math]\vec{b}=\left(b_x,b_y\right)[/math] is a fixed vector in the plane.[br][br]Experiment with the applet below. You can change the slope vector or the fixed point. Animate will show you how the line is traced out as the parameter increases from 0 to 1.
It's important to remember that when we analyze the curve that results from some path, we are analyzing the [i]image[/i] of that path. While the parameterization I gave you above is one way to parameterize a directed line, there are many others. It is [i]not[/i] a requirement that the component functions be linear! For example, show that the image curve of the following path is a segment of a directed line:[br][br][math]\vec{l}\left(t\right)=\left(2\cos t,3\cos t\right),t\in\mathbb{R}[/math][br][br]What is the slope of the line? If [math]y=f\left(x\right)[/math] is a function whose graph is the image curve of [math]\vec{r}\left(t\right)[/math], what is the domain of [math]f[/math]?
We can eliminate the parameter as follows:[br][math]x=2\cos t[/math][br][math]\frac{x}{2}=\cos t[/math][br]Subbing into the y-component:[br][math]y=3\cos t=3\left(\frac{x}{2}\right)=\frac{3}{2}x[/math][br]So the line that contains the image curve of [math]\vec{l}\left(t\right)[/math] is [math]f\left(x\right)=\frac{3}{2}x[/math] and has slope [math]\frac{3}{2}[/math]. However since we know [math]-1\le\cos t\le1[/math], the domain of [math]f[/math] must be [math]-2\le x\le2[/math]
The procedure I modeled in the last question to describe the image curve as [math]y=\frac{3}{2}x[/math] is called [b][color=#ff0000]eliminating the parameter[/color][/b] and the resulting equation directly relating the variables [math]x[/math] and [math]y[/math] is called a [b][color=#ff0000]rectangular equation [/color][/b](rectangular because the grid created by the [math]x[/math] and [math]y[/math] axes divides the plane into lots of little rectangles).