Anonymous at Sat, 19 Oct 2024 01:27:17 UTC No. 16439073

>>16438962
The Desmos graph is correct. There's multiple variables, so the change with respect to x would be [math]3\left(x^{2}-2y\right)=0[/math]. You probably haven't read vector calc yet -- look up total and partial derivatives(or read them after you're done with univariate calc).

Anonymous at Sat, 19 Oct 2024 01:35:50 UTC No. 16439091

>>16439073
Dunno why the equation wasn't rendered, preview showed fine. Maybe it can't render Desmos syntax. Let's see:
Text: 3(x^2-2y)=0
Inline: [math]3(x^2-2y)=0[/math]
Block: [eqn]3(x^2-2y)=0[/eqn]
Desmos-style: [math]3\left(x^{2}-2y\right)=0[/math]

Anonymous at Sat, 19 Oct 2024 01:38:25 UTC No. 16439099

Anonymous at Sat, 19 Oct 2024 02:14:53 UTC No. 16439141

>>16438962
the reason why your plot will always appear nonsensical with this approach is that your equation represents a continuous implicit surface in three dimensions and you are taking a cross section at z = 0. This cross section is very bad, clearly, because it introduces a discontinuity at y = 0, x =0. If you plotted the implicit surface then it would be continuously differentiable and smooth and you could get the total derivative in three dimensions which would be f'(x, y, z) = 0i + 0j + k. Notice that the only non-zero component of the derivative is in the z dimension.

Anonymous at Sat, 19 Oct 2024 02:18:02 UTC No. 16439144

>>16439141
correction
>f'(x, y, z) = 0i + 0j + k.
should be
f'(0, 0, z) = 0i + 0j + k.

Anonymous at Sun, 20 Oct 2024 07:50:48 UTC No. 16440903

https://www.desmos.com/calculator/wfeaaku1aw

New desmos feature- "complex mode"

Does this accurately graph the derivative? I understand that neither of these are functions but does it not give you the rate of change for any point on x^3 + y^3 = 6xy?

folium of descartes

Anonymous at Mon, 21 Oct 2024 09:34:10 UTC No. 16442330

>>16438962
x^2 + y^2 y' = 2(y + x y')

y' = (y^2 - 2x) / (2y - x^2)

Anonymous at Mon, 21 Oct 2024 16:15:08 UTC No. 16442690

>>16438962
The implicite function [math] x^3 + y^3 = 6xy [/math] given.

Let's suppose, we move a little (by dx) to the right.

[math] (x+dx)^3 + y^3 = 6(x+dx)y [/math]

Subtract the second from the first equation to get:

[math] 3x^2dx+3xdx^2+dx^3 = 6ydx [/math]

For infinitesimal dx, the equation is identical to:

[math] 3x^2dx = 6ydx [/math]

The same with dy instead of dx results in:

[math] 3y^2dy = 6xdy [/math]

Yet, we haven't found an y' that way because we can arbitrarily move by dy and dx. However, when moving by dx, we will also generally move by a very specific amount dy on the implicitly given curve.
This means that, after having moved by (dx, dy): [math] (x+dx)^3 + (y+dy)^3 = 6(x+dx)(y+dy) [/math] must hold true.

Upon subtracting the first equation from the last one, the result will be [math] 3x^2dx+3y^2dy=6ydx+6xdy [/math]
Arranging for dx and dy, we get
[math] (3x^2-6y)dx = (6x-3y^2)dy [/math] or:
[math] -(3x^2-6y)/(3y^2-6x) = dy/dx = y' [/math]