Solve the given problem that involves a space curve

[chwala]

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...

 

[Mark44]

Looks fine to me...

 

[MidgetDwarf]

A greater way to think of this, is the relationship between distance, velocity, and acceleration.
Have you seen the derivation?

 

[chwala]

2. Curvature;

$κ=|\dfrac{dT}{ds}|$

$\dfrac{dT}{ds}=\dfrac{dT}{dt} ⋅\dfrac{dt}{ds}=\dfrac{\sqrt{2}(-2ti+(1-t^2)j)}{(1+t^2)^2}⋅\dfrac{1}{\sqrt{2}(1+t^2)}$

$\dfrac{dT}{ds}=\dfrac{-2ti+(1-t^2)j}{(1+t^2)^3}$=$\dfrac{-2t}{(1+t^2)^3} i +\dfrac{1-t^2}{(1+t^2)^3}j$

$|\dfrac{dT}{ds}|=\sqrt{\dfrac {4t^2}{(1+t^2)^6}+\dfrac {(1-t^2)^2}{(1+t^2)^6}}=\sqrt{\dfrac{t^4+2t^2+1}{(1+t^2)^6}}=\sqrt{\dfrac{(t^2+1)^2}{(1+t^2)^6}}=\dfrac{1+t^2}{(1+t^2)^3}=\dfrac{1}{(1+t^2)^2}$

 

[chwala]

As i attempt this i would be interested in how the given space curve looks like on 3D graph... I will also be looking at that...i need Mathematica to do this?

 

[chwala]

3. Principal Normal;

$N=\dfrac{1}{κ} ⋅\dfrac{dT}{ds}=(1+t^2)^2⋅\dfrac{-2ti+(1-t^2)j}{(1+t^2)^3}=\dfrac{-2ti+(1-t^2)j}{1+t^2}$

 

[chwala]

4. Binormal

$B= T ×N$

i​	j​	k​
$\dfrac{1-t^2}{\sqrt{2} (1+t^2)}$​	$\dfrac{2t}{\sqrt{2} (1+t^2)}$	$\dfrac{1}{\sqrt{2}}$
$\dfrac{-2t}{ 1+t^2}$	$\dfrac{1-t^2}{ 1+t^2}$	0​

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

 

[chwala]

5. Torsion $τ$

$\dfrac{dB}{dt}=\dfrac{\sqrt{2} (t^2+1)[2ti-2j+2tk]-2t\sqrt{2}[(t^2-1)i-2tj+(t^2+1)k]}{2(1+t^2)^2}$

$\dfrac{dB}{ds}=\dfrac{dB}{dt}⋅\dfrac{dt}{ds}=\dfrac{\sqrt{2} (t^2+1)[2ti-2j+2tk]-2t\sqrt{2}[(t^2-1)i-2tj+(t^2+1)k]}{2(1+t^2)^2} ⋅\dfrac{1}{\sqrt{2}(1+t^2)}$

$\dfrac{dB}{ds}=\dfrac{dB}{dt}⋅\dfrac{dt}{ds}=\left[\dfrac{4t} {\sqrt{2}(t^2+1)^2}i+\dfrac{2t^2-2} {\sqrt{2}(t^2+1)^2}j\right]⋅\dfrac{1}{\sqrt{2}(1+t^2)}$

$=\left[\dfrac{2t} {(t^2+1)^3}i+\dfrac{t^2-1} {(t^2+1)^3}j\right]=-τN$

$=\left[\dfrac{2t} {(t^2+1)^3}i+\dfrac{t^3-1} {(t^2+1)^3}j\right]=-τ \left[\dfrac{-2ti+(1-t^2)j}{1+t^2}\right]$

$=\left[\dfrac{2t} {(t^2+1)^3}i+\dfrac{t^2-1} {(t^2+1)^3}j\right]=τ \left[\dfrac{2ti+(t^2-1)j}{1+t^2}\right]$

$\dfrac{τ}{1+t^2}=\dfrac{1}{(1+t^2)^3}$

$τ =\dfrac{1+t^2}{(1+t^2)^3}=\dfrac{1}{(1+t^2)^2}$

Bingo!

 

[MidgetDwarf]

in general, these problems become trivial when one knows the definition. Ie., the derivation of acceleration, starting with distance.

I do not see the point of this thread, since it is apparent you do not need any help.

 

[chwala]

Noted...I agree...i will move to other problems...meanwhile let me know how I can go about plotting the space curve in 3d.

 

[MidgetDwarf]

Noted...I agree...i will move to other problems...meanwhile let me know how I can go about plotting the space curve in 3d.

ummm you cannot demand anything from me...