Question

Find the equation of the tangent line to the curve y=sec(x)-2cos(x) at the point (pi/3,1).

Answer 1

Where are you stuck?

Answer 2

i got the derv part which is sec(x)tan(x)+2sin(x)

Answer 3

how do i put in the points

Answer 4

just plug pi/3 in

Answer 5

remember y' is slope of the line, and at pi/3, y' = something, not y. so that, ignore 1 from the point at this time.

Answer 6

is this right y=sec(pi/3)tan(pi/3)+2sin(pi/3)

Answer 7

y' , not y , Again :)

Answer 8

and yes, that is. y' =??

Answer 9

that where i dont get to solve it

Answer 10

\[y' = sec (\pi/3)tan(\pi/3)+2sin(\pi/3)\]
sec (pi/3) = 1/cos (pi/3) =1/1/2 =2
tan (pi/3) = sqrt 3
sin pi/3 = sqrt3/2
just plug number in

Answer 11

I messed up tan (pi/3) , correct it

Answer 12

\[y'= 2(this ~is~ sec (pi/3) * \dfrac{\sqrt3}{3}(this~ is~ tan(pi/3) +2\dfrac{1}{2}(this~ is~ sin (pi/3))\]

Answer 13

y'= \(2\sqrt 3/3+1\)
got it??

Answer 14

oh okay thx

Answer 15

its not right