Question

Find an equation of the tangent line to the given curve at the given point.
y= 2sinx at (pi/6 , 1)

Answer 1

I found that y' = 2cosx

y=mx+b
b= y-mx
= 1 - 2cosx (pi/6)

am i right or wrong?
what do i do next?
i'm so off from the answer

Answer 2

or in slope intercept form

y = f'(a)x -f'(a)a+f(a)

Answer 3

i see a typo in my first getup :)

y - f(a) = f'(a) (x-a)

Answer 4

where is b in here?
or..is this some other formulas？

Answer 5

cuz the only slope formula i know is y=mx+b..

Answer 6

b = -f'(a)a+f(a)

Answer 7

ohh.. i see..it's like the same
so what i did was right..?

Answer 8

b is a constant and can be derived from the point slope form of a line

y - yo = m(x-xo) ; and solve for y

y - yo = mx -mxo

y = mx -mxo + yo

Answer 9

you did good, lets check it tho

Answer 10

but then my answer looks nothing similar to my textbook answer..

Answer 11

y = 2sin(x)

y' = 2cos(x)

y = 2cos(30)x-2cos(30)pi/6+1

Answer 12

cos 30 = sqrt(3)/2 right?

Answer 13

yes

Answer 14

then

y = sqrt(3)x -sqrt(3)pi/6 + 1

Answer 15

how does the book look, cause theres different ways to simplify

Answer 16

6 sqrt (3x) - 6y - pi sqrt (3) +6 =0

Answer 17

ok; we can try to algebrate this to that

Answer 18

y = sqrt(3)x -sqrt(3)pi/6 + 1 ; multiply it all by 6

6y = 6 sqrt(3)x -sqrt(3)pi + 6 ; bringit all to one side

0 = 6 sqrt(3)x -6y -sqrt(3)pi + 6

Answer 19

wow..thanks! very helpful. it will take some time for my brain to process all this stuff :) but thanks! i get what u are saying :)

Answer 20

good :)