Finding the unit normal of a vector field

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Homework Help Overview

The discussion revolves around finding the unit normal vector of a vector field, specifically in the context of calculus. Participants are exploring the implications of the unit normal being equal to k and its relation to curves parallel to the xy-plane.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method for finding the unit normal vector, questioning why it equals k in certain cases. There are inquiries about the meaning of d sigma and its relation to area differentials.

Discussion Status

Some participants have provided insights regarding the conditions under which the unit normal vector is k or -k, depending on the orientation of the surface. There is ongoing exploration of the meaning of d sigma and its calculation.

Contextual Notes

Participants mention that the problem states n = k, which raises questions about the assumptions made regarding the surface orientation. The discussion also touches on the implications of the curve being parallel to the xy-plane.

robertjford80
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I'm on the last chapter of a 1200 page calc book, I'm really psyched.

Homework Statement



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The Attempt at a Solution



The method I learned for finding the unit normal of a vector field, n, is take the derivative of the equation and divide that by the magnitude of the derivative. This technique works for questions 3 and 4 above, but for 1,2 and 5, the solution manual just says n = k. I can't figure out why n = k. What are they talking about?
 
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I also need to know what d sigma = dxdy means.
 
robertjford80 said:
I also need to know what d sigma = dxdy means.

dσ=dx dy simply means that the differential of area is equal to the differential of x multiplied by the differential of y.
 
robertjford80 said:
I'm on the last chapter of a 1200 page calc book, I'm really psyched.

Homework Statement



Screenshot2012-05-31at41711AM.png


Screenshot2012-05-31at41418AM.png




The Attempt at a Solution



The method I learned for finding the unit normal of a vector field, n, is take the derivative of the equation and divide that by the magnitude of the derivative. This technique works for questions 3 and 4 above, but for 1,2 and 5, the solution manual just says n = k. I can't figure out why n = k. What are they talking about?

Because the curve is parallel to the xy plane.
 
robertjford80 said:
The method I learned for finding the unit normal of a vector field, n, is take the derivative of the equation and divide that by the magnitude of the derivative. This technique works for questions 3 and 4 above, but for 1,2 and 5, the solution manual just says n = k. I can't figure out why n = k. What are they talking about?

\hat n = \vec k means that the unit normal vector is parallel to the z-axis, in this case, it is in the vertically upward direction.
 
what about d sigma?

how do you find n if the curve is parallel to the xy plane, does it just equal k?
 
robertjford80 said:
how do you find n if the curve is parallel to the xy plane, does it just equal k?

Yes.
 
robertjford80 said:
what about d sigma?

d sigma is dx dy.
 
sharks said:
\hat n = \vec k means that the unit normal vector is parallel to the z-axis, in this case, it is in the vertically upward direction.

Or in other words, the curve is parallel to the xy plane.
 
  • #10
robertjford80 said:
how do you find n if the curve is parallel to the xy plane, does it just equal k?

n could be k or -k, depending on the orientation of the surface. n is the outward unit normal vector, and if the problem states "counterclockwise from above" then n is pointing upward, otherwise if it states ""counterclockwise from below"" then n = -k. If the surface lies in any plane (x=0, y=0 or z=0), then there are only 2 possible orientations for n, without any need for calculations, as you can deduce n directly.

If the surface lies in:
plane z=0, n is either k or -k
plane y=0, n is either j or -j
plane x=0, n is either i or -i

In all 3 cases, the positive value denotes moving along the positive direction parallel the respective axis, and similarly the negative value denotes moving in the negative direction parallel to the axis.
 
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  • #11
sharks said:
n could be k or -k, depending on the orientation of the surface. n is the outward unit normal vector, and if the problem states "counterclockwise from above" then n is pointing upward, otherwise if it states ""counterclockwise from below"" then n = -k. If the surface lies in any plane (x=0, y=0 or z=0), then there are only 2 possible orientations for n, without any need for calculations, as you can deduce n directly.

If the surface lies in:
plane z=0, n is either k or -k
plane y=0, n is either j or -j
plane x=0, n is either i or -i

In all 3 cases, the positive value denotes moving along the positive direction parallel the respective axis, and similarly the negative value denotes moving in the negative direction parallel to the axis.

The OP said that the problem says n=k, so n is just k.
 
  • #12
dimension10 said:
d sigma is dx dy.

but what does d sigma refer to? how do you calculate it? what does it mean?
 
  • #13
##d\sigma## refers to the differential area, which when expressed in terms of Cartesian coordinates, converts to dxdy, as you are projecting the surface onto the x-y plane.
 
  • #14
robertjford80 said:
but what does d sigma refer to? how do you calculate it? what does it mean?

As sharks has said, it is the differential of area.

{\rm{d}}\sigma = \left\| {\frac{{\partial \vec r}}{{\partial s}} \times \frac{{\partial \vec r}}{{\partial t}}} \right\|{\rm{d}}s{\rm{ d}}t

In your case, \left\| {\frac{{\partial \vec r}}{{\partial s}} \times \frac{{\partial \vec r}}{{\partial t}}} \right\| = 1 and ##s=x,t=y##
 

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