Parametric representation of curves

[jegues]

Homework Statement

See figure.

Homework Equations

N/A

The Attempt at a Solution

I've dealt with parametric equations for lines before in my linear algebra class but I'm not sure how I'm suppose to model two curves with one.

Anyone have any suggestions?

 

[gomunkul51]

Those equations are two quadratic surfaces (I'm assuming we are in 3D).
you can try to combine them into one y=f(x) equation witch will give you a curve (assuming you want the curve of the intersection).
Moreover, if you need, you can model it with an independent variable (t) :)

 

[jegues]

I'm still pretty confused on how I'm suppose to get parametric equations from this. I could rewrite the two equations in one as,

$$1 = x^{2}-xy+y$$

but I don't see how that helps me...

 

[Raskolnikov]

$$z = 1 - xy$$

$$1 = x^2 - xy + y \ \Rightarrow \ y = \frac{1 - x^2}{1 - x} = 1 + x$$

So we can rewrite z as

$$z = 1 - x(1 + x)$$.

Now can you think of a good variable to use for the parameterization from here?

Hint: it's extremely simple ^^

 

[jegues]

Now can you think of a good variable to use for the parameterization from here?

I'm not really sure... Maybe,

$$x=t$$?

If so then,

$$x(t) = t$$

$$y(t) = 1 + t$$

$$z(t)=1 - t(1+t)$$

I'm used to seeing parametric equations in the following form(straight lines),

$$z = A + Bt$$ where $$A&B\in Z$$

So this is sort of new for me

 

[Raskolnikov]

I'm not really sure... Maybe,

$$x=t$$?

I'm used to seeing parametric equations in the following form,

$$z = A + Bt$$ where $$A&B\in Z$$

So this is sort of new for me

yep, x=t works well! The way to write it is...

x(t) = ... \
y(t) = ... |- t $$\in \textbf{R}$$
z(t) = ... /

...where that thing in the middle is supposed to be a curly bracket...I'm too lazy to make it work in LaTeX haha

 

[jegues]

...where that thing in the middle is supposed to be a curly bracket...I'm too lazy to make it work in LaTeX haha

It's all good! I edited my post above with the equations of x(t), y(t), and z(t).

Thanks again for your help!