Purpose of parametric equations

[xyz3003]

I am confused myself, so I post the Q.

when we talk about "definite integral of area" in rectangular or polar coordinates, the "area" is quite clear, at least people do it in this way in general:

rectangular coordinate: area between locus y=f(x) and x axis.
polar coordinate: sector area from original point to locus between start and end angles.

parametric equations use third param, such as t, to describle (x,y).

when we use parametric equations in real world (such as physics), is the "definite integral of area" similar to rectangular coordinate or polar coordinate in general or in most of cases?

any samples or explanations are highly appreciated.

thanks.

 

[Mark44]

I am confused myself, so I post the Q.

when we talk about "definite integral of area" in rectangular or polar coordinates, the "area" is quite clear, at least people do it in this way in general:

rectangular coordinate: area between locus y=f(x) and x axis.
polar coordinate: sector area from original point to locus between start and end angles.

parametric equations use third param, such as t, to describle (x,y).

when we use parametric equations in real world (such as physics), is the "definite integral of area" similar to rectangular coordinate or polar coordinate in general or in most of cases?

In the context of your question, "definite integral of area" doesn't make sense. You are calculating a definite integral to find the area of some region -- you aren't taking the definite integral of an area.

With parametric equations, we can represent the x and y coordinates as functions of a parameter, say t.

Here's an example of how they might be used. Consider the portion of the unit circle in the first quadrant. We know that the area of this quarter-circle is $\frac \pi 4$.

We can parametrize the circle by the equations $x = \cos(\theta), y = \sin(\theta), 0 \le \theta \le \frac \pi 2$.
Divide the quarter circle into thin "pie slices," each of area $\Delta A = \frac 1 2 r^2 d\theta$. This is a formula you probably learned when you were doing integration with polar coordinates. Each of these slices is formed by two rays extending from the origin out to the circle, where the angle between the rays is $\Delta \theta$.

The integral for the area of this quarter circle is $$\int_0^{\pi/2} dA = \int_0^{\pi/2} \frac 1 2 r^2 d\theta$$
Since r = 1 for the circle, the integral becomes $\frac 1 2 \int_0^{\pi/2} d\theta = \frac 1 2 (\frac \pi 2 - 0) = \frac \pi 4$, which is what we expected.