Vector functions: solving for curves of intersection

  • Thread starter kylera
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Homework Statement


Solve for the vector function that represents the curve of intersection in the following two surface: z = sqrt(x^2 + y^2) and z = 1 + y


Homework Equations





The Attempt at a Solution


Through blind trial and error, I managed to get the book-specified answer of x = t, y = 0.5(t^2-1) and z = 0.5(t^2+1). What I'm curious about is why using parametric equations wouldn't work in this case. Which leads to what kind of right I have to using seemingly random functions when parameterizing a vector function.

I first worked with x = cos(t) and y = sin(t) to get z = 1, only to get something like 1 = 1 + sin(t), which got me stuck. Only after making some wild assumption x = t did I get the exact same book answer.
 

Answers and Replies

  • #2
Defennder
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I don't understand what you mean by "using parametric equations". To solve this, all you need do is to equate the variables of both equations of surface. In this case, since they are both written as z = (something), simply equate the two of them. The resulting is the equation of curve you need.

Now this is where the parametric form comes in. We let x be the parameter, and we express y in terms of the parameter. This is why we assume x = t, and obtain the same answer as that of the book. Now of course no one says that we can't use some fancy parametric representation like x = 1-t or some other x = f(t) so long as f(t) isn't bounded above or below by real number, the way sin t is. The key thing is that the parameter must be allowed to vary across all possible real numbers for some values of t.
 
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Now of course no one says that we can't use some fancy parametric representation like x = 1-t or some other x = f(t) so long as f(t) isn't bounded above or below by real number, the way sin t is. The key thing is that the parameter must be allowed to vary across all possible real numbers for some values of t.

In other words, when I try to obtain a function, it has to be able to return any real number value for some value t? Hence, x = t would work?
 
  • #4
Defennder
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x = t works because we usually assume [tex]t \in R[/tex] without having to write that our explictly (though you should). The "parameter function" f(t) has to be able to take on the all possible values of x.

For example, consider the unit circle at the origin. x^2+ y^2 = 1. A parametric representation of this is:
x = cos t
y = sin t.

We know that this works because cos t can take on any value betwee 1 and -1, just like sin t. But more importantly we know that his parametric representation satisfies the equation of the circle. Of course no one says you can't use a parametric representation like:

[tex]x = \sin (2t)[/tex]
[tex]y = \cos (2t)[/tex], just that in this case t isn't the angle between the x-axis and the line from the point (x,y) and the origin. Both these parametric representations satisfy the equation of the circle as well.
 

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