anonymous
  • anonymous
Medal****Please Help Given the vectorfield F(x,y)= (x+2xsiny) i + (x^2cosy +2y) j Determine whether F is conservative. If it is, find a potential function
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@ganeshie8
ganeshie8
  • ganeshie8
Hint : curl must be 0 for the vector field to be conservative
anonymous
  • anonymous
what formula would i apply here?

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ganeshie8
  • ganeshie8
\(F = Mi + Nj\) \(\text{curl(F)} = N_x - M_y\)
ganeshie8
  • ganeshie8
F(x,y)= (x+2xsiny) i + (x^2cosy +2y) j M = x + 2xsiny N = x^2cosy + 2y N_x = ? M_y = ?
anonymous
  • anonymous
i take the derivative?
ganeshie8
  • ganeshie8
N_x means the partial of N with respect to x
anonymous
  • anonymous
2xcosy?
ganeshie8
  • ganeshie8
what is 2xcosy ?
anonymous
  • anonymous
Nx
ganeshie8
  • ganeshie8
OK, work the curl
anonymous
  • anonymous
2xcosy-My....i dont know the My
Loser66
  • Loser66
.
ganeshie8
  • ganeshie8
You need to work M_y too
anonymous
  • anonymous
-2xsiny+2 for My?
ganeshie8
  • ganeshie8
M = x + 2xsiny M_y = ?
anonymous
  • anonymous
1+2cosy
ganeshie8
  • ganeshie8
try again
anonymous
  • anonymous
2xcosy
ganeshie8
  • ganeshie8
Yes, curl = ?
anonymous
  • anonymous
2xcosy-2xcosy=0
ganeshie8
  • ganeshie8
Since the curl is 0, the given vector field is conservative and a potential function exists
ganeshie8
  • ganeshie8
Let \(f(x,y)\) be the potential function, then this must satisfy : \(f_x = M\) \(f_y = N\)
anonymous
  • anonymous
what if it wasnt 0, it would be non-conservative but still a potential function?
ganeshie8
  • ganeshie8
If the curl is not 0, then the vector field is not conservative and consequently there will not be a potential function. You cannot find a potential function.
ganeshie8
  • ganeshie8
potential function for a vector field exists if and only if the curl of vector field is 0
anonymous
  • anonymous
oki! from fx=m fy=n....whats the next step?
ganeshie8
  • ganeshie8
plug in m and n
ganeshie8
  • ganeshie8
\[f_x = x + 2x\sin y\] simply integrate both sides with respect to \(x\) to get \(f\)
anonymous
  • anonymous
x^2/2 +2x^3/3siny?
ganeshie8
  • ganeshie8
try again
ganeshie8
  • ganeshie8
and what about the integration constant ?
anonymous
  • anonymous
x.... x^2/2+x^2siny
anonymous
  • anonymous
x^2/2+2x^2/2siny
ganeshie8
  • ganeshie8
Easy, \[f_x = x + 2x\sin y\] integrating both sides with respect to \(x\) gives \[f = \frac{x^2}{2} + x\sin y + \color{red}{g(y)}\]
ganeshie8
  • ganeshie8
that \(\color{red}{g(y)}\) is the arbitrary constant, it shows up everytime you integrate
ganeshie8
  • ganeshie8
you need to find \(\color{red}{g(y)}\)
anonymous
  • anonymous
xsiny why isnt it x^2siny
ganeshie8
  • ganeshie8
Oops! my mistake... see if below looks fine \[f_x = x + 2x\sin y\] integrating both sides with respect to \(x\) gives \[f = \frac{x^2}{2} + x^2\sin y + \color{red}{g(y)}\]
anonymous
  • anonymous
i take the third derviative to find g(Y)
ganeshie8
  • ganeshie8
There is an easy way take the derivative of \(f\) with respect to \(y\) and compare it with the equation \(f_y = N\)
ganeshie8
  • ganeshie8
\[f = \frac{x^2}{2} + x^2\sin y + \color{red}{g(y)}\] \[f_y = ?\]
anonymous
  • anonymous
Y^2
ganeshie8
  • ganeshie8
how
anonymous
  • anonymous
nevermind...is it x^2cosy
ganeshie8
  • ganeshie8
\[f = \frac{x^2}{2} + x^2\sin y + \color{red}{g(y)}\] \[f_y = x^2\cos y + \color{red}{g'(y)}\]
ganeshie8
  • ganeshie8
compare this with the other equation \(f_y = N\)
anonymous
  • anonymous
g'(y) is 2y?
ganeshie8
  • ganeshie8
Yes, integrate and solve \(g(y)\)
anonymous
  • anonymous
y^2
ganeshie8
  • ganeshie8
Looks good! plug that in the potential function and you're done!
ganeshie8
  • ganeshie8
\[f = \frac{x^2}{2} + x^2\sin y + \color{red}{g(y)} = \frac{x^2}{2} + x^2\sin y + \color{red}{y^2} \]
anonymous
  • anonymous
thank you so much! x^2/2+x^2siny+Y^2
ganeshie8
  • ganeshie8
just a sanity check : find the partials and see if you really get M and N
ganeshie8
  • ganeshie8
you must get \(f_x = M\) and \(f_y = N\)
anonymous
  • anonymous
yes i do
anonymous
  • anonymous
thank you again ganeshie8!!
IrishBoy123
  • IrishBoy123
for completeness, there's a really handy way around the final step that avoids fiddling around with the integration constants. because the field is conservative, it is path independent so you can do a line integral. \(W = \int \vec F \bullet \vec dr = \int \bullet \) from (0,0) to (x,y) using, for example, these steps: |dw:1436724170339:dw| the line integral is straightforward
ganeshie8
  • ganeshie8
I prefer the line integral too (ofcourse when eyeballing method fails :))

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