Solve the problem involving space curve

  • #1
chwala
Gold Member
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Homework Statement
see attached.
Relevant Equations
Vector differentiation
Refreshing... i'll start with part (a).

1709640703990.png


Just sharing in case there is more insight...

In my working i have,

##T = \dfrac{dr}{ds}=\dfrac{dx}{ds}i + \dfrac{dy}{ds}j + \dfrac{dz}{ds}k##

and

##x=\tan^{-1} s, y = \dfrac{\sqrt2}{2} \ln (s^2+1), z=\tan^{-1} s##

##\dfrac{ds}{dx} = \sec^2 x = 1 +\tan^2x ##

##\dfrac{dx}{ds}= \dfrac{1}{1+s^2}##.

similarly,

##\dfrac{dy}{ds}=\dfrac{\sqrt 2}{2}⋅ \dfrac{1}{s^2+1}⋅2s = \dfrac{\sqrt 2}{s^2+1}s##

...
thus,

##T=\dfrac{1}{1+s^2} i + \dfrac{\sqrt 2}{s^2+1}sj + \left(1-\dfrac{1}{1+s^2}\right)##
##T=\dfrac{1}{1+s^2} i + \dfrac{\sqrt 2}{s^2+1}sj + \dfrac{s^2k}{1+s^2}##

For (d), curvature

My lines are

##\dfrac{dT}{ds} = \dfrac{-2s}{(1+s^2)^2} i + \dfrac{\sqrt 2(1-s^2)}{(1+s^2)^2}j +\dfrac{2s}{(1+s^2)^2}k##

##k=\dfrac{|dT|}{|ds|}= \dfrac{4s^2+2(1-s^2)^2 +4s^2}{(1+s^2)^4}##

##k=\sqrt{\dfrac{2s^4+4s^2+2}{(1+s^2)^4}}=\sqrt{\dfrac{2(s^2+1)^2}{(1+s^2)^4}}=\dfrac{\sqrt2⋅ (s^2+1)}{(1+s^2)^2}=\dfrac{\sqrt2}{1+s^2}##

...involves some bit of working...cheers ...rest of questions can be solved similarly as long as one knows the formula and how to differentiate...any insight is welcome. bye.
 
Last edited:
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  • #3
jedishrfu said:
You could continue and solve N and B and then show that T,N and B are all perpendicular.

https://en.wikipedia.org/wiki/Frenet–Serret_formulas
Yes I'll do that later...

done already for ##N## and ##B##... Not difficult ...had to use cross product... let me post my working later. Cheers man!
 
Last edited:

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