Suppose I have this contour described by the equation the root $z$ of this equation $$ \frac{1}{x^2 + y^2} + \frac{1}{xz} = 2y $$
Now suppose the equation is tweaked slightly, with an addition of $w$, where $w$ is a number: $$w + \frac{1}{x^2 + y^2} + \frac{1}{xz} = 2y$$
eq1[x_, y_, z_] := 1/(x + y^2) + 1/(x*z) == 2 y
ContourPlot[
z /. FindRoot[eq1[x, y, z], {z, 1}], {x, -5, 5}, {y, -5, 5}]
eq2[x_, y_, z_] := 2 + 1/(x + y^2) + 1/(x*z) == 2 y
ContourPlot[
z /. FindRoot[eq2[x, y, z], {z, 1}], {x, -5, 5}, {y, -5, 5}]
eq3[x_, y_, z_] := 4 + 1/(x + y^2) + 1/(x*z) == 2 y
ContourPlot[
z /. FindRoot[eq3[x, y, z], {z, 1}], {x, -5, 5}, {y, -5, 5}]
By visual inspection we see that the curves shift upwards as $w$ increases.
However, I need a better way to describe it. Is there a way to "differentiate the contour" to find the contour $\frac{\partial z}{\partial w}$?
Now suppose the equation is tweaked slightly, with an addition of $w$, where $w$ is a number.
Is there a way to "differentiate the contour" to find the contour $\frac{\partial z}{\partial w}$?
I tried "subtracting" the roots then plotting the contour, but mathematica can't seem to execute this:
eq1[x_, y_, z_] := 1/(x + y^2) + 1/(x*z) == 2 y
ContourPlot[
z /. FindRoot[eq1[x, y, z], {z, 1}], {x, -5, 5}, {y, -5, 5}]
eq2[x_, y_, z_] := 2 + 1/(x + y^2) + 1/(x*z) == 2 y
ContourPlot[
z /. FindRoot[eq2[x, y, z], {z, 1}], {x, -5, 5}, {y, -5, 5}]
sol[x_,y_] := FindRoot[eq2[x, y, z], {z, 1}] - FindRoot[eq1[x, y, z], {z, 1}]
ContourPlot[ sol[x,y], {x, -5, 5}, {y, -5, 5}]
sol[..]
tosol[x_, y_] := (z /. FindRoot[eq2[x, y, z], {z, 1}]) - (z /. FindRoot[eq1[x, y, z], {z, 1}])
? $\endgroup$