How can I determine the direction of parametrization for a curve?

[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





Can someone please help me with this example and shed some light on parametrization?

 

[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.

 

[cytochrome]

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.

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?

 

[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.

 

[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.