Question

Find an equation of the tangent line to given curve at the given point:

y=xsecx at x=0 and x=-pi/6;

I'm not sure hot find the equation of the tangent line when there are two x's given..

Answer 1

well I think you'll have 2 separate equations...

1 at x = 0

1 wat x = -pi/6

Answer 2

If that's true, then that means I need to use the x, y, and m for its own relative eqn? I got \[(\pi +\sqrt{3})/9\] as the slope for when x = -pi/6, but is that really alright to use that for the point slope form?

Answer 3

have you done calculus...?

Answer 4

yea, but im using 2 variables for m:
y-y1 = m(x-x1)
Is that really fine to use (pi/9 + 2/sqrt(3)) for m?

Answer 5

well a tangent has 1 point of contact... so if you are using 2 points you have a secant...


I would have thought

1. find the 1st derivative

2. substitute the point to get the slope at that point...

3. find the equation of the line...

the points are (0,0) and \[(\frac{-\pi}{6},\frac{\pi}{12})\]

Answer 6

the derivative that I got is secx(xtanx+1), but I try plugging in the x of -pi/6, and thats why I'm confused.

Answer 7

oops the 2nd point should be

\[(\frac{-\pi}{6}, \frac{-\pi \sqrt{3}}{12})\]

Answer 8

so use 1/cos(-pi/6)....

Answer 9

as the slope?

For instance:
y-pi/6 = sec(pi/6)(x-pi/6)?

Answer 10

Also, wouldn't the points for x = -pi/6 be \[(\sqrt{3}/2 , -1/2)?\]