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 <b>circular</b> 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.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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.<br /> <br /> $x + y + z + (- x - y) = 1 + ( - x - y)$ ---> $z = - x - y + 1$ ---> $z = - (2cost) - (2sint) + 1$<br /> <br /> 3c. Last step would be to combine everything into parametric equations.<br /> <br /> $x = 2cos(t)$ , $y = 2sin(t)$ , $z = - 2sin(t) - 2cos(t) + 1$.<br /> <br /> I think I have the right idea, but any suggestions or guidance would be appreciated. </div> </div> </blockquote> The plane, x+ y+ z= 1 can be written 2cos(t)+ 2sin(t)+z= 1 so that z= 1- 2cos(t)- 2sin(t).<br /> And that gives the parametric equations describing the intersection:<br /> x= 2cos(t)<br /> y= 2sin(t)<br /> z= 1- 2cos(t)- 2sin(t).[/tex]

 

[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?