# Intersection of cylinder and plane

• Feb 18th 2013, 10:04 PM
bchemmath
Intersection of cylinder and plane
"Let S denote the elliptical cylinder given by the equation 4y2+ z2=4, and let C be the curve obtained by intersecting S with the plane y=x.

Parameterize C

I am not sure how to go about this. I tried solving the equations for each other

4y2= 4-z
y2= 1- z2/4

y= sqrt(1-z2/4)

and then do i do some sort of parameterization of sqrt(1-z2/4) -x=0 or am i supposed to parameterize the cylinder and plane separately? I know the cylinder parameterizes to
y=cos(theta)
z=2sin(theta)
but i don't know what to do from here. If anyone could hit me with some tips i would greatly appreciate it

EDIT:

would it simply be <cos (t), cos (t), 2 sin (t)> as x=y and y is equal to cos(t)? is it really this simple?
• Feb 19th 2013, 06:30 AM
hollywood
Re: Intersection of cylinder and plane
Quote:

Originally Posted by bchemmath
would it simply be <cos (t), cos (t), 2 sin (t)> as x=y and y is equal to cos(t)? is it really this simple?

Yes, that's correct.

- Hollywood
• Feb 19th 2013, 06:37 PM
bchemmath
Re: Intersection of cylinder and plane
The next part says:
" Use the parametrization above to compute the unit tangent vector, the principal normal vector, and the binormal vector at each of the two points where C intersects the xy-plane"

I know how to solve for unit tangent vector/ principal normal vector/binormal vector, etc but it says at each of the two points where C intersects the xy plane. Doesn't C intersect it in its entire domain? Should i have parameterized this differently without using sin/cos? or is it just at 0 and 2pi since that is the edge of the domain? i am slightly confused if anyone could please explain to me
• Feb 19th 2013, 11:23 PM
hollywood
Re: Intersection of cylinder and plane
The xy plane is where z is zero. Since in your parameterization of the curve, $\displaystyle z=2\sin{t}$, it crosses the xy plane where $\displaystyle \sin{t}=0$, i.e. at $\displaystyle t=0$ and $\displaystyle t=\pi$.

- Hollywood