Finding Equations of Normal Lines Parallel to a Given Line

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SUMMARY

The discussion focuses on finding the equations of normal lines to the curve defined by the equation y = x^3 - 4x that are parallel to the line represented by x + 8y - 8 = 0. The slope of the tangent line is determined by differentiating the curve, yielding dy/dx = 3x^2 - 4. The correct slope of the given line is -1/8, not -8, which leads to the correct evaluation of x for the normal line's slope. The error in identifying the slope of the given line is identified as the source of confusion in the calculations.

PREREQUISITES
  • Understanding of calculus, specifically differentiation.
  • Familiarity with the concept of normal lines in relation to curves.
  • Knowledge of linear equations and slope-intercept form.
  • Ability to manipulate algebraic expressions to solve for variables.
NEXT STEPS
  • Review the principles of differentiation and its applications in finding slopes.
  • Study the relationship between tangent and normal lines in calculus.
  • Learn how to convert linear equations into slope-intercept form for easier analysis.
  • Practice solving problems involving normal lines to various curves.
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and their applications in geometry, as well as educators looking for examples of common mistakes in slope calculations.

SherlockOhms
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Homework Statement


Find the equation of each of the normal lines to the curve y = x^3 - 4x that is parallel to the line x + 8y - 8 = 0.


Homework Equations


Differentiation, y - y1 = m(x - x1)


The Attempt at a Solution


Well, clearly I start by differentiating y = x^3 - 4x to gey dy/dx = 3x^2 - 4. Then, this is the slope of the tangent to the graph y. This slope is equal to the slope of the line x + 8y - 8 = 0. Which is -8. So, I then evaluate for x and sub back into dy/dx, giving me the slope of the tangent. The slope of the normal line is the negative reciprocal of this. If you solve for x using
3x^2 - 4 = -8, you get an imaginary value though. Where have I gone wrong?
 
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You have the gradient of the given line wrong.
 
CAF123 said:
You have the gradient of the given line wrong.
Cool. See where I went wrong now. Stupid slip.
 

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