How do you find the equation of the tangent line to the curve y = sec(x) at the point (p/3, 2)?

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To find the equation of the tangent line to the curve y = sec(x) at the point (p/3, 2), we need to follow several steps:

  1. Find the derivative of the function: The derivative of y = sec(x) can be calculated using the chain rule. The derivative is given by:

    2. dy/dx = sec(x) * tan(x)

  2. Evaluate the derivative at the given point: We need to substitute x = p/3 into the derivative to find the slope of the tangent line at that point.

    3. dy/dx at x = p/3 = sec(p/3) * tan(p/3)

  3. Calculate the values: Using known values, we have:
    • sec(p/3) = 2
    • tan(p/3) = sqrt(3)

    Therefore, the slope at the point (p/3, 2) is:

    m = 2 * sqrt(3)

  4. Use the point-slope form of the line: The point-slope form of a line is given by:

    5. y - y1 = m(x - x1)

    Where (x1, y1) is the point on the curve. Here, (x1, y1) = (p/3, 2).

    Substituting these values, we have:

    y - 2 = 2 * sqrt(3) (x - p/3)

  5. Simplify the equation: Rearranging this will give us the equation of the tangent line:

    6. y = 2 * sqrt(3) * x - (2 * sqrt(3) * p/3) + 2

Hence, the equation of the tangent line to the curve y = sec(x) at the point (p/3, 2) is:

y = 2 * sqrt(3) * x - (2 * sqrt(3) * p/3) + 2

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