# Solve the given problem that involves a space curve

• chwala
In summary, the conversation discusses the unit tangent, curvature, principal normal, binormal, and torsion of a space curve. The equations for each of these components are derived and the discussion also touches on plotting the space curve in 3D. There is also mention of the importance of understanding the definitions in order to solve these types of problems.
chwala
Gold Member
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...

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Looks fine to me...

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

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}##

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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?

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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}##

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)}##

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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!

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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.

chwala said:
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...

## 1. How do you define a space curve?

A space curve is a curve that exists in three-dimensional space and can be described by a set of parametric equations. It is a continuous path that is defined by a set of points in space.

## 2. What is the difference between a space curve and a plane curve?

A space curve exists in three-dimensional space, while a plane curve exists in a two-dimensional plane. This means that a space curve can move in all three dimensions, while a plane curve can only move in two dimensions.

## 3. How do you find the length of a space curve?

The length of a space curve can be found using the arc length formula, which involves integrating the magnitude of the derivative of the parametric equations over the given interval. This gives the total distance traveled along the curve.

## 4. What is the curvature of a space curve?

The curvature of a space curve is a measure of how much the curve deviates from being a straight line at a given point. It is calculated using the second derivative of the parametric equations and can be used to determine the shape and behavior of the curve.

## 5. How do you plot a space curve?

To plot a space curve, you can use a parametric plotter or graphing calculator. You will need to input the parametric equations for the curve and specify the range of the parameter. The plot will show the curve in three-dimensional space.

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