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Determine whether or not FF is a conservative vector field. If it is, find a function ff such that F=fF=f. (If the vector field is not conservative, enter DNE.\newline{F(x,y)=exsin(y)i+excos(y)j\begin{cases} F(x,y)=e^{x}\sin(y)i+e^{x}\cos(y)j \end{cases},f(x,y)=f(x,y)=

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Q. Determine whether or not FF is a conservative vector field. If it is, find a function ff such that F=fF=f. (If the vector field is not conservative, enter DNE.\newline{F(x,y)=exsin(y)i+excos(y)j\begin{cases} F(x,y)=e^{x}\sin(y)i+e^{x}\cos(y)j \end{cases},f(x,y)=f(x,y)=
  1. Check Curl of F: Determine if F is conservative by checking if the curl of F is zero. \newlineCurl F = (y(exsin(y))x(excos(y)))k(\frac{\partial}{\partial y}(e^x \sin(y)) - \frac{\partial}{\partial x}(e^x \cos(y)))k \newline = (excos(y)excos(y))k(e^x \cos(y) - e^x \cos(y))k \newline = 0k0k
  2. Verify Conservativity: Since the curl of FF is zero, FF is a conservative vector field.
  3. Find Potential Function: Find a potential function ff such that F = \(\newlineabla f\). We need fx=exsin(y)\frac{\partial f}{\partial x} = e^x \sin(y) and fy=excos(y)\frac{\partial f}{\partial y} = e^x \cos(y).
  4. Integrate fx\frac{\partial f}{\partial x}: Integrate fx=exsin(y)\frac{\partial f}{\partial x} = e^x \sin(y) with respect to xx.
    f(x,y)=exsin(y)dxf(x, y) = \int e^x \sin(y) \, dx
    =exsin(y)+g(y)\quad\quad= e^x \sin(y) + g(y), where g(y)g(y) is a function of yy only.
  5. Differentiate to Find g(y)g'(y): Differentiate f(x,y)=exsin(y)+g(y)f(x, y) = e^x \sin(y) + g(y) with respect to yy to find g(y)g'(y).
    y(exsin(y)+g(y))=excos(y)+g(y)\frac{\partial}{\partial y}(e^x \sin(y) + g(y)) = e^x \cos(y) + g'(y)
    Since fy=excos(y)\frac{\partial f}{\partial y} = e^x \cos(y), we have g(y)=0g'(y) = 0.
  6. Integrate g(y)g'(y): Integrate g(y)=0g'(y) = 0 to find g(y)g(y).\newlineg(y)=Cg(y) = C, where CC is a constant.
  7. Substitute g(y)g(y): Substitute g(y)g(y) back into f(x,y)f(x, y).f(x,y)=exsin(y)+Cf(x, y) = e^x \sin(y) + C

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