Question

find the equation of a tangent line y=arcsec x , x=2 i need the y value

Answer 1

\[\frac{d}{dx}arcsec(x)=\frac{1}{x^2 \sqrt{x^2-1}} \implies y'(2)=\frac{1}{4*\sqrt{4-1}}=\frac{\sqrt3}{12}\]

Answer 2

?

Answer 3

That's the slope.

Answer 4

nope u r wrong the slope is

Answer 5

Yeah. The derivative should be:
\[\frac{1}{x \sqrt{x^2-1}} \implies \frac{\sqrt3}{6}\]

Answer 6

So the arcsec(2) is pi/3 so:
\[y-\frac{\pi}{3}=\frac{\sqrt3}{6}(x-2)\]
Solve for y.

Answer 7

I realize...
\[\frac{1}{2*\sqrt{3}}=\frac{\sqrt{3}}{2*\sqrt3*\sqrt3}=\frac{\sqrt3}{2*3}=\frac{\sqrt3}{6}\]

Answer 8

ok wait a second i dont under stand that part arcsec(2) pi/3 ???

Answer 9

So to find the equation of a line at a point (x_0,y_0) with slope m you need:
\[y-y_0=m(x-x_0)\]

Answer 10

So you need your "y_0" with x_0=2. So you say y=arcsec(2) implies sec(y)=2 implies 1/cos(y)=2 or cos(y)=1/2 implies y=pi/3

Answer 11

i know but how to get the y value first without the slope

Answer 12

could u plz draw because i cant understand

Answer 13

You don't need the slope to get the y value. If you have:
\[y=x^2; x=2 \implies y=2^2=4\]

Answer 14

yes i know but i did not the arcsec(2) how to do it (unit circle)

Answer 15

Well I did a "trick" so if you have:
\[y=arcsec(2) \implies \sec(y)=2\]
\[use: \sec(y)=\frac{1}{\cos(y)}\]
This gives:
\[\frac{1}{\cos(y)}=2 \implies \cos(y)=\frac{1}{2} \implies y = \frac{\pi}{3}\]

Answer 16

ohhh i got it now thx u