- #1

- 108

- 0

## Homework Statement

Find the equation of a tangent line to the curve y = x³ - 1, that is perpendicular to y = -x (I mean the tangent line should be perpendicular to y=-x, sorry for my bad english).

## Homework Equations

## The Attempt at a Solution

[tex]y = x^3 - 1; \;\; y' = 3x^2[/tex]

If the tangent line must be perpendicular to y = -x then its slope must be +1, right? So we need to know at what value of x the slope is +1:

[tex]1 = 3x^2; \;\; x = \pm \frac{\sqrt{3}}{3}[/tex]

[tex]y - \left ( \frac{\sqrt{3}^3}{3^3} -1 \right ) = \left ( x - \frac{\sqrt{3}}{3} \right )[/tex]

[tex]27y - 3\sqrt{3} + 27 = x - 9\sqrt{3}[/tex]

[tex]x - 27y - 9\sqrt{3} + 3\sqrt{3} + 27 = 0[/tex]

[tex]x - 27y +6\sqrt{3} + 27 = 0[/tex]

The correct answer is

[tex]3\sqrt{3} x - 3\sqrt{3} y - 3\sqrt{3} -2 = 0; \;\; 3\sqrt{3} x - 3\sqrt{3} y - 3\sqrt{3} +2 = 0[/tex]

I hope this was not an error in arithmetics...

Thank you for the help...