Vector-Valued Functions

[roam]

Hello!

In vector valued functions, I don't know how to find a curve's cartesian equation by inspecting its parametric ones...

For example I know from a worked example that if f: R^2 \rightarrow R is given by f(x,y) = xy, and r(t) = \left[\begin{array}{ccccc} sin(t) \\ cos(t) \end{array}\right], then the Cartesian equation for this curve r is: x²+y²=1 (which is just the unit circle).

But what if we had f: R^2 \rightarrow R is given by f(x,y) = x²y, and r(t) = \left[\begin{array}{ccccc} sin(t) \\ cos^2(t) \end{array}\right], (t \in [0, \pi/2)?

How do can I try to find its Cartesian equation?

 

[HallsofIvy]

Hello!

In vector valued functions, I don't know how to find a curve's cartesian equation by inspecting its parametric ones...

For example I know from a worked example that if f: R^2 \rightarrow R is given by f(x,y) = xy, and r(t) = \left[\begin{array}{ccccc} sin(t) \\ cos(t) \end{array}\right], then the Cartesian equation for this curve r is: x²+y²=1 (which is just the unit circle).

I don't understand what you are saying here. Certainly, since sin^2(t)= cos^2(t)= 1, x= sin(t), y= cos(t) is the same as x^2+ y^2= 1, but what does that have to do with f:R^2\rightarrow R?

But what if we had f: R^2 \rightarrow R is given by f(x,y) = x²y, and r(t) = \left[\begin{array}{ccccc} sin(t) \\ cos^2(t) \end{array}\right], (t \in [0, \pi/2)?

How do can I try to find its Cartesian equation?

If x= sin(t) and y= cos^2(t) then x^2+ y= sin^2(t)+ cos^2(t)= 1 so x^2+ y= 1 or y= 1- x^2, a parabola. Again, that has nothing to do with f.