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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!
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!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0by 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
 4 years ago
Best ResponseYou've already chosen the best response.0that'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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\text{r'(t)=2sin(2t)i2cos(2t)j+3k}\] \[\text{T'(t)}=\frac{\text{(2sin(2t)i2cos(2t)j+3k)}}{\text{sqrt((2sin(2t))^2+(2cos(2t))^2+3^2)}}\] 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

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0inside the sin and cos there is a negative not that it matters for the cos part.

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0((4 cos(pi/3))/sqrt(13)) for the x component is wrong.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0given vector equation r r' = tangent vector r'' = normal vector

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0or simply cross r and r'

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0.... thats a stray thought that might be erroneous :)

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0which i think is different from what you said

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0just a bit, I was thinking 2d; not 3d

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0cant see the img you posted to be sure

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0remove the comma at the end of the link and it should work

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0better lol, ), was messing it up

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[k=\frac{r'xr"}{r'^3}\] if memory serves, but id have to dbl chk that

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0i checked myself to test :)

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0ya that's correct. k = v x a/v^3

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0but why isn't my way working?

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0my book and the video confirm my formula.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0r = < cos(2t), sin(2t), 3t > r' = < 2sin(2t), 2cos(2t), 3 > r'' = < 4cos(2t), 4sin(2t), 0 >

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0r' = tangent vector r'' = normal vector

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0is 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?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0we can unit them up afterwards I believe

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0http://tutorial.math.lamar.edu/Classes/CalcIII/TangentNormalVectors.aspx

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0r' = < 2sin(2t), 2cos(2t), 3 > r' = \(\sqrt{4sin^2(2t)+4cos^2(2t)+9}=\sqrt{13}\) right?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0r'/r' = unit tangent then in my view

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0but 2sin(pi/3)/sqrt(13) is wrong for the x component.

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0i did. 2*t = 2 * pi/6 = pi/3

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0does sin(2t) = sin(2t)?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0r = < 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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0given r; that has to be the tangent to it ..... at any rate :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0unit still give us a /sqrt(13) tho

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0x = 2sin(pi/3)/sqrt(13) = 2*1/2*1/sqrt(13) = sqrt(13)/13

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0sqrt(3)/2 can never really do that in my head right lol sqrt(3/13) would be tangent x

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0wait. i'm looking for N not T.

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0o i found my mistake. 4cos(pi/3)/sqrt(16) is the answer for the x component.

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0my mistake was basically i factored the denominator wrong and got 2*16=32 when i should have gotten 16*1
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