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Find a vector function that represents the curve of intersection of the two surfaces

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Litcyb
#1
Jan31-13, 02:00 AM
P: 36
1. The problem statement, all variables and given/known data

Find a vector function that represents the curve of intersection of the two surfaces:
The cone z = sqrt( x^2 + y^2) and the plane z = 1+y.

2. Relevant equations

z = sqrt( x^2 + y^2) and the plane z = 1+y.

3. The attempt at a solution
This problem can be solved as following using x as the parameter.
x^2+y^2 = z^2 = (1+y)^2 = 1+2y+y^2. => x^2 = 1 + 2y.

x=t; y = (t^2-1)/2; z = 1+(t^2-1)/2 = (t^2+1)/2

My question is, what if we use y as the parameter,

i get ,

y=t, x=(2t+1)^(1/2) z=t+1,

is this answer also correct?
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Dick
#2
Jan31-13, 09:03 AM
Sci Advisor
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P: 25,228
Quote Quote by Litcyb View Post
1. The problem statement, all variables and given/known data

Find a vector function that represents the curve of intersection of the two surfaces:
The cone z = sqrt( x^2 + y^2) and the plane z = 1+y.

2. Relevant equations

z = sqrt( x^2 + y^2) and the plane z = 1+y.

3. The attempt at a solution
This problem can be solved as following using x as the parameter.
x^2+y^2 = z^2 = (1+y)^2 = 1+2y+y^2. => x^2 = 1 + 2y.

x=t; y = (t^2-1)/2; z = 1+(t^2-1)/2 = (t^2+1)/2

My question is, what if we use y as the parameter,

i get ,

y=t, x=(2t+1)^(1/2) z=t+1,

is this answer also correct?
Sure, with the proviso that it only represents part of the curve. For example, points with negative values of x won't appear in the second parametrization (where you also should specify t>=(-1/2)). You'd need the x=(-(2t+1)^(1/2)) solution as well to get them all.


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