What Is the Correct Parametrization for the Intersection of Two Surfaces?

[Weave]

Homework Statement

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

Homework Equations

z^2=x^2+y^2 with plane z=1+y

The Attempt at a Solution

So shouldn't it be
r(t)=<cos(t), sin(t), 1+sin(t)>
since x=cos(t), y=sin(t), and z= 1+sin(t)?
The book gives a wacky answer

 

[NateTG]

What kind of a shape do you think
z^2=x^2+y^2
is?

 

[Weave]

I know z^2=x^2+y^2 is a cone.

 

[Weave]

Can anyone help me, I have got probably an hour

 

[Weave]

 

[NateTG]

The shape you parametrized is an elipse, but the intersection of the double-cone with side slope 1, and a plane with slope 1 is going to be a parabola.

 

[Weave]

So how would I go about coming to a vector equation?

 

[NateTG]

So how would I go about coming to a vector equation?

Since you already have:
z=1+y
I'd substitute, simplify, and see what happens:

z^2=x^2+y^2
(1+y)^2=x^2+y^2
...

Which should work out reasonably well.

 

[jambaugh]

...
So shouldn't it be
r(t)=<cos(t), sin(t), 1+sin(t)>
since x=cos(t), y=sin(t), and z= 1+sin(t)?
The book gives a wacky answer

There are more than one way to parameterize a curve, so your answer and the books answer needn't agree.
[edit:]
But I note an error. Since you've chosen x=cos(t), y=sin(t) then x^2+y^2=1 you've implicitly added then another constraint and you cannot satisfy z^2 = x^2+y^2. Rather try x = z\cdot\cos(t) and y=z\cdot\sin(t)

[end edit:]
With regard to vectorizing you have already done that:
\mathbf{r}(t)=<x(t), y(t), z(t)> = x(t)\hat{\imath}+y(t)\hat{\jmath} + z(t)\hat{k}
These are just two ways of writing the same vector. The basis is:
\langle 1,0,0\rangle=\hat{\imath}
\langle 0,1,0\rangle = \hat{\jmath}
\langle 0,0,1\rangle = \hat{k}

 

[HallsofIvy]

Homework Statement

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

Homework Equations

z^2=x^2+y^2 with plane z=1+y

The Attempt at a Solution

So shouldn't it be
r(t)=<cos(t), sin(t), 1+sin(t)>
since x=cos(t), y=sin(t), and z= 1+sin(t)?
The book gives a wacky answer

Since z= 1+ y, z^2= (1+y)^2= 1+ 2y+ y^2= x^2+ y^2. Cancelling the y^2 on each side, 1+ 2y= x^2 or y= (x^2- 1)/2. Taking x itself to be the parameter, we have x= t (of course, y= (t^2- 1)/2, z= 1+ y= 1+ (t^2- 1)/2