Find an equation of the tangent line to the curve at the given point. y = sec(x), (π/3, 2)?

Solution:

Consider the point (π/3, 2) as A

The line passes through A (π/3, 2)

The slope can be found by performing y’ at x = π/3

Now by differentiating y

y’ = (sec x)’ = tan x sec x

At x = π/3, the slope of tangent line is

\( y'_{x=\frac{\pi }{3}}=tan(\frac{\pi}{3})sec(\frac{\pi}{3}) \)= √3 × 2 = 2 √3

Equation of a tangent line passing through A (π/3, 2) and slope 2√3 is

\( y - y_{A}=y_{x_{a}}(x-x_{A}) \)

Substituting the values

y - 2 = 2 √3 (x - π/3)

By further calculation

y = 2√3x - 2π√3/3 + 2

Therefore, the equation of the tangent line is y = 2√3x - 2π√3/3 + 2.

Find an equation of the tangent line to the curve at the given point. y = sec(x), (π/3, 2)?

Summary:

The equation of the tangent line to the curve at the given point. y = sec(x), (π/3, 2) is y = 2√3x - 2π√3/3 + 2.