Trajectory as a sufrace intersection

[Telemachus]

Homework Statement

Well, I must express this trajectory: [tex]\vec{r}=(t^2,2t,t^2)[/tex] as an intersection of two surfaces. I really don't know how to work this. It seems to be some kind of parabola, but I'd really like to see some step by step for solving this.

Bye, and thanks off course.

 

[gabbagabbahey]

Homework Statement

Well, I must express this trajectory: [tex]\vec{r}=(t^2,2t,t^2)[/tex] as an intersection of two surfaces. I really don't know how to work this. It seems to be some kind of parabola, but I'd really like to see some step by step for solving this.

Bye, and thanks off course.

Hint: How are [itex]x(t)[/itex] and [itex]y(t)[/itex] related? How are [itex]z(t)[/itex] and [itex]y(t)[/itex] related?

 

[Telemachus]

Mmm let's see. I got that [tex]x=t^2[/tex], [tex]y=2t[/tex], then [tex](\frac{y}{2})^2=x[/tex]?
And [tex]z=x[/tex]

Is that right? so its the intersection between the plane [tex]z=x[/tex] and the parabolic cylinder...?

I got one more question. I have another exercise where it asks me to get the parametric form for a line given as the intersection of two planes. So, to define the parametric form I need one point that satisfies the equation for the two planes, that is a point over the line. And I find the work of finding such a point really tedious. Is there any technique to find a point that satisfies two equations with three variables? I mean, I'm not much intuitive on that sense, I've been looking for a single point that satisfies the equations for two planes for many minutes and couldn't get it. Is there any procedure to follow?

Bye, and thank you gabba.

 

[HallsofIvy]

I see no reason to find a specific point if you are looking for the equation of the line itself. If you are given two planes, then you are given two linear equations:
Ax+ By+ Cz= D and Px+ Qy+ Rz= S. You can solve those two equations for two of the variables in terms of the other one. Use that one as parameter.

For example, if the planes are given by 2x- y+ z= 4 and 3x+ y- 2z= 3, then adding the two equations eliminates y and leaves 5x- z= 4. We can solve that for z in terms of x: z= 5x- 4.

Putting that into the first equation, 2x- y+ z= 2x- y+ 5x- 4= 4 so that y= 8- 5x.

Take x as parameter: x= t, y= 8- 5t, z= 5t- 4 are the parametric equations for the line of intersection.