Finding a function in conservative field

[greenfrog]

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!

 

[gabbagabbahey]

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...

 

[greenfrog]

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).

 

[gabbagabbahey]

= [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?

 

[greenfrog]

thanks! i got it.