Solve the problem involving space curve

Refreshing... i'll start with part (a).



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.

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

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!