can somebody please check the following for me. many thanks
Find an Expression for DY/DX (I use D as i will use d for partial derivatives)
When: x^3 + y^3 - 2x^2 y = 0
i) Assign;
z = x^3 + y^3 - 2x^2 y = 0 (eqn 1)
we also have Ez = dz/dx . Ex + dz/dy . Ey ( Where E = Delta/Change)
ii) Divide Both Sides By Ex
which then equates to ;
Ez/Ex = dz/dx + dz/dy . Ey / Ex (eqn 2)
and when Ex approaches zero eqn1 becomes;
DZ/DX = dz/dx + dz/dy . DY/DX (eqn 3)
iii) we can now find the partial derivatives;
dz/dx = 3x^2 - 4xy
dz/dy = 3y^2 - 2x^2
and since z = zero we can now substitue these back into eqn 3 to get;
0 = 3x^2 - 4xy + (3y^2 - 2x^2) . DY/DX
- 3x^2 + 4xy = (3y^2 - 2x^2) . DY/DX
DY/DX = (-3x^2 + 4xy) / (3y^2 - 2x^2)
Think thats not far off, sorry for changing the script and making it confursing.
Thanks again


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