Vector-Valued Functions

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

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

How do can I try to find its Cartesian equation?
 

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  • #2
HallsofIvy
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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 [tex]f: R^2 \rightarrow R[/tex] is given by f(x,y) = xy, and [tex]r(t) = \left[\begin{array}{ccccc} sin(t) \\ cos(t) \end{array}\right][/tex], 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 [itex]sin^2(t)= cos^2(t)= 1[/itex], x= sin(t), y= cos(t) is the same as [itex]x^2+ y^2= 1[/itex], but what does that have to do with [itex]f:R^2\rightarrow R[/itex]?

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

How do can I try to find its Cartesian equation?
If x= sin(t) and [itex]y= cos^2(t)[/itex] then [itex]x^2+ y= sin^2(t)+ cos^2(t)= 1[/itex] so [itex]x^2+ y= 1[/itex] or [itex]y= 1- x^2[/itex], a parabola. Again, that has nothing to do with f.
 

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