Homework Help

Find an equation of the tangent line to the curve y = sec x at the at the point `(pi/6,...

user profile pic

lebstar3 | Student, Grade 11 | Honors

Posted March 2, 2012 at 1:04 PM via web

dislike 0 like

Find an equation of the tangent line to the curve y = sec x at the at the point `(pi/6, 2/sqrt3)`

1 Answer | Add Yours

user profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted March 2, 2012 at 1:47 PM (Answer #1)

dislike 1 like

The equation of the tangent to the curve y = sec x at the point `(pi/6, 2/sqrt 3)` has to be determined.

y' = sec x*tan x

The slope of the line tangent to y = sec x at any point is equal to the value of sec x*tan x at that point.

y' at `pi/6` = `sec (pi/6)*tan (pi/6)` = `2/3`

The equation of the tangent is `(y - 2/sqrt 3)/(x - pi/6) = 2/3`

=> `3y - 6/sqrt 3 = 2x - pi/3`

=> `2x - 3y - pi/3 + 2*sqrt 3 = 0`

The required tangent is `2x - 3y - pi/3 + 2*sqrt 3 = 0`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes