Finding a function in conservative field

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Homework Help Overview

The discussion revolves around finding a non-zero function h(x) that makes the vector field F(x,y) conservative. The field is defined in terms of h(x) and involves trigonometric functions of x and y.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the condition for the curl of the vector field to be zero, with attempts to derive relationships involving h(x). There is a focus on the implications of the curl components and the use of the product rule in differentiation.

Discussion Status

Some participants have provided insights into the necessary conditions for h(x) and the implications of setting the curl to zero. There is an ongoing exploration of the relationship between x and y in the context of the function h(x), with no explicit consensus reached on the form of h(x).

Contextual Notes

Participants note that h(x) must be a non-zero function, and there is discussion about the implications of choosing specific values for x and y in relation to the equations derived from the curl condition.

greenfrog
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Homework Statement



Find the non-zero function h(x) for which:

field F(x,y) = h(x) [xsiny + ycosy] i + h(x) [xcosy - ysiny] j

is conservative.

The Attempt at a Solution



curlF=0
d/dx [h(x) [xcosy - ysiny] ] - d/dy [h(x) [xsiny + y cos y] ] = 0

xcosy = ysiny ?

I have no idea!
 
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greenfrog said:
curlF=0
d/dx [h(x) [xcosy - ysiny] ] - d/dy [h(x) [xsiny + y cos y] ] = 0

That is only the z-component of curlF...all 3 components will have to be zero.

You will need to use the product rule when computing the derivatives...
 
Thanks for the reply but I'm not sure I understand... do u mean this? ...

curlF= {d/dy[0]-d/dz[ h(x)[xcosy - ysiny]]} - {d/dx[0]-d/dz[ h(x)[xsiny + ysiny]]} + {d/dx[ h(x)[xcosy - ysiny]] - d/dy[ h(x)[xsiny + ycosy]]}

= 0 - 0 + d/dx[ h(x)[xcosy - ysiny]] - d/dy[ h(x)[xsiny + ycosy]]}

= [ h`(x)xcosy + h(x)cosy - h`(x)ysiny] - [ h(x)xcosy - h(x)ysiny + h(x)cosy ]

= h`(x)xcosy + h(x)cosy - h`(x)ysiny - h(x)xcosy + h(x)ysiny - h(x)cosy

= h`(x)xcosy - h`(x)ysiny - h(x)xcosy + h(x)ysiny

= xcosy [h`(x) - h(x)] + ysiny [h(x) - h`(x)]

= xcosy [h`(x) - h(x)] - ysiny [h`(x) - h(x)]

= [h`(x) - h(x)] [xcosy - ysiny] = 0

as h(x) is a non-zero vector then xcosy - ysiny = 0

xcosy = ysiny


And the no matter what I do I can't seem to get h(x).
 
greenfrog said:
= [h`(x) - h(x)] [xcosy - ysiny] = 0

as h(x) is a non-zero vector then xcosy - ysiny = 0

xcosy = ysiny

No, you can't pick and choose x and y values such that xcosy = ysin y , you are looking to choose an h(x) that makes [h`(x) - h(x)] [xcosy - ysiny] = 0 for all x and y...The only way that can happen is if
[h`(x) - h(x)]=0...right? What kind of non-trivial function accomplishes that?:wink:
 
thanks! i got it.
 

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