# Parametrizing a curve

1. Jan 7, 2013

### cytochrome

Is there a general way to find a vector valued function that parametrizes a curve??? I'm reading through a textbook and it says nothing in depth about parametrization and suddenly there's a question...

Find a vector valued function f that parametrizes the curve in the direction indicated.

4x^2 + 9y^2 = 36

2. Jan 7, 2013

### HallsofIvy

There is no general "algorithm" for finding a parameterization- it usually results from some geometric or physical insight into the problem. Here, for example, I would recognise the graph of $4x^2+ 9y^2= 36$ as an ellipse which, in "standard form", would be $x^2/9+ y^2/4= 1$. In that form I can see that the ellipse has center (0, 0) and has vertices at (3, 0), (-3, 0), (0, 2), and (0, -2).

And that I can think of as a "warped circle". Standard parametric equations for a circle with center at (0, 0) and radius R are $x= R cos(\theta)$ and $y= R sin(\theta)$. Then it should be easy to see that choosing different values for "R" will give $x= 3 cos(\theta)$ and $y= 2sin(\theta)$ will give the desired ellipse.

3. Jan 7, 2013

### cytochrome

That was extremely helpful, thank you so much.

One more thing - when looking for a vector valued function, can I just assign the unit vectors to the x and y equations so that f(t) = 3cos(t)i + 2sin(t)j, where i and j are both unit vectors and t is the parameter?

4. Jan 7, 2013

### HallsofIvy

Yes, of course, I should have said that: If x= f(t) and y= g(t) then the "vector form" is v(t)= xi+ yj= f(t)i+ g(t)j.

5. Jan 11, 2013

### HallsofIvy

By the way, in your the post it says "in the direction indicated" but there was no "direction indicated". My parameterization "goes around" the ellipse as $\theta$ increases- in the counterclockwise direction. If, instead we use $x= 3cos(t)$, $y= -2sin(t)$, or v(t)= 3cos(t)i-2 sin(t)j goes around the ellipse in the clockwise direction.