Solve Calculus Exercise: Find Perpendicular Tangent Line to y=x^3-1

[Taturana]

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

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

The correct answer is
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...

 

[Mark44]

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

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:

Right.

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 )

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.

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

The correct answer is
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...

 

[Mark44]

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.

 

[Taturana]

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.

Ah, yes, arithmetic error again ;( haha

Thank you for the help...

 

[HallsofIvy]

Ah, yes, arithmetic error again ;( haha

Thank you for the help...

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