How to parametrize inersection of 2 paraboloids?

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Homework Help Overview

The discussion revolves around the parametrization of the intersection of two paraboloids defined by the equations z = x^2 and x = y^2. Participants are exploring how to derive the unit tangent vector, normal vector, and binormal vector from these parametrizations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to set x = t and derive y and z accordingly, leading to different parameterizations. Some participants question the completeness of the first parameterization due to its inability to account for negative y values. Others suggest that the second parameterization may be more promising and inquire about the calculations for tangent and normal vectors.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's attempts. There is a focus on evaluating the correctness of the derived vectors and the need to assess calculations at specific points. Some guidance has been offered regarding the evaluation of the unit tangent vector at t = 1.

Contextual Notes

Participants note that the question specifically asks for results at the point <1,1,1>, which may limit the exploration of negative values for y. There is also mention of discrepancies between the poster's results and those found in the textbook.

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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 = [itex]\sqrt{}t[/itex] and then z = t^2.
Using that equation r(t) = < t, [itex]\sqrt{}t[/itex] , 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, [itex]\sqrt{}t[/itex] , t^2> I got that

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

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, [itex]\sqrt{}t[/itex] , t^2> I got that

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

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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