Question

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

Answer 1

is curvature gonna be binormal as well?

Answer 2

or just a k

Answer 3

we are trying to find k or curvature

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

r = <x,y,z>
=<cos(t), 2sin(t), t>

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

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

Answer 8

okay, okay, that what i got...

Answer 9

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?

Answer 10

so did you cross r prime and r double prime

Answer 11

i hope so :)

Answer 12

hmm, , so the magnitude of prime is..

Answer 13

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

Answer 14

now how do we simplify that

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

thats seems like the best option

Answer 19

|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

Answer 20

oh we take the magnitude of the cross product as well

Answer 21

thats seems impossible to simplify

Answer 22

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

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

Answer 23

what happened to the denominator how did you factor that?

Answer 24

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

Answer 25

ah, yes good man

Answer 26

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

Answer 27

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

Answer 28

r'' is the normal; so just unit it

Answer 29

r''/|r''|

Answer 30

reall????thats is krazy

Answer 31

are you doubly sure?

Answer 32

yep

Answer 33

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

Answer 34

they would have to =0

Answer 35

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

hold on

Answer 36

yeah, man, this is going haywire

Answer 37

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

so far so good

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

so, we are saying thats the curvature

Answer 42

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

Answer 43

as far as I can tell, thats good

Answer 44

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

Answer 45

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

Answer 46

r'/|r'| = unit T

T'/|T'| = unit N

Answer 47

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

Answer 48

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

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

Answer 49

thanks for your help peppy

Answer 50

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

Answer 51

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

Answer 52

I hear the Lorax is in theatres ;)

Answer 53

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

Answer 54

sure thing

Answer 55

you still here?

Answer 56

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

Answer 57

@amistre64

Answer 58

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

Answer 59

http://home.iitk.ac.in/~psraj/mth101/lecture_notes/lecture25.pdf

the principle Normal is defined as T'\|T'|

Answer 60

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