Find an equation of the tangent line to the curve at the given point.
y = sec x, (π/6, 2

3
/3)

Question

Find an equation of the tangent line to the curve at the given point.
y = sec x,    (π/6, 2

3
/3)

hartnn · Answer

y= sec x
can you first find dy/dx ?

anonymous · Answer

y=secx tanx

hartnn · Answer

you mean dy/dx = sec x tan x, right ?

now just plug in values of your point
x= pi/6

anonymous · Answer

I got 2x+2rt3/3-pi/3 but it is incorrect apparently

hartnn · Answer

dy/dx = slope = sec (pi/6) tan (pi/6) = 2/3
did you get this first ?

anonymous · Answer

Shouldn't it be 2?

hartnn · Answer

sec pi/6 = 2/sqrt 3 
tan pi/6 = 1/sqrt 3

so, its 2/ sqrt3  * (1/sqrt 3) = 2/3

hartnn · Answer

still confused ?

anonymous · Answer

Yes because I tried that and it still did not work

hartnn · Answer

thats just slope, we still have to find the equation of line...

hartnn · Answer

we have slope= m= 2/3 now 
and a point, (pi/6, 2 3/3) = (x1,y1)

use the slope point form
$y- y_1 = m (x-x_1)$

anonymous · Answer

I know and I plugged that in and tried that answer which did not work

hartnn · Answer

whats your y-co-ordinate in the question ?

anonymous · Answer

2rt3/3

hartnn · Answer

ok, 

so its $2\sqrt 3/3$

y-  $2\sqrt 3/3 = (2/3) (x -\pi/6)$ 

$3y -2\sqrt 3 =  2x - \pi/3$

hartnn · Answer

so,
$3y -2x = 2\sqrt3 -\pi/3$
try this.

anonymous · Answer

It has to be y=mx+b

hartnn · Answer

okk,
$3y = 2x +2\sqrt 3 -\pi/3 \ y= (2/3
)x + (2/\sqrt 3 -\pi/9)$

anonymous · Answer

Just got it. Thank you.

hartnn · Answer

welcome ^_^