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inside the sin and cos there is a negative not that it matters for the cos part.

(-(4 cos(-pi/3))/sqrt(13)) for the x component is wrong.

given vector equation r
r' = tangent vector
r'' = normal vector

or simply cross r and r'

.... thats a stray thought that might be erroneous :)

N = B x T

which i think is different from what you said

just a bit, I was thinking 2d; not 3d

cant see the img you posted to be sure

o that's crucial

wait

remove the comma at the end of the link and it should work

better lol, ), was messing it up

\[k=\frac{|r'xr"|}{|r'|^3}\]
if memory serves, but id have to dbl chk that

i checked myself to test :)

ya that's correct. k = |v x a|/v^3

but why isn't my way working?

my book and the video confirm my formula.

r = < cos(-2t), sin(-2t), 3t >
r' = < 2sin(-2t), -2cos(2t), 3 >
r'' = < -4cos(-2t), 4sin(2t), 0 >

r' = tangent vector
r'' = normal vector

we can unit them up afterwards I believe

http://tutorial.math.lamar.edu/Classes/CalcIII/TangentNormalVectors.aspx

r' = < 2sin(-2t), -2cos(2t), 3 >
|r'| = \(\sqrt{4sin^2(-2t)+4cos^2(-2t)+9}=\sqrt{13}\)
right?

sry it froze. 2 secs.

yes.

r'/|r'| = unit tangent then in my view

but 2sin(-pi/3)/sqrt(13) is wrong for the x component.

-pi/3 ?

-2 * pi/6 = -pi/3

compute at t = pi/6

i did. -2*t = -2 * pi/6 = -pi/3

does sin(-2t) = -sin(2t)?

ya. also cos(x) = cos(-x)

given r; that has to be the tangent to it ..... at any rate :)

unit still give us a /sqrt(13) tho

x = -2sin(pi/3)/sqrt(13) = -2*1/2*1/sqrt(13) = sqrt(13)/13

sqrt(3)/2 can never really do that in my head right lol
-sqrt(3/13) would be tangent x

wait. i'm looking for N not T.

o i found my mistake. -4cos(pi/3)/sqrt(16) is the answer for the x component.

thanks for the help.