What is the unit normal for a cylinder?

In summary, the normal for a sphere is defined as, \hat{a}_{n} = x\hat{x} + y\hat{y} + z\hat{z} where \sqrt{3} is the magnitude. However, if you compute the norm, it's not a unit vector. The top equation is for a general sphere of radius r = Sqrt(x^2+y^2+z^2), but the second one is specific for a unit sphere.
  • #1
jeff1evesque
312
0

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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  • #2
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)?
 
  • #3
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.
 
  • #4
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?
 
  • #5
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.
 
  • #6
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].
 
  • #7
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:
  • #8
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?
 
  • #9
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
 

What is a unit normal to a sphere?

A unit normal to a sphere is a vector that is perpendicular to the surface of the sphere at a specific point. It has a magnitude of 1 and is used to determine the orientation of the surface at that point.

How is a unit normal to a sphere calculated?

A unit normal to a sphere can be calculated using the gradient vector of the equation that defines the sphere. The gradient vector is a vector that points in the direction of the greatest rate of change of the function at a given point.

Why is a unit normal to a sphere important?

A unit normal to a sphere is important because it is used to calculate the curvature of the sphere at a given point. It also helps in determining the direction of light or force on the surface of the sphere.

What is the relationship between the unit normal and the surface normal of a sphere?

The unit normal and the surface normal of a sphere are equivalent as they both represent a vector perpendicular to the surface of the sphere at a specific point. However, the unit normal has a magnitude of 1, while the surface normal can have any magnitude.

Can a unit normal to a sphere change at different points on the surface?

Yes, the unit normal to a sphere can change at different points on the surface. This is because the orientation of the surface changes at each point, and the unit normal is always perpendicular to the surface at that point.

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