# Thread: Equation For The Line Tangent To The Graph

1. ## Equation For The Line Tangent To The Graph

Hey everyone,

I have this question,
“Find an equation for the line tangent to the graph y = sec x at (pi/3 , 2)”

So far I have that,
f(x) = sec x
f '(x) = sec x . tan x

I then subbed in x = (pi/3) into f '(x) to get the gradient which I found was 0, is this right?

I wanted to make I was approaching this question correctly and if I have that just means I sub m into the equation,
y - y1 = m(x - x1)
and finish it off.

Thanks!

2. Yep sounds right if all your calculations are right

3. Originally Posted by 11rdc11
Yep sounds right if all you calculations are right
Awesome, thanks ^^

5. Originally Posted by 11rdc11
Really? What did I do wrong.

6. $f'\left(\frac{\pi}{3}\right) = \sec \left(\frac{\pi}{3}\right) \cdot \tan \left(\frac{\pi}{3}\right) = 2 \cdot \sqrt{3}$

7. $sec\bigg(\frac{\pi}{3}\bigg) = 2$

$tan\bigg(\frac{\pi}{3}\bigg) = \sqrt{3}$

8. Originally Posted by o_O
$f'\left(\frac{\pi}{3}\right) = \sec \left(\frac{\pi}{3}\right) \cdot \tan \left(\frac{\pi}{3}\right) = 2 \cdot \sqrt{3}$
Originally Posted by 11rdc11
$sec\bigg(\frac{\pi}{3}\bigg) = 2$

$tan\bigg(\frac{\pi}{3}\bigg) = \sqrt{3}$

Ohhh I understand now for some reason I had that Tan pi/3 = 0

Thanks for the help!