Curve of intersection with two surfaces

**kenshinofkin** · May 10th 2008, 08:47 PM

Show that the curve with vector equation $r(t) = <2 cos^2(t), sin(2t), 2sin(t)>$ is the curve of intersection of the surfaces $(x-1)^2+y^2 = 1$ and $x^2 + y^2 + z^2 = 4$ . Use this fact to sketch the curve.

Im completely lost here!

**mr fantastic** · May 10th 2008, 09:41 PM

Originally Posted by kenshinofkin

Show that the curve with vector equation $r(t) = <2 cos^2(t), sin(2t), 2sin(t)>$ is the curve of intersection of the surfaces $(x-1)^2+y^2 = 1$ and $x^2 + y^2 + z^2 = 4$ . Use this fact to sketch the curve.

Im completely lost here!

You know that there's two equations to be solved simultaneously

$(x-1)^2+y^2 = 1 \Rightarrow y^2 = 1 - (x-1)^2$ .... (1)

$x^2 + y^2 + z^2 = 4$ .... (2)

and three unknowns x, y and z. So clearly there's gonna be a parameter. Choose the parameter to be real convenient. $z = 2 \sin t$ is an obvious choice (see below *) because when you substitute it into (2) you get:

$x^2 + y^2 = 4 \cos^2 t$ .... (3),

a nice easy-on-the-eye equation.

Solve (1) and (3) simultaneously for x and y in terms of t:

Substitute (1) into (3):

$x^2 + 1 - (x-1)^2 = 4 \cos^2 t$ .

Expand, simplify and solve for x. $x = 2 \cos^2 t$ . Now solve for y.

The curve is the intersection of the infinite cylinder $(x-1)^2+y^2 = 1$ and the sphere $x^2 + y^2 + z^2 = 4$ . I leave the visualisation of this to you.

* You should see what solution you get if you choose $z = 2 \cos t$ ......

**kenshinofkin** · May 10th 2008, 10:30 PM

Im having a problem getting y. If I solve $(x-1)^2 +y^2 = 1$ for x I get $((1-y^2)+1)^(1/2)$ this can't be right. Also once I have y. What do I do with x,y,z to show that they intersect?

**mr fantastic** · May 10th 2008, 10:49 PM

Originally Posted by kenshinofkin

Im having a problem getting y. If I solve $(x-1)^2 +y^2 = 1$ for x I get $((1-y^2)+1)^(1/2)$ this can't be right. Also once I have y. What do I do with x,y,z to show that they intersect?

I have already shown you how to get the solution for z and x!

To get the solution for y, substitute $x = 2 \cos^2 t$ into equation (3) and solve for y!

Then the solution to equations (1) and (2) is $x = 2 \cos^2 t, ~ y = \sin (2t), ~ z = 2 \sin t$ . This defines a curve parametrically. It's the curve of intersection of the cylinder and the sphere, as explained above. And as you can see from the given position vector that you're asked to get.

**kenshinofkin** · May 10th 2008, 10:54 PM

Ya after I posted I just realized how stupid I was being. I solved and saw that it was the same as the vector. Thanks

Thread: Curve of intersection with two surfaces

LinkBack

Thread Tools

Display

Curve of intersection with two surfaces

Similar Math Help Forum Discussions

Intersection of surfaces

Intersection of two surfaces

intersection between two surfaces

Curve of intersection of surfaces

Parametrize curve of intersection of surfaces

Search Tags