# Vector-Valued Functions

## Main Question or Discussion Point

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
Homework Helper
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.