- #1

chwala

Gold Member

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- Homework Statement
- see attached.

- Relevant Equations
- Vector differentiation

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.

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.

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