Solve Shortest Distance to Origin on Quadric Surface

[jesuslovesu]

[SOLVED] shortest distance

Homework Statement

Find the shortest distance to the origin given the quadric surface x^2 + y^2 - 2zx = 4

Homework Equations

The Attempt at a Solution

F = x^2 + y^2 + z^2
g = x^2 + y^2 - 2zx = 4

Well I initially substituted y^2 = 4 + 2zx - x^2 into F

F = 4 + 2zx + z^2
which leads to z = 0 x = 0 and so y = 2, however this is not the shortest distance. I know this is because of the boundaries of the surface; however I don't know in what way to modify my analysis so that I can take into account the boundaries and get the shortest distance.

 

[Dick]

I think you've got a good picture of why this solution is going wrong. Taking a partial derivative wrt to z assumes x and y can be held fixed, which takes you off the surface. The 'neat' way to enforce a constraint like this is to do it using a Lagrange multiplier. If you don't know what that is try to express the surface in parametric form (in terms of two truly independent variables). Notice you can write the surface as (x-z)^2+y^2=z^2+4, which might suggest a nice parametrization.

 

[bobbyk]

Laplace self transform

Dick, you seem to be a very good mathematician! Have you ever seen a function of
t whose Laplace Transform is the same function of s? Besides 0.
Bob

 

[arildno]

Dick, you seem to be a very good mathematician! Have you ever seen a function of
t whose Laplace Transform is the same function of s? Besides 0.
Bob

Try Gaussians.

 

[bobbyk]

I did. Gaussians don't work.

 

[Rainbow Child]

Dick, you seem to be a very good mathematician! Have you ever seen a function of
t whose Laplace Transform is the same function of s? Besides 0.
Bob

No!

But is this the right thread to post a reply??

 

[OrderOfThings]

Calculate \nabla F and \nabla g and find those x,y,z that make them parallel. That is, solve

\nabla F = \lambda \nabla g

Make sure you can visualise why the gradients must be parallel.

 

[Dick]

Calculate \nabla F and \nabla g and find those x,y,z that make them parallel. That is, solve

\nabla F = \lambda \nabla g

Make sure you can visualise why the gradients must be parallel.

Nice way to explain it. It's mathematically the same content as 'lagrange multiplier' without the fancy name.

 

[jesuslovesu]

Hey guys, thanks for your replies. Unfortunately, I'm still having a problem getting the answer. I am familiar with lagrange multipliers, however it's been a while since I've done them because I took Calc 3 2 semesters ago. Anyway, I've done a few sample problems from my calculus book to refresh my memory.

I think I'm close to the answer because it's similar, but when I plug in what I get for x^2 and z^2 into f i get 0 and 12/(1 +- sqrt(5)) but I'm not sure where I messed up... I think I'm correct in saying that y = 0 because the other option where lambda = 1 gives 2 = 4.
My work is here http://img169.imageshack.us/img169/9062/ohrl6.th.jpg

 

[HallsofIvy]

From 2y= \lambda 2y, either y= 0 or \lambda= 1. You seem to have rejected \lambda= 1 because it "leads to 2x= 4x". Yes, it does. What's wrong with that? It does not lead to 2= 4 but rather to x= 0. If \lambda= 1, then we have 2x= 2x- 2z and 2z= -2x. From the first, cancelling the "2x" terms on each side, z= 0 and then x= 0. From requirement that x^2+ y^2- 2xz= 4, with x= z= 0, we have y^2= 4 so y= \pm 2. The points on x^2+ y^2- 2xz= 4 closest to the origin are (0, 2, 0) and (0, -2, 0). The distance from either of those points to the origin, and so the shortest distance from the hyperboloid to the origin, is 2.

 

[Dick]

From 2y= \lambda 2y, either y= 0 or \lambda= 1. You seem to have rejected \lambda= 1 because it "leads to 2x= 4x". Yes, it does. What's wrong with that? It does not lead to 2= 4 but rather to x= 0. If \lambda= 1, then we have 2x= 2x- 2z and 2z= -2x. From the first, cancelling the "2x" terms on each side, z= 0 and then x= 0. From requirement that x^2+ y^2- 2xz= 4, with x= z= 0, we have y^2= 4 so y= \pm 2. The points on x^2+ y^2- 2xz= 4 closest to the origin are (0, 2, 0) and (0, -2, 0). The distance from either of those points to the origin, and so the shortest distance from the hyperboloid to the origin, is 2.

That solution is a saddle point. If z=0, then ALL of the points on the circle x^2+y^2=4 are distance 2 from the origin. And the OP already has figured out that there are closer points. And if the y=0 track is fixed, will find a pair. Hint: look at the line just before you concluded x^2(1+lambda)=4.