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Calculus exercise

  • Thread starter Taturana
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  • #1
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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...
 

Answers and Replies

  • #2
33,627
5,284

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:
Right.
[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]
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.
[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...
 
  • #3
33,627
5,284
Starting from here:
[tex]y - \left ( \frac{\sqrt{3}^3}{3^3} -1 \right ) = \left ( x - \frac{\sqrt{3}}{3} \right )[/tex]

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

There is a mistake in your work. In this equation -
[tex]27y - 3\sqrt{3} + 27 = x - 9\sqrt{3}[/tex]
you forgot to multiply the x on the right side by 27.
 
  • #4
108
0
Starting from here:
[tex]y - \left ( \frac{\sqrt{3}^3}{3^3} -1 \right ) = \left ( x - \frac{\sqrt{3}}{3} \right )[/tex]

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

There is a mistake in your work. In this equation -
[tex]27y - 3\sqrt{3} + 27 = x - 9\sqrt{3}[/tex]
you forgot to multiply the x on the right side by 27.
Ah, yes, arithmetic error again ;( haha

Thank you for the help...
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
955
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!":rofl:
 

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