Given #x^3 - x y^2 + y^3 - 1=0# determine the points where #dy/dx = 0# ?

1 Answer
Feb 13, 2018

See below.

Explanation:

As we know given

#f(x,y)=x^3 - x y^2 + y^3 - 1=0# we have

#(dy)/(dx) = - (f_x)/(f_y) = (y^2-3x^2)/(3y^2-2x y)#

with the companion condition #f(x,y)=0#

Now if we need to compute the coordinates for which #(dy)/(dx) = 0# in the first quadrant, we should solve

#{(x^3 - x y^2 + y^3 - 1=0),(y^2-3x^2=0):}#

obtaining the real solutions

#(-0.517964,0.897141)# and #(0.678877, 1.17585)#

so the solution is for

#(x,y)=(0.678877, 1.17585)#

Attached the plot for #f(x,y)=0# in blue and the point found in red.

enter image source here

NOTE

To solve the system

#{(x^3 - x y^2 + y^3 - 1=0),(y^2-3x^2=0):}#

making #y = lambda x# we have

#{(x^3(1-lambda^2+lambda^3)=1),(lambda^2=3):}#

giving after some considerations

#lambda = sqrt3# and #x = (2/23 + (3 sqrt[3])/23)^(1/3)#

and consequently

#y = sqrt[3] (2/23 + (3 sqrt[3])/23)^(1/3)#