# Homework Help: Calculus exercise

1. Aug 17, 2010

### Taturana

1. The problem statement, all variables and given/known data

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).

2. Relevant equations
3. The attempt at a solution

$$y = x^3 - 1; \;\; y' = 3x^2$$

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:

$$1 = 3x^2; \;\; x = \pm \frac{\sqrt{3}}{3}$$

$$y - \left ( \frac{\sqrt{3}^3}{3^3} -1 \right ) = \left ( x - \frac{\sqrt{3}}{3} \right )$$

$$27y - 3\sqrt{3} + 27 = x - 9\sqrt{3}$$

$$x - 27y - 9\sqrt{3} + 3\sqrt{3} + 27 = 0$$

$$x - 27y +6\sqrt{3} + 27 = 0$$

$$3\sqrt{3} x - 3\sqrt{3} y - 3\sqrt{3} -2 = 0; \;\; 3\sqrt{3} x - 3\sqrt{3} y - 3\sqrt{3} +2 = 0$$

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

Thank you for the help...

2. Aug 17, 2010

### Staff: Mentor

Right.
What you have above looks fine, but there are two points at which the slope of the curve y = x3 - 1 is 1. You need to find the normal line at each of these points.

3. Aug 17, 2010

### Staff: Mentor

Starting from here:
$$y - \left ( \frac{\sqrt{3}^3}{3^3} -1 \right ) = \left ( x - \frac{\sqrt{3}}{3} \right )$$

and rewriting as:
$$y - \left ( \frac{1 }{3\sqrt{3}} -1 \right ) = \left ( x - \frac{1}{\sqrt{3}} \right )$$
Just multiply both sides by 3 sqrt(3).

There is a mistake in your work. In this equation -
$$27y - 3\sqrt{3} + 27 = x - 9\sqrt{3}$$
you forgot to multiply the x on the right side by 27.

4. Aug 17, 2010

### Taturana

Ah, yes, arithmetic error again ;( haha

Thank you for the help...

5. Aug 18, 2010

### HallsofIvy

I always say "I am only in Advanced Calculus", I haven't taken arithmetic yet!":rofl: