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

In summary, the homework statement is that you 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). The Attempt at a Solution solves this problem by translating the equation into a system of equations, and finding the solution that corresponds to r=1/2.
  • #1
Jamin2112
986
12

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.
 
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  • #2
Jamin2112 said:
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: [tex]z = x^2 + y^2 - 1[/tex], not [tex]z = x^2 + y^2 + 1[/tex].

Jamin2112 said:
<-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 [tex]x[/tex], the second by [tex]y[/tex], 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 [tex]\langle -2x, -2y, 1\rangle = \beta \langle x, y, x^2 + y^2 - 1 \rangle[/tex] translates into the system of equations [tex]\begin{cases}-2x = \beta x & \\ -2y = \beta y & \\ 1 = \beta(x^2 + y^2 - 1) & \end{cases}[/tex]. Proceed from there.
 
  • #3
ystael said:
This is a reasonable approach, yes. Note that I corrected your sign above: [tex]z = x^2 + y^2 - 1[/tex], not [tex]z = x^2 + y^2 + 1[/tex].



How did you get from the first line to the second? It looks like you divided the first component of your vector equation by [tex]x[/tex], the second by [tex]y[/tex], 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 [tex]\langle -2x, -2y, 1\rangle = \beta \langle x, y, x^2 + y^2 - 1 \rangle[/tex] translates into the system of equations [tex]\begin{cases}-2x = \beta x & \\ -2y = \beta y & \\ 1 = \beta(x^2 + y^2 - 1) & \end{cases}[/tex]. Proceed from there.



I just tried to get the components to match up; for example, -2xi=ßxi. But now I see what's happenin'.
 
  • #4
Jamin2112 said:
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.
 
  • #5
Jamin2112 said:
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, [tex]\beta = -2[/tex] is not the only possible solution to the first two equations.

Jamin2112 said:
Explain how to solve that system.

That's not very polite.
 
  • #6
Jamin2112 said:
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.
 
  • #7
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.
 

What is the concept of "Points where normal at point on surface equals line from origin to that point"?

The concept refers to a point on a surface where the normal vector (perpendicular to the surface) is equal to the line from the origin of the coordinate system to that point. This relationship is important in the study of surfaces and their curvature.

How are these points determined?

These points can be determined through mathematical calculations using the equations of the surface and the normal vector. They can also be visually identified by plotting the surface and the normal vector on a graph.

What is the significance of these points?

These points are significant because they indicate where the curvature of the surface is changing, and can help in understanding the overall shape and behavior of the surface.

Can these points exist on any type of surface?

Yes, these points can exist on any type of surface, including curved and flat surfaces. However, the equations and calculations for determining these points may vary depending on the type of surface.

Are there any real-world applications of this concept?

Yes, this concept has various applications in fields such as computer graphics, engineering, and physics. It can be used to analyze and design surfaces, as well as to understand the behavior of light and sound waves on different surfaces.

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