Find points nearest/farthest from origin on the intersection of a plane and a parabol

In summary: So to find the points nearest and furthest from the origin, we just have to solve for the minimum and maximum values of x^2 + y^2 + z^2, subject to the two given constraints.
  • #1
skyturnred
118
0

Homework Statement



Find the points nearest and furthest from the origin on the intersection of a plane with a paraboloid.

Plane:
x+y+2z=30

Paraboloid:
z=x[itex]^{2}[/itex]+y[itex]^{2}[/itex]

Homework Equations





The Attempt at a Solution



Obviously the first step is to find the equation of the ellipse that is formed

I do that by plugging the equation of the paraboloid into the equation of the plane and get:

x+y+2x[itex]^{2}[/itex]+2y[itex]^{2}[/itex]=30

But my problem is that the equation above isn't really the equation I need. The equation above gives the shape of the ellipse *on the plane* but I need an equation for that shape on the original 3D Cartesian plane. How do I go about doing that?

Also, once I find said equation, would the following way to go about finding the max/min be correct?

Choose the equation of the ellipse as the constraint, choose f(x,y,z)=[itex]\sqrt{x^{2}+y^{2}+z^{2}}[/itex] and then just treat it like any straight forward max/min question?

Thanks
 
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  • #2


skyturnred said:

Homework Statement



Find the points nearest and furthest from the origin on the intersection of a plane with a paraboloid.

Plane:
x+y+2z=30

Paraboloid:
z=x[itex]^{2}[/itex]+y[itex]^{2}[/itex]

Homework Equations





The Attempt at a Solution



Obviously the first step is to find the equation of the ellipse that is formed

I do that by plugging the equation of the paraboloid into the equation of the plane and get:

x+y+2x[itex]^{2}[/itex]+2y[itex]^{2}[/itex]=30

But my problem is that the equation above isn't really the equation I need. The equation above gives the shape of the ellipse *on the plane* but I need an equation for that shape on the original 3D Cartesian plane. How do I go about doing that?

Also, once I find said equation, would the following way to go about finding the max/min be correct?

Choose the equation of the ellipse as the constraint, choose f(x,y,z)=[itex]\sqrt{x^{2}+y^{2}+z^{2}}[/itex] and then just treat it like any straight forward max/min question?

Thanks

You don't need to find the ellipse of intersection; you just need to minimize (or maximize) x^2 + y^2 + z^2, subject to the two equality constraints x+y+2z = 30 and z = x^2 + y^2. You could, for example, use Lagrange multipliers to solve this problem; you would have two Lagrange multipliers, one for each constraint. Of course, you COULD find the intersection ellipse and proceed that way, but it is not the only way and maybe not even the easiest way.

Note: min/max ##x^2+y^2+z^2## has exactly the same solutions as min/max ##\sqrt{x^2 + y^2 + z^2}## but is much easier to work with.

RGV
 
  • #3


Ray Vickson said:
You don't need to find the ellipse of intersection; you just need to minimize (or maximize) x^2 + y^2 + z^2, subject to the two equality constraints x+y+2z = 30 and z = x^2 + y^2. You could, for example, use Lagrange multipliers to solve this problem; you would have two Lagrange multipliers, one for each constraint. Of course, you COULD find the intersection ellipse and proceed that way, but it is not the only way and maybe not even the easiest way.

Note: min/max ##x^2+y^2+z^2## has exactly the same solutions as min/max ##\sqrt{x^2 + y^2 + z^2}## but is much easier to work with.

RGV

Thanks! That simplifies things a lot.
 

1. What is the equation for a parabola?

The equation for a parabola is y = ax^2 + bx + c, where a, b, and c are constants that determine the shape and position of the parabola.

2. How do you find the points nearest to the origin on the intersection of a plane and a parabola?

To find the points nearest to the origin, we need to find the shortest distance from the origin to the intersection of the plane and the parabola. This can be done by setting up a system of equations and using the distance formula to solve for the values of x and y that minimize the distance.

3. How do you find the points farthest from the origin on the intersection of a plane and a parabola?

To find the points farthest from the origin, we need to find the longest distance from the origin to the intersection of the plane and the parabola. This can also be done by setting up a system of equations and using the distance formula to solve for the values of x and y that maximize the distance.

4. Can the points on the intersection of a plane and a parabola be imaginary?

Yes, it is possible for the points on the intersection of a plane and a parabola to be imaginary. This can happen when the plane and the parabola do not intersect at any real points, but instead intersect at a complex number solution.

5. Are there any real-world applications for finding points on the intersection of a plane and a parabola?

Yes, there are many real-world applications for finding points on the intersection of a plane and a parabola. For example, this concept is used in optics to find the optimal position for a lens to focus light to a specific point, and in engineering to determine the trajectory of a projectile launched from a given point with a specific initial velocity.

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