## s3a 4 years ago 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!

1. anonymous

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

2. 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?

3. anonymous

$\text{r'(t)=-2sin(2t)i-2cos(2t)j+3k}$ $\text{T'(t)}=\frac{\text{(-2sin(2t)i-2cos(2t)j+3k)}}{\text{sqrt((-2sin(2t))^2+(-2cos(2t))^2+3^2)}}$ 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

4. s3a

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

5. s3a

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

6. amistre64

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

7. amistre64

or simply cross r and r'

8. amistre64

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

9. s3a

N = B x T

10. s3a

which i think is different from what you said

11. amistre64

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

12. amistre64

cant see the img you posted to be sure

13. s3a

o that's crucial

14. s3a

wait

15. s3a

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

16. amistre64

better lol, ), was messing it up

17. amistre64

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

18. s3a

i checked myself to test :)

19. s3a

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

20. s3a

but why isn't my way working?

21. s3a

my book and the video confirm my formula.

22. amistre64

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

23. amistre64

r' = tangent vector r'' = normal vector

24. 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?

25. amistre64

we can unit them up afterwards I believe

26. amistre64
27. amistre64

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

28. s3a

sry it froze. 2 secs.

29. s3a

yes.

30. amistre64

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

31. s3a

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

32. amistre64

-pi/3 ?

33. s3a

-2 * pi/6 = -pi/3

34. amistre64

compute at t = pi/6

35. s3a

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

36. amistre64

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

37. s3a

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

38. amistre64

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

39. amistre64

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

40. amistre64

unit still give us a /sqrt(13) tho

41. amistre64

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

42. amistre64

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

43. s3a

wait. i'm looking for N not T.

44. s3a

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

45. s3a

my mistake was basically i factored the denominator wrong and got 2*16=32 when i should have gotten 16*1

46. s3a

thanks for the help.