• s3a

UNIT NORMAL VECTOR Question: For this question (http://f.imgtmp.com/BIMPQ.jpg), given that I have the unit tangent, I tried dT/dt/|dT/dt| and that does not yield the correct answer for the normal vector. For the x components (to be brief): I get 2sin(-pi/3)/sqrt(13) (for unit tangent - correct answer) and -4cos(-pi/3)/sqrt(32) (for unit normal - wrong answer). but my book and this video (http://www.youtube.com/watch?v=vsKVV5_Hchs) at 1:20 say N = (dT/dt)/|dT/dt| (each vector is a unit vector here). Any help would be greatly appreciated! Thanks in advance!

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  • s3a

UNIT NORMAL VECTOR Question: For this question (http://f.imgtmp.com/BIMPQ.jpg), given that I have the unit tangent, I tried dT/dt/|dT/dt| and that does not yield the correct answer for the normal vector. For the x components (to be brief): I get 2sin(-pi/3)/sqrt(13) (for unit tangent - correct answer) and -4cos(-pi/3)/sqrt(32) (for unit normal - wrong answer). but my book and this video (http://www.youtube.com/watch?v=vsKVV5_Hchs) at 1:20 say N = (dT/dt)/|dT/dt| (each vector is a unit vector here). Any help would be greatly appreciated! Thanks in advance!

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by definition T'(t)/|T'(t)| is the normal vector of r(t)=x(t)i+y(t)j+k(t)k and N(t) = T'(t)/|T'(t)| so to compute normal vector first compute Tangent vector and calculate its unit vector then evaluate the tangent and normal vector at pi/6 NOTE than tangent vector should be in vector form when taking its unit vector to calculate normal vector
  • s3a
that's exactly what i did or think i did. so I guess i should show you more specific work for you to find a mistake?
\[\text{r'(t)=-2sin(2t)i-2cos(2t)j+3k}\] \[\text{T'(t)}=\frac{\text{(-2sin(2t)i-2cos(2t)j+3k)}}{\text{sqrt((-2sin(2t))^2+(-2cos(2t))^2+3^2)}}\] N(t) =(-(4 cos(2 t))/sqrt(13)+(sin(2 t))/sqrt(13))/sqrt(16/13 cos^2(2 t)+1/13 sin^2(2 t)) \[\href{http://www.wolframalpha.com/input/?i=%28-%284+cos%282+t%29%29%2Fsqrt%2813%29%2B%28sin%282+t%29%29%2Fsqrt%2813%29%29%2Fsqrt%2816%2F13+cos%5E2%282+t%29%2B1%2F13+sin%5E2%282+t%29%29&lk=1&a=ClashPrefs_*Math-}{wolfram result}\] just add i and j and evaluate at pi/6

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  • s3a
inside the sin and cos there is a negative not that it matters for the cos part.
  • s3a
(-(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 :)
  • s3a
N = B x T
  • s3a
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
  • s3a
o that's crucial
  • s3a
wait
  • s3a
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
  • s3a
i checked myself to test :)
  • s3a
ya that's correct. k = |v x a|/v^3
  • s3a
but why isn't my way working?
  • s3a
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
  • s3a
is r'' really the normal vector or is it just the acceleration vector which just happens to be parallel and is only the same when it's a unit 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?
  • s3a
sry it froze. 2 secs.
  • s3a
yes.
r'/|r'| = unit tangent then in my view
  • s3a
but 2sin(-pi/3)/sqrt(13) is wrong for the x component.
-pi/3 ?
  • s3a
-2 * pi/6 = -pi/3
compute at t = pi/6
  • s3a
i did. -2*t = -2 * pi/6 = -pi/3
does sin(-2t) = -sin(2t)?
  • s3a
ya. also cos(x) = cos(-x)
r = < cos(-2t), sin(-2t), 3t > r = < cos(2t), -sin(2t), 3t > r' = < -2sin(2t), -2cos(2t), 3 > would amout to the same then, dunno if that improves the results tho
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
  • s3a
wait. i'm looking for N not T.
  • s3a
o i found my mistake. -4cos(pi/3)/sqrt(16) is the answer for the x component.
  • s3a
my mistake was basically i factored the denominator wrong and got 2*16=32 when i should have gotten 16*1
  • s3a
thanks for the help.

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