Intersection of a Quadric Surface and a Plane in 3-D

[Onionknight]

1. "Find the equation that describes the intersection of the quadric given by $x^2 + y^2 = 4$ with the plane $x + y + z = 1$."


2. Parametric equations for elliptic curve: $x = a cos(t)$ , $y = b sin(t)$ , z = ?


3. Surface is an [EDIT: right circular] cylinder. Plane is not parallel to xy plane, has some increasing z and will not cut a circle in the cylinder. Curve of intersection of plane and elliptic cylinder will result in an elliptic curve.

3a. My first approach would be to plug in $a cos(t)$ and $b sin(t)$ for x and y respectively in the $x^2 + y^2 = 4$ equation.


$(a cos(t))^2 + (b sin(t))^2 = 4$ ---> $a^2(cost)^2 + b^2(sint)^2 = 4$ ---> $\frac{a^2(cost)^2}{4} + \frac{b^2(sint)^2}{4} = 1$ ---> a = b = 2 --> $x = 2cost$ and $y = sint$

This step confuses me since I would think that because the curve is elliptic, the a and b coefficients would differ.


3b. My second step would be to rearrange the equation of the plane to be some z = x + y + c to find the parametric equation for z.

$x + y + z + (- x - y) = 1 + ( - x - y)$ ---> $z = - x - y + 1$ ---> $z = - (2cost) - (2sint) + 1$

3c. Last step would be to combine everything into parametric equations.

$x = 2cos(t)$ , $y = 2sin(t)$ , $z = - 2sin(t) - 2cos(t) + 1$.

I think I have the right idea, but any suggestions or guidance would be appreciated.

 

[HallsofIvy]

1. "Find the equation that describes the intersection of the quadric given by $x^2 + y^2 = 4$ with the plane $x + y + z = 1$."


2. Parametric equations for elliptic curve: $x = a cos(t)$ , $y = b sin(t)$ , z = ?


3. Surface is an elliptic cylinder. Plane is not parallel to xy plane, has some increasing z and will not cut a circle in the cylinder. Curve of intersection of plane and elliptic cylinder will result in an elliptic curve.

3a. My first approach would be to plug in $a cos(t)$ and $b sin(t)$ for x and y respectively in the $x^2 + y^2 = 4$ equation.


$(a cos(t))^2 + (b sin(t))^2 = 4$ ---> $a^2(cost)^2 + b^2(sint)^2 = 4$ ---> $\frac{a^2(cost)^2}{4} + \frac{b^2(sint)^2}{4} = 1$ ---> a = b = 2 --> $x = 2cost$ and $y = sint$

This step confuses me since I would think that because the curve is elliptic, the a and b coefficients would differ.

This is wrong. [tex]x^2+ y^2= 4[/itex] is the equation of a circular cylinder. It can be written as x= 2cos(t), y= 2 sin(t). Yes, its intersection with the tilted plane will be an ellipse but that is not relevant here.

3b. My second step would be to rearrange the equation of the plane to be some z = x + y + c to find the parametric equation for z.

$x + y + z + (- x - y) = 1 + ( - x - y)$ ---> $z = - x - y + 1$ ---> $z = - (2cost) - (2sint) + 1$

3c. Last step would be to combine everything into parametric equations.

$x = 2cos(t)$ , $y = 2sin(t)$ , $z = - 2sin(t) - 2cos(t) + 1$.

I think I have the right idea, but any suggestions or guidance would be appreciated.

The plane, x+ y+ z= 1 can be written 2cos(t)+ 2sin(t)+z= 1 so that z= 1- 2cos(t)- 2sin(t).
And that gives the parametric equations describing the intersection:
x= 2cos(t)
y= 2sin(t)
z= 1- 2cos(t)- 2sin(t).

 

[Onionknight]

Thank you. Don't know why I typed elliptic cylinder. I think I understand why it doesn't matter that a doesn't equal b. Since the curve of intersection is tilted by the z component anyways, it's going to be an ellipse. If the z parametric equation wasn't in effect, the curve would be a circle. I think this makes sense. Is my overall logic and procedure sound?