How to parametrize inersection of 2 paraboloids?

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Homework Statement



I need to parametrize the equations z = x^2 and x = y^2 in order to find the unit tangent vector, normal vector, and binormal vector.

Homework Equations



z = x^2 and x = y^2

The Attempt at a Solution



I tried setting x = t, so then y = \sqrt{}t and then z = t^2.
Using that equation r(t) = < t, \sqrt{}t , t^2> I would get completely wrong answers from the book, and if I tried another function like r(t) = < t^2 , t , t^4> , I would get different results from my previous vector. Any ideas?
 
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You can see that your first parameterization is not going to give you the full intersection, because there won't be any points with negative y values.

The second parameterization looks more promising. Can you show your calculations for the tangent and normal vectors?
 
jbunniii said:
You can see that your first parameterization is not going to give you the full intersection, because there won't be any points with negative y values.

The second parameterization looks more promising. Can you show your calculations for the tangent and normal vectors?

Well I needed to find the unit tangent and normal vectors to find the normal and osculating plane equations, and the question only asked for it at the point < 1,1,1> so y isn't negative. However, with either function it looks like I am not getting the correct answer.

When I used r(t) = < t, \sqrt{}t , t^2> I got that

r'(t) = <1 , 1/2sqrt(t) , 2t> and if I found the magnitude of that, I get \sqrt{}5.25

From that, my unit tangent vector is 1/sqrt(5.25) <1 , 1/2sqrt(t) , 2t >

Something already looks wrong, so when I kept going, I obviously didn't get the right answer
 
bobhi123 said:
Well I needed to find the unit tangent and normal vectors to find the normal and osculating plane equations, and the question only asked for it at the point < 1,1,1> so y isn't negative. However, with either function it looks like I am not getting the correct answer.

When I used r(t) = < t, \sqrt{}t , t^2> I got that

r'(t) = <1 , 1/2sqrt(t) , 2t> and if I found the magnitude of that, I get \sqrt{}5.25

From that, my unit tangent vector is 1/sqrt(5.25) <1 , 1/2sqrt(t) , 2t >

Well, you have to evaluate this at t = 1, but aside from that, I don't see anything wrong. What answer does the book give?

By the way, I get the same answer using either of your two parameterizations.
 
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