Hi again, I need to find the the vector function that represents the curve of intersection of the two surfaces listed below.

Paraboloid: Z = 4x^2 + y^2
Parabolic Cylinder: y = x^2

I had a very difficult time following my instructor and office hours with the TA next Weds but i really want to understand this before then to start my homework. Please help thanks again!

2. Originally Posted by multivariablecalc
Hi again, I need to find the the vector function that represents the curve of intersection of the two surfaces listed below.

Paraboloid: Z = 4x^2 + y^2
Parabolic Cylinder: y = x^2

I had a very difficult time following my instructor and office hours with the TA next Weds but i really want to understand this before then to start my homework. Please help thanks again!
If you let $y = x^2$ then $z = 4x^2 + y^2 = 4x^2 + x^4$. If you let $x = t$ then $y = t^2$ and $z = 4t^2 + t^4$. This is the curve of intersection. If you want the vector which represents this space curve then

$
\vec{r} = \,< t, t^2, 4t^2+t^4>.
$

3. Thanks for the response Danny. I figured it out! but i just stumbled on to one more of the same problem.

Hyperboloid z = x^2-y^2
Cylinder x^2+y^2 = 1. I understand the fx and y part of the answer but i do not understand how Z comes out to be cos2t if you can help out with this extra one i would really appreciate it. Thanks a bunch!

4. Originally Posted by multivariablecalc
Thanks for the response Danny. I figured it out! but i just stumbled on to one more of the same problem.

Hyperboloid z = x^2-y^2
Cylinder x^2+y^2 = 1. I understand the fx and y part of the answer but i do not understand how Z comes out to be cos2t if you can help out with this extra one i would really appreciate it. Thanks a bunch!
It you let $x = \cos t$ and $y = \sin t$ then

$x^2+y^2 = \cos^2 t + \sin^2 t = 1$ so the cylinder is automatically satisfied. Now substituting these into your surface gives

$
z = x^2 - y^2 = \cos^2 t - \sin^2 t = \cos 2t.
$

5. Danny thanks again!