# Parametrization homework help

1. Jul 2, 2007

### Weave

1. The problem statement, all variables and given/known data
Find a vector function that represents the curve of the intersection of two surfaces.

2. Relevant equations
$$z^2=x^2+y^2$$ with plane $$z=1+y$$

3. 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

2. Jul 2, 2007

### NateTG

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

3. Jul 2, 2007

### Weave

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

4. Jul 2, 2007

### Weave

Can anyone help me, I have got probably an hour

5. Jul 2, 2007

:uhh:

6. Jul 2, 2007

### 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.

7. Jul 2, 2007

### Weave

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

8. Jul 2, 2007

### NateTG

$$z=1+y$$
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.

9. Jul 2, 2007

### jambaugh

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:
$$\langle 1,0,0\rangle=\hat{\imath}$$
$$\langle 0,1,0\rangle = \hat{\jmath}$$
$$\langle 0,0,1\rangle = \hat{k}$$

Last edited: Jul 2, 2007
10. Jul 2, 2007

### HallsofIvy

Staff Emeritus
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