- #1

chwala

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- Homework Statement
- For the space curve; ##\left[x=t-\dfrac{t^3}{3}, y=t^2,

z=t+\dfrac{t^3}{3} \right]##

Find;

1. Unit tangent

2. Curvature

3. Principal Normal

4. Binormal

5. Torsion

- Relevant Equations
- vector differentiation

Relatively new area to me; will solve one -at- time as i enjoy the weekend with coffee.

1. Unit tangent

##r=xi+yj+zk##

##r=(t-\dfrac{t^3}{3})i+t^2j+(t+\dfrac{t^3}{3})k##

##T=\dfrac{dr}{dt} ⋅\dfrac{dt}{ds}##

##\dfrac{dr}{dt}=(1-t^2)i+2tj+(1+t^2)k##

##\dfrac{ds}{dt}=\sqrt{(1-t^2)^2+4t^2+(1+t^2)^2}##

##=\sqrt{1-2t^2+t^4+4t^2+1+2t^2+t^4}##

##=\sqrt{2t^4+4t^2+2}##

## =\sqrt{2}(1+t^2)##

##T=\dfrac{(1-t^2)i+2tj+(1+t^2)k}{\sqrt{2} (1+t^2)}##

of course you may chip in with your insight/cheers. ..will look at rest later...

1. Unit tangent

##r=xi+yj+zk##

##r=(t-\dfrac{t^3}{3})i+t^2j+(t+\dfrac{t^3}{3})k##

##T=\dfrac{dr}{dt} ⋅\dfrac{dt}{ds}##

##\dfrac{dr}{dt}=(1-t^2)i+2tj+(1+t^2)k##

##\dfrac{ds}{dt}=\sqrt{(1-t^2)^2+4t^2+(1+t^2)^2}##

##=\sqrt{1-2t^2+t^4+4t^2+1+2t^2+t^4}##

##=\sqrt{2t^4+4t^2+2}##

## =\sqrt{2}(1+t^2)##

##T=\dfrac{(1-t^2)i+2tj+(1+t^2)k}{\sqrt{2} (1+t^2)}##

of course you may chip in with your insight/cheers. ..will look at rest later...

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