anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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amistre64
  • amistre64
is curvature gonna be binormal as well?
amistre64
  • amistre64
or just a k
anonymous
  • anonymous
we are trying to find k or curvature

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amistre64
  • amistre64
\[k=\frac{|r'xr''|}{|r'|^3}\] if i recall it correctly
anonymous
  • anonymous
or 1/magnitude of V *magniutde of Tangent unit vector
anonymous
  • anonymous
but your formula looks cool, so lets try taht one please
amistre64
  • amistre64
r = = r' = <-sin(t), 2cos(t),1> r'' = <-cos(t), -2sin(t),0>
anonymous
  • anonymous
okay, okay, that what i got...
amistre64
  • 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
  • anonymous
so did you cross r prime and r double prime
amistre64
  • amistre64
i hope so :)
anonymous
  • anonymous
hmm, , so the magnitude of prime is..
anonymous
  • anonymous
sqrt(sin^2t+4cos^2t)
anonymous
  • anonymous
now how do we simplify that
amistre64
  • amistre64
|r'| = |<-sin(t), 2cos(t),1>| = sqrt(sin^2(t)+4cos^2(t)+1)
anonymous
  • anonymous
oh i missed that 1, but anyway how do we simplify that
amistre64
  • 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
  • anonymous
thats seems like the best option
amistre64
  • 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
  • anonymous
oh we take the magnitude of the cross product as well
anonymous
  • anonymous
thats seems impossible to simplify
amistre64
  • 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
  • anonymous
what happened to the denominator how did you factor that?
amistre64
  • amistre64
\[(\sqrt{a})^3=\sqrt{a}\sqrt{a}\sqrt{a}=a\sqrt{a}\]
anonymous
  • anonymous
ah, yes good man
amistre64
  • amistre64
now the radical covers top to bottom and we might be able to play inside it
anonymous
  • anonymous
do you have a fancy formula for computing the principal unit normal?
amistre64
  • amistre64
r'' is the normal; so just unit it
amistre64
  • amistre64
r''/|r''|
anonymous
  • anonymous
reall????thats is krazy
anonymous
  • anonymous
are you doubly sure?
amistre64
  • amistre64
yep
amistre64
  • amistre64
if you dot r' and r''; are they perped?
anonymous
  • anonymous
they would have to =0
amistre64
  • 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
  • anonymous
yeah, man, this is going haywire
amistre64
  • amistre64
http://tutorial.math.lamar.edu/Classes/CalcIII/Curvature.aspx so far so good
amistre64
  • amistre64
lets finish up k, then well worry about unit normals ...
anonymous
  • anonymous
okay, yeah lets finish up K, at least i need that cause then i can try and find the pricipal
amistre64
  • 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
  • anonymous
so, we are saying thats the curvature
amistre64
  • amistre64
yes, but im checking the math with the wolf first :)
amistre64
  • amistre64
as far as I can tell, thats good
anonymous
  • anonymous
alright,so then for the principal unit......
amistre64
  • amistre64
http://tutorial.math.lamar.edu/Classes/CalcIII/TangentNormalVectors.aspx
amistre64
  • amistre64
r'/|r'| = unit T T'/|T'| = unit N
amistre64
  • amistre64
since T' = r'' /////
amistre64
  • 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
  • anonymous
thanks for your help peppy
amistre64
  • amistre64
i dont spose thats a common little nickname all the young kids are throwing about these days :)
anonymous
  • 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
  • amistre64
I hear the Lorax is in theatres ;)
amistre64
  • amistre64
make sure to go thru the math and adjust for errors that pervade
anonymous
  • anonymous
sure thing
anonymous
  • anonymous
you still here?
anonymous
  • anonymous
witch one is the principla unit normal, i think you wrote down two of them?
anonymous
  • anonymous
@amistre64
amistre64
  • amistre64
id have to read up on what defines a principle; im sure its the one that heads into the curve ....
amistre64
  • amistre64
http://home.iitk.ac.in/~psraj/mth101/lecture_notes/lecture25.pdf the principle Normal is defined as T'\|T'|
amistre64
  • amistre64
except for a usual / for division and not that \ that gets in the way :)

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