Equation of the tangent to a curve

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SUMMARY

The discussion focuses on finding the equation of the tangent line to the curve defined by the function y = sec(x) at the points x = -9π/4 and x = -7π/4. The derivative of the function, sec(x)tan(x), is used to determine the slope of the tangent line at these points. Participants confirm that substituting the x-values into the derivative provides the slope, and once the slope is known, the y-intercept can be calculated using the point-slope form of the equation.

PREREQUISITES
  • Understanding of trigonometric functions, specifically secant and tangent.
  • Knowledge of calculus, particularly derivatives and their applications.
  • Familiarity with the point-slope form of a linear equation.
  • Ability to perform substitutions in mathematical expressions.
NEXT STEPS
  • Study the properties of the secant and tangent functions in trigonometry.
  • Learn how to apply the derivative to find tangent lines for various functions.
  • Explore the point-slope form of linear equations in detail.
  • Practice finding tangent lines for different trigonometric functions at various points.
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and tangent lines, as well as educators looking for examples of applying calculus concepts to trigonometric functions.

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Homework Statement



y = secx
find the equation of the tangent to the curve at x = -9pi/4 and -7pi/4


Homework Equations



the derivative is secxtanx. do I just subsitute in the values -9pi/4 and then -7pi/4 to get the answer to the slope then what?


The Attempt at a Solution

 
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If you know the slope of the line, you can find the intercept once you know at least one point on that line.
 

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