Could someone please help with this:
Find the coordinates of the point in the first quadrant at which the tangent line to the curve (x)^3-xy+(y)^3 =0 is parallel to the x axis.

Question

Could someone please help with this:
Find the coordinates of the point in the first quadrant at which the tangent line to the curve (x)^3-xy+(y)^3 =0 is parallel to the x axis.

turingtest · Answer

differentiate implicitely:$$3x^2-y-xy'+3y^2y'=0$$solve for y'$$y'(3y^2-x)=y-3x^2$$$$y'={y-3x^2\over3y^2-x}$$when the tangent is parallel to the x-axis we have y'=0, so we must solve$$y'={y-3x^2\over3y^2-x}=0\implies y=3x^2$$to find the actual value of x we plug this expression for y into the original equation$$x^3-3x^3+27x^6=0$$$$x^3(27x^3-2)=0\implies x=\{0,{\sqrt[3]2\over3}\}$$plugging this into the formula for y above gives the points$$(0,0)\text{ and }({\sqrt[3]2\over3},{\sqrt[3]4\over3})$$which is where our tangent will be parallel to the x-axis.