Question

Find an equation of the tangent line to the curve y=arccos(x/2) at the point (1,4pi)

Answer 1

Any ideas?

Answer 2

well, the point (1,4pi) is not on the line

Answer 3

Oh sorry, it was y=3arccos(x/2)

Answer 4

the slope of the line at any given point is the derivative of the function

Answer 5

is the equation spose to be in point slope or slope intercept?

Answer 6

either way, the point is not on the line,
the derivative of y=3arccos(x/2) is y'=-3/(sqrt(4-x^2))
subbing x=1 we get the slope we get the slope of -sqrt(3)

Answer 7

so it narrows down to 2 answers

Answer 8

cos(60)=1/2 so (1,pi/3) is on it i believe

Answer 9

-sqrt(3)=(y-1,x-4pi)

Answer 10

oops, i mean -sqrt(3)=(y-1/x-4pi)

Answer 11

oops, i mean -sqrt(3)=(y-1)/(x-4pi)

Answer 12

So the answer would be?

Answer 13

funny thing is none of them match, so i must have made a mistake somewhere

Answer 14

Got any ideas amistre?

Answer 15

i got it

Answer 16

-sqrt(3)=(y-4pi)/(x-1

Answer 17

mixed up the coordinates

Answer 18

answer is the 3rd one

Answer 19

-sqrt(3)=(y-4pi)/(x-1) isolate y and you'll get the answer

Answer 20

Thanks man, I never would have got that one, I guess I need more studying :P

Answer 21

no prob, by math skills are getting rusty