Question

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!

Answer 1

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

Answer 2

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?

Answer 3

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

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

Answer 4

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

Answer 5

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

Answer 6

given vector equation r

r' = tangent vector

r'' = normal vector

Answer 7

or simply cross r and r'

Answer 8

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

Answer 9

N = B x T

Answer 10

which i think is different from what you said

Answer 11

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

Answer 12

cant see the img you posted to be sure

Answer 13

o that's crucial

Answer 14

wait

Answer 15

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

Answer 16

better lol, ), was messing it up

Answer 17

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

Answer 18

i checked myself to test :)

Answer 19

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

Answer 20

but why isn't my way working?

Answer 21

my book and the video confirm my formula.

Answer 22

r = < cos(-2t), sin(-2t), 3t >

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

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

Answer 23

r' = tangent vector

r'' = normal vector

Answer 24

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?

Answer 25

we can unit them up afterwards I believe

Answer 26

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

Answer 27

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

|r'| = \(\sqrt{4sin^2(-2t)+4cos^2(-2t)+9}=\sqrt{13}\)

right?

Answer 28

sry it froze. 2 secs.

Answer 29

yes.

Answer 30

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

Answer 31

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

Answer 32

-pi/3 ?

Answer 33

-2 * pi/6 = -pi/3

Answer 34

compute at t = pi/6

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

unit still give us a /sqrt(13) tho

Answer 41

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

Answer 42

sqrt(3)/2 can never really do that in my head right lol

-sqrt(3/13) would be tangent x

Answer 43

wait. i'm looking for N not T.

Answer 44

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

Answer 45

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

Answer 46

thanks for the help.