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DETAILS
MY NOTES
SCALCET9 16.3.007.
Determine whether or not 
F is a coliservative vector fleld. If it is, find a function 
f such that 
F=grad f. (If the vector field is not conservative, enter DNE.)


F(x,y)=(ye^(x)+sin(y))i+(e^(x)+x cos(y))j

f(x,y)=

44. [0/1 [0 / 1 Points ] ] \newlineDETAILS\newlineMY NOTES\newlineSCALCET99 1616.33.007007.\newlineDetermine whether or not F \mathrm{F} is a coliservative vector fleld. If it is, find a function f f such that \mathrm{F}=\(\newlineabla f \). (If the vector field is not conservative, enter DNE.)\newlineF(x,y)=(yex+sin(y))i+(ex+xcos(y))j \mathbf{F}(x, y)=\left(y e^{x}+\sin (y)\right) \mathbf{i}+\left(e^{x}+x \cos (y)\right) \mathbf{j} \newlinef(x,y)= f(x, y)=

Full solution

Q. 44. [0/1 [0 / 1 Points ] ] \newlineDETAILS\newlineMY NOTES\newlineSCALCET99 1616.33.007007.\newlineDetermine whether or not F \mathrm{F} is a coliservative vector fleld. If it is, find a function f f such that \mathrm{F}=\(\newlineabla f \). (If the vector field is not conservative, enter DNE.)\newlineF(x,y)=(yex+sin(y))i+(ex+xcos(y))j \mathbf{F}(x, y)=\left(y e^{x}+\sin (y)\right) \mathbf{i}+\left(e^{x}+x \cos (y)\right) \mathbf{j} \newlinef(x,y)= f(x, y)=
  1. Check Curl of F: To check if FF is conservative, we need to see if the curl of FF is zero. The vector field F(x,y)=(yex+sin(y))i+(ex+xcos(y))jF(x,y) = (y e^x + \sin(y))i + (e^x + x \cos(y))j. Calculate the partial derivatives: y\frac{\partial}{\partial y} of the first component: y(yex+sin(y))=ex+cos(y)\frac{\partial}{\partial y} (y e^x + \sin(y)) = e^x + \cos(y). x\frac{\partial}{\partial x} of the second component: x(ex+xcos(y))=ex+cos(y)\frac{\partial}{\partial x} (e^x + x \cos(y)) = e^x + \cos(y).
  2. Verify Conservative Field: Since the partial derivatives are equal, the curl of FF is zero, indicating that FF is a conservative vector field.
  3. Find Potential Function: Now, find a potential function ff such that F = \(\newlineabla f\). Start by integrating the first component of FF with respect to xx:f(x,y)=(yex+sin(y))dx=yex+xsin(y)+g(y)f(x,y) = \int (y e^x + \sin(y)) \, dx = y e^x + x \sin(y) + g(y), where g(y)g(y) is a function of yy only.
  4. Differentiate with Respect to y: Next, differentiate this potential function with respect to yy to match the second component of FF:fy=ex+xcos(y)+g(y).\frac{\partial f}{\partial y} = e^x + x \cos(y) + g'(y).Since fy\frac{\partial f}{\partial y} should equal the second component of FF, which is ex+xcos(y)e^x + x \cos(y), we set g(y)=0g'(y) = 0.
  5. Integrate g(y)=0g'(y) = 0: Integrate g(y)=0g'(y) = 0 to find g(y)g(y):g(y)=Cg(y) = C, where CC is a constant.
  6. Substitute g(y)=Cg(y) = C: Substitute g(y)=Cg(y) = C into the potential function:\newlinef(x,y)=yex+xsin(y)+Cf(x,y) = y e^x + x \sin(y) + C.\newlineSince the constant CC does not affect the gradient, we can set C=0C = 0 for simplicity.

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