Question

Find the points on the graph of
y= sec x, 0<=x<=2pi,
where the tangent is parallel to the line
3y-2x=4

Answer 1

have you found y'?

Answer 2

you need to find the slope of 3y-2x=4
and set it equal to the d/dx sec(x)
and the solve

Answer 3

that is,
\[\frac{d}{dx}\sec(x)=\text{ find slope of } (3y-2x=4)\]

Answer 4

the solve that equation

Answer 5

alright so
y= 2/3x +4/3
y' = 2/3
z= sec x
z' = sec x tan x
sec x tan x = 2/3
so i should find the value of x so that z' = 2/3?

Answer 6

in this case it should be pi/6, correct?

Answer 7

@freckles

Answer 8

yes we are looking for when sec(x)tan(x)=2/3
we are doing that because the slopes need to be the same
parallel lines have same slopes (different y-intercept though )

Answer 9

ah, but do we need to solve for x then?

Answer 10

yes

Answer 11

or should we rewrite it to 3tanxsecx = 2

Answer 12

you can rewrite it

Answer 13

o wait, it says find the point, so pi/6 is the answer then, right?

Answer 14

hmm well tan(pi/6)=1/sqrt(3)
and sec(pi/6)=2

Answer 15

i'm thinking of turning it into an algebraic equation

Answer 16

let tan(x)=u

sec(x)=sqrt(u^2+1)