Respuesta :
[Note: for whatever reason "f(x,y,z)" isn't allowed in this answer, so I've replaced [tex]f[/tex] with [tex]\varphi[/tex].]
Find the curl of [tex]\vec\varphi[/tex]. If it's [tex]\vec0[/tex], then [tex]\vec\varphi[/tex] is conservative.
So there is a scalar function [tex]\varphi[/tex] that satisfies [tex]\nabla\varphi=\vec\varphi[/tex]. We have
[tex]D_x\varphi(x,y,z)=y\cos xy[/tex]
[tex]\implies \varphi(x,y,z)=\sin xy+g(y,z)[/tex]
[tex]D_y\varphi(x,y,z)=D_y(\sin xy+g(y,z))[/tex]
[tex]\implies x\cos xy=x\cos xy+D_yg(y,z)[/tex]
[tex]D_yg(y,z)=0[/tex]
[tex]\implies g(y,z)=h(z)[/tex]
[tex]D_z\varphi(x,y,z)=D_zh(z)[/tex]
[tex]\implies-3\sin z=h'(z)[/tex]
[tex]\implies h(z)=3\cos z+C[/tex]
[tex]\varphi(x,y,z)=\sin xy+3\cos z+C[/tex]
Find the curl of [tex]\vec\varphi[/tex]. If it's [tex]\vec0[/tex], then [tex]\vec\varphi[/tex] is conservative.
So there is a scalar function [tex]\varphi[/tex] that satisfies [tex]\nabla\varphi=\vec\varphi[/tex]. We have
[tex]D_x\varphi(x,y,z)=y\cos xy[/tex]
[tex]\implies \varphi(x,y,z)=\sin xy+g(y,z)[/tex]
[tex]D_y\varphi(x,y,z)=D_y(\sin xy+g(y,z))[/tex]
[tex]\implies x\cos xy=x\cos xy+D_yg(y,z)[/tex]
[tex]D_yg(y,z)=0[/tex]
[tex]\implies g(y,z)=h(z)[/tex]
[tex]D_z\varphi(x,y,z)=D_zh(z)[/tex]
[tex]\implies-3\sin z=h'(z)[/tex]
[tex]\implies h(z)=3\cos z+C[/tex]
[tex]\varphi(x,y,z)=\sin xy+3\cos z+C[/tex]
Following are the calculation to the given function:
Function:
[tex]\bold{F(x, y, z)= y \cos(xy)i + x \cos(xy)j - \sin(z)k}.[/tex]
The object is to find a function such that [tex]F= \bigtriangledown f[/tex].
That is,
[tex]F= \bigtriangledown f[/tex].
[tex]y \cos(xy) i + x \cos(xy) j- \sin (z)k = f_x i+f_y j+f_z k[/tex]
Then,
[tex]f_x = y \cos(xy), f_y = x \cos(xy), f_z =- \sin (z). \\\\Integrate\\\\ f_x = y \cos(xy) \\\\ \int f_x =\int y \cos (xy) dx\\\\f =y\int \cos (xy) dx\\\\\ \ \ \ = y \frac{\sin (xy)}{y} + C (x , y) \ \ \ \ \ Use \int \cos(xy) dx= \frac{sin(xy)}{y}\\\\f=sin (xy)+C(y,z)[/tex]
Differentiate on both sides with respect to y partially.
[tex]f_z=\frac{\delta }{\delta y} (\sin(xy)+C(y,z)) \\\\ = \cos(xy) \frac{\delta}{\delta y}(xy)+C_y(y,z) \\\\f_y =x\cos (xy) +C_y (y,z) \\\\[/tex]
Then,
[tex]f_y = x \cos( xy)+C_y (y,z) \\\\x \cos(xy) = x \cos(xy) + C_y (y,z)[/tex]
Use
[tex]\ f_y = x \cos(xy)\\\\ C_y (y,z)=0[/tex]
Integrate on both sides with respect to y.
[tex]\int C_y (y,z)=0 \\\\C(y, z)= D(z)[/tex]
Plug
[tex]\ C(y, z)= D(z) into f = \sin(xy)+C(y,z)\\\\f = \sin(xy)+D(z) \\\\[/tex]
Differentiate on both sides with respect to z partially.
[tex]f_z= \frac{\delta }{\delta z} (\sin (xy) +D(z)) \\\\[/tex]
[tex]= \frac{\delta }{\delta z} (\sin(xy)+ \frac{\delta }{\delta z} (D(z))\\\\=0+D'(z)\\\\ =D' (z)\\\\[/tex]
Plug
[tex]f_z =- \sin (z) into f_z = D'(z). \\\\-\sin z = D' (z)[/tex]
Integrate with respect to z.
[tex]\int -\sin z dz= \int D'(z)\\\\ -(- \cos z)+E= D(z)\\\\ \cos Z + E = D(z) \\\\[/tex]
Here, E is the integration constant.
Plug
[tex]D(z)= \COS Z + E into f = \sin(xy)+D(z).\\\\ f = \sin(xy ) + \cos z + E.[/tex]
Therefore, the desired function is, " [tex]\bold{f = \sin(xy) + \cos Z+E }[/tex]"
Learn more:
brainly.com/question/12492083
