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A good introduction to parametric curves is given by ballistics. If we shoot a bullet into the air with speed v horizontally, and we neglect all forces but gravity, then the bullet will trace out a parabola (some bullets are larger than others):
Now how might we characterize the path of the bullet? The answer is a parametric curve, of the form $C(t)=(x(t), y(t))$.
$x(t)=vt$ | $y(t)=h-\frac{1}{2}gt^2$ | |
or | $c(t)=(vt,h-\frac{1}{2}gt^2)$ | |
If we wish, we can solve for t in the equation for x and use that to eliminate the parameter t from the equation for y, hence getting an equation for the parabola traced out: | ||
$y=h-\frac{1}{2}g\left(\frac{x}{v}\right)^2$ |
According to this parameterization, where is the "bullet" at time t=0? In which direction is the motion occuring -- left to right, or right to left? I guess it depends on $v$, doesn't it?
(You probably have a parametric graphing mode on your calculator.)