Points where normal at point on surface equals line from origin to that point

[Jamin2112]

Homework Statement

Find the points (x,y,z) on the paraboloid z=x²+y²-1 at which the normal line to the surface is the line from the origin to the point (x,y,z).

Homework Equations

normal line means take the gradient, <∂F/∂x, ∂F,∂y, ∂F,∂z>, and evaluate at the point

The Attempt at a Solution

n= -2xi + -2yj + 1k, where i=(1 0 0)^T, j=(0 1 0)^T, k=(0 0 1)^T.

So we want an (x,y,z) that such that the normal vector n is a scalar multiple of xi + yj + (z=x²+y²+1)k.

<-2x,-2y,1> = ß<x,y,x²+y²+1>

<-2,-2,0> = <ß, ß, ß(x²+y²)>

Am I doing this right? Something ain't working.

 

[ystael]

So we want an (x,y,z) that such that the normal vector n is a scalar multiple of xi + yj + (z=x²+y²-1)k.

This is a reasonable approach, yes. Note that I corrected your sign above: z = x^2 + y^2 - 1, not z = x^2 + y^2 + 1.

<-2x,-2y,1> = ß<x,y,x²+y²-1>

<-2,-2,0> = <ß, ß, ß(x²+y²)>

Am I doing this right? Something ain't working.

How did you get from the first line to the second? It looks like you divided the first component of your vector equation by x, the second by y, and then tried to add something or other to the third. While in theory you could organize your manipulations that way (except that whatever you did to the third component is wrong), in practice it's a bad idea.

The equation \langle -2x, -2y, 1\rangle = \beta \langle x, y, x^2 + y^2 - 1 \rangle translates into the system of equations \begin{cases}-2x = \beta x &amp; \\ -2y = \beta y &amp; \\ 1 = \beta(x^2 + y^2 - 1) &amp; \end{cases}. Proceed from there.

 

[Jamin2112]

This is a reasonable approach, yes. Note that I corrected your sign above: z = x^2 + y^2 - 1, not z = x^2 + y^2 + 1.



How did you get from the first line to the second? It looks like you divided the first component of your vector equation by x, the second by y, and then tried to add something or other to the third. While in theory you could organize your manipulations that way (except that whatever you did to the third component is wrong), in practice it's a bad idea.

The equation \langle -2x, -2y, 1\rangle = \beta \langle x, y, x^2 + y^2 - 1 \rangle translates into the system of equations \begin{cases}-2x = \beta x &amp; \\ -2y = \beta y &amp; \\ 1 = \beta(x^2 + y^2 - 1) &amp; \end{cases}. Proceed from there.

I just tried to get the components to match up; for example, -2xi=ßxi. But now I see what's happenin'.

 

[Jamin2112]

I just tried to get the components to match up; for example, -2xi=ßxi. But now I see what's happenin'.

Actually, I don't see what's happening. ß would be -2, looking at the first two equations, but then plugging that into the third gives the equation x² + y² = -1. Explain how to solve that system.

 

[ystael]

Actually, I don't see what's happening. ß would be -2, looking at the first two equations, but then plugging that into the third gives the equation x² + y² = -1.

No, it doesn't; check your algebra. Also, \beta = -2 is not the only possible solution to the first two equations.

Explain how to solve that system.

That's not very polite.

 

[D H]

ß would be -2, looking at the first two equations

Correct.

but then plugging that into the third gives the equation x² + y² = -1

Incorrect. Check your math.

 

[Jamin2112]

For some reason, I'm not getting this problem. The wires in my brain must be crossed.

So what I did is just say z = r² - 1, where r is the radius of the cross section of a plane z=c that cuts through the paraboloid. Then dz/dr=2r, meaning that at r=1/2, a scalar multiple of the normal line will pass through the origin.