Can someone assist me with this curvature problem:
Compute the curvature and the principal unit normal of the ellpitical hellix described by the equations:
x=cost, y=2sint, and z=t

- anonymous

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- amistre64

is curvature gonna be binormal as well?

- amistre64

or just a k

- anonymous

we are trying to find k or curvature

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## More answers

- amistre64

\[k=\frac{|r'xr''|}{|r'|^3}\]
if i recall it correctly

- anonymous

or 1/magnitude of V *magniutde of Tangent unit vector

- anonymous

but your formula looks cool, so lets try taht one please

- amistre64

r =
=
r' = <-sin(t), 2cos(t),1>
r'' = <-cos(t), -2sin(t),0>

- anonymous

okay, okay, that what i got...

- amistre64

r' x r''
x -sin(t) -cos(t) x=2sin(t) = 2sin(t)
y 2cos(t) -2sin(t) -y=cos(t) = -cos(t)
z 1 0 z=2sin^2(t) + 2cos^2(t) = 2
r' x r'' = <2sin(t), -cos(t),2>
whats our magnitude? of that?

- anonymous

so did you cross r prime and r double prime

- amistre64

i hope so :)

- anonymous

hmm, , so the magnitude of prime is..

- anonymous

sqrt(sin^2t+4cos^2t)

- anonymous

now how do we simplify that

- amistre64

|r'| = |<-sin(t), 2cos(t),1>| = sqrt(sin^2(t)+4cos^2(t)+1)

- anonymous

oh i missed that 1, but anyway how do we simplify that

- amistre64

well, sin^2 = 1-cos^2
sqrt(sin^2(t)+4cos^2(t)+1)
sqrt(1-cos^2(t)+4cos^2(t)+1)
sqrt(3cos^2(t)+2) is one option

- anonymous

thats seems like the best option

- amistre64

|r' x r''| = sqrt(4sin^2+cos^2+4)
sooo, if my memory serves me .... IF .....
\[\frac{\sqrt{4sin^2(t)+cos^(t)+4}}{(\sqrt{3cos^2(t)+2})^3}\]
is something to play with

- anonymous

oh we take the magnitude of the cross product as well

- anonymous

thats seems impossible to simplify

- amistre64

\[\frac{\sqrt{4-4cos^2(t)+cos^2(t)+4}}{(\sqrt{3cos^2(t)+2})^3}\]
\[\frac{\sqrt{-3cos^2(t)+8}}{(3cos^2(t)+2) \sqrt{3cos^2(t)+2}}\]

- anonymous

what happened to the denominator how did you factor that?

- amistre64

\[(\sqrt{a})^3=\sqrt{a}\sqrt{a}\sqrt{a}=a\sqrt{a}\]

- anonymous

ah, yes good man

- amistre64

now the radical covers top to bottom and we might be able to play inside it

- anonymous

do you have a fancy formula for computing the principal unit normal?

- amistre64

r'' is the normal; so just unit it

- amistre64

r''/|r''|

- anonymous

reall????thats is krazy

- anonymous

are you doubly sure?

- amistre64

yep

- amistre64

if you dot r' and r''; are they perped?

- anonymous

they would have to =0

- amistre64

r' = <-sin(t), 2cos(t),1>
r'' = <-cos(t), -2sin(t),0>
----------------------
sincos -4sincos +0 great, now I gotta wonder lol
hold on

- anonymous

yeah, man, this is going haywire

- amistre64

http://tutorial.math.lamar.edu/Classes/CalcIII/Curvature.aspx
so far so good

- amistre64

lets finish up k, then well worry about unit normals ...

- anonymous

okay, yeah lets finish up K, at least i need that cause then i can try and find the pricipal

- amistre64

http://www.wolframalpha.com/input/?i=sqrt%284sin%5E2%28t%29%2Bcos%28t%29%2B4%29%2F%28sqrt%283cos%5E2%28t%29%2B2%29%5E3+
unless we messed up a math someplse this looks to be it

- anonymous

so, we are saying thats the curvature

- amistre64

yes, but im checking the math with the wolf first :)

- amistre64

as far as I can tell, thats good

- anonymous

alright,so then for the principal unit......

- amistre64

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

- amistre64

r'/|r'| = unit T
T'/|T'| = unit N

- amistre64

since T' = r'' /////

- amistre64

\[N=\frac{r''}{|r''|}=\frac{1}{\sqrt{cos^2+4sin^2}}<-cos(t), -2sin(t),0> \]
\[N=\frac{r''}{|r''|}=\frac{1}{\sqrt{4-3cos^2}}<-cos(t), -2sin(t),0> \]

- anonymous

thanks for your help peppy

- amistre64

i dont spose thats a common little nickname all the young kids are throwing about these days :)

- anonymous

no but seriously thanks for helping me, my teacher will be so happy,now i have to go write my bookreport on Dr. Seuss

- amistre64

I hear the Lorax is in theatres ;)

- amistre64

make sure to go thru the math and adjust for errors that pervade

- anonymous

sure thing

- anonymous

you still here?

- anonymous

witch one is the principla unit normal, i think you wrote down two of them?

- anonymous

@amistre64

- amistre64

id have to read up on what defines a principle; im sure its the one that heads into the curve ....

- amistre64

http://home.iitk.ac.in/~psraj/mth101/lecture_notes/lecture25.pdf
the principle Normal is defined as T'\|T'|

- amistre64

except for a usual / for division and not that \ that gets in the way :)

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