What is the unit normal for a cylinder?

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Homework Help Overview

The discussion revolves around the concept of unit normals, specifically in the context of spheres and potentially extending to cylinders. The original poster questions the definition of the unit normal vector and its relation to the sphere's radius.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definition of unit normals, questioning the conditions under which a vector qualifies as a unit vector. There is a discussion about the magnitude of vectors and how it relates to their classification as unit normals.

Discussion Status

Participants are actively engaging with the definitions and properties of unit normals, with some providing clarifications about the geometric interpretation of normals to surfaces. The conversation is evolving towards understanding how these concepts apply to different shapes, such as cylinders.

Contextual Notes

There is an indication that the original poster is considering a problem involving a cylinder, which introduces additional complexity to the discussion of unit normals. The transition from spheres to cylinders suggests a broader exploration of the topic.

jeff1evesque
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Homework Statement


The unit normal to a sphere is defined as, [tex]\hat{a}_{n} = x\hat{x} + y\hat{y} + z\hat{z}[/tex]

But I thought it would be defined as, [tex]\hat{a}_{n} = 1\hat{x} + 1\hat{y} + 1\hat{z}[/tex]

Could someone explain to me why am I thinking incorrectly?

Thanks,


JL
 
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If you compute the norm a.k.a. magnitude a.k.a. length of [itex]1\hat{x} + 1\hat{y} + 1\hat{z}[/itex], it's [itex]\sqrt{3}[/itex]. So that's not a unit vector.

Is the sphere you're talking about a unit sphere (radius 1)?
 
diazona said:
If you compute the norm a.k.a. magnitude a.k.a. length of [itex]1\hat{x} + 1\hat{y} + 1\hat{z}[/itex], it's [itex]\sqrt{3}[/itex]. So that's not a unit vector.

Is the sphere you're talking about a unit sphere (radius 1)?

Yes sir.
 
diazona said:
If you compute the norm a.k.a. magnitude a.k.a. length of [itex]1\hat{x} + 1\hat{y} + 1\hat{z}[/itex], it's [itex]\sqrt{3}[/itex]. So that's not a unit vector.

Is the sphere you're talking about a unit sphere (radius 1)?

Yes sir. But I thought the magnitude of the unit vector is a scalar, in which case [itex]\sqrt{3}[/itex] is fine?
 
I believe the top equation is the normal for a general sphere of radius r = Sqrt(x^2+y^2+z^2). The second one is specifically for a unit sphere.

Also, the top equation is not necessarily of unit length. Imagine you had a sphere of radius of 5. The sphere would include the pointing (5,0,0), giving the above vector a magnitude of 5.
 
jeff1evesque said:
Yes sir. But I thought the magnitude of the unit vector is a scalar, in which case [itex]\sqrt{3}[/itex] is fine?
The magnitude of a vector is a scalar. BUT: the magnitude of a unit vector is 1. That's the definition of a unit vector, a vector that has magnitude 1. A vector with any other magnitude is not a unit vector.

As flatmaster said, the magnitude of [itex]x\hat{x} + y\hat{y} + z\hat{z}[/itex] is [itex]\sqrt{(x^2+y^2+z^2)}[/itex]. So [itex]x\hat{x} + y\hat{y} + z\hat{z}[/itex] is going to be a unit normal if and only if [itex]\sqrt{(x^2+y^2+z^2)} = 1[/itex].
 
diazona said:
The magnitude of a vector is a scalar. BUT: the magnitude of a unit vector is 1. That's the definition of a unit vector, a vector that has magnitude 1. A vector with any other magnitude is not a unit vector.

As flatmaster said, the magnitude of [itex]x\hat{x} + y\hat{y} + z\hat{z}[/itex] is [itex]\sqrt{(x^2+y^2+z^2)}[/itex]. So [itex]x\hat{x} + y\hat{y} + z\hat{z}[/itex] is going to be a unit normal if and only if [itex]\sqrt{(x^2+y^2+z^2)} = 1[/itex].

what if [tex] \hat{a}_{n} = 1\hat{x} + 1\hat{y} + 1\hat{z}[/tex]? I don't understand why that would be wrong? Is it because if we take the magnitude then it's not equal to 1? I think i understand now.
 
Last edited:
I guess my question is, how do you get the normal vector [tex]\hat{a}_{n}[/tex] From a sphere with a radius of 1?
 
jeff1evesque said:
what if [tex] \hat{a}_{n} = 1\hat{x} + 1\hat{y} + 1\hat{z}[/tex]? I don't understand why that would be wrong? Is it because if we take the magnitude then it's not equal to 1?
Yes, exactly.

EDIT: Also, it's not normal to the sphere's surface everywhere.
 
  • #10
Redbelly98 said:
Yes, exactly.

EDIT: Also, it's not normal to the sphere's surface everywhere.

But how do you find the normal? I am reading some stuff, and I notice that the gradient is normal to a surface/curve. If I can find the gradient of our sphere then I can find the normal, and thus the "unit normal". But I am not sure how to find the gradient.
 
  • #11
For this one, we just have to think about the geometry of a sphere. No gradients are necessary.

Any vector that is directed from the origin to some point (x,y,z) on the sphere will be directed along the normal at that point.
 
  • #12
Redbelly98 said:
For this one, we just have to think about the geometry of a sphere. No gradients are necessary.

Any vector that is directed from the origin to some point (x,y,z) on the sphere will be directed along the normal at that point.

That makes much more sense. I totally forgot that the unit normal vector [tex]\hat{a}_{n}[/tex] is a vector beginning from the origin. What if we had a different curve, perhaps a cylinder? Say a problem gave us an equation for a cylinder [namely [tex]x^2 + y^2 = 9[/tex]], along with some arbitrary Force vector. How would we find this unit normal.
Thanks,JL
 

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