course Mth 275
Resubmitted copied from Notepad.
Section 12.5: This section took some thinking about to fully understand, but I’m comfortable with all the material now. I did have a problem with a couple of the questions.
#51 gives z = xy + f(x^2-y^2) and says to show that y(dz/dx) – x(dz/dy) = y^2 – x. Using w = x^2 – y^2, I can get to x(dz/dy) + y(dz/dx) = 0, but no further.
If z = xy + f(x^2-y^2) then
z_y = x - 2 y f ' (x^2 - y^2) and
z_x = y + 2 x f ' ( x^2 - y^2).
Here we use the chain rule on the expression f(x^2 - y^2), in which the 'inner' function has derivative -2 y when take with respect to y, 2x if take with respect to x.
Then -x z_y + y z_x = -x (x - 2 y f ' (x^2 - y^2)) + y (y + 2 x f ' ( x^2 - y^2) ) = y^2 - x^2.
#60 gives F(x,y,z) = 0 and G(x,y,z) = 0 with y=y(x) and z = z(x), and asks for dy/dx and dz/dx in terms of the partial derivatives of F and G. I came up with dy/dx = (dF/dx)/( dF/dy) = (dG/dx)/( dG/dy) and dz/dx = (dF/dx)/( dF/dz) = (dG/dx)/( dG/dz), but this seems too simple.
dF = F_x dx + F_y dy + F_z dz = F_x dx + F_y dy/dx dx + F_z dz/dx dx = 0 so that
F_x + F_y dy/dx + F_z dz/dx = 0 and similarly
G_x + G_y dy/dx + G_z dz/dx = 0.
Rearranging we have
F_y dy/dx + F_z dz/dx = F_x and
G_y dy/dx + G_z dz/dx = G_x.
By Cramer's Rule or any of a variety of other methods the solution is
dy/dx = (F_x G_z - F_z G_x) / (F_y G_z - F_z G_y) and
dz/dx = (F_y G_x - F_x G_y) / (F_y G_z - F_z G_y).
Note that the same result would follow as long as F(x,y,z) and G(x,y,z) are both constant. The only thing that is required is that the differentials dF and dG both be zero.
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This looks good. Let me know if you have questions.