Question

find the unit tangent and curvature for the space curve x=t-t^3/3, y=t^2 , z=t+t^3/3

Answer 1

The unit tangent vector defines a function that calculates the tangent vector with a magnitude of 1 (a unit vector) of a particular vector function.

The equation for the unit tangent vector is:
\[T=\frac{V}{\left| V \right|}\]

In this case:
\[T=\frac{<t-\frac{t^3}{3},t^2,t+\frac{t^3}{3}>}{\sqrt{({t-\frac{t^3}{3})^2+(t^2)^2+(t+\frac{t^3}{3})^2}}}\]

I'll type another post about the curvature

Answer 2

ohk the whole problem carries 14marks

Answer 3

thet^2/3 is a fraction not exponent

Answer 4

The curvature is defined as the rate at which a curve changes direction, and can be defined multiple ways.

\[k=\frac{dT}{ds}=\frac{1}{\left| V \right|}\frac{|dT|}{|dt|}=\frac{\left| A \times V \right|}{\left| V \right|}\]

Where ;
T=unit tangent vector
s=position vector function
V=velocity vector function
A= acceleration vector function

In most cases you should use the first form.

Answer 5

ohkay... but in class when we looked for the curvature we took the tangent into tne quotient formular

Answer 6

like now m confused y do we have to use the quotient rule when we got the curvaure formular

Answer 7

hold on there, the unit tangent for a vector function \(\vec r(t)\) is\[\vec T={\vec r'(t)\over\|\vec r'(t)\|}\]so you gotta take the derivative first

Answer 8

yep... i agree with turningtest so when we look for the tangent we first look for the derivative and the derivative of the magnitude

Answer 9

the curvature is\[\kappa={\|\vec T'(t)\|\over\|\vec r'(t)\|}\]

Answer 10

so once you have the unit tangent just take the derivative, find the magnitude of that at the point in question, and divide by the magnitude of the derivative of the parametric curve, which we already found

Answer 11

The curvature is just the derivative the unit tangent vector divided by r'(t)=v(t).

And yes, what turingtest said is the same thing as what I said r'(t)=v(t), I also don't know how to put the arrow on top.

Answer 12

yeap... that is clear to me so what is the answer you get

Answer 13

or okay, no point needed, I guess they just want the general formula

Answer 14

@abstracted
\vec{whatever}
=
\[\vec{whatever}\]

Answer 15

but yours doesn't take the derivative is the problem

Answer 16

guys ur making to get confused what is the answer u get on the unit tangent

Answer 17

\[\vec T={\vec V'\over\|\vec V'\|}\]

Answer 18

In the second equation of my first post, I did forget to actually differentiate it, so ignore that one.

But,
\[\vec{T}=\frac{\vec{r'(t)}}{|\vec{r'(t)}|}=\frac{\vec{v(t)}}{|\vec{v(t)}|}\]

Answer 19

ok I just figured out you meant that, sorry for the confusion
you just didn't say that at first, and I didn't read every post :P

Answer 20

Yea, it would've been simpler if I said r'(t) instead, I just copied straight from my old book.

Answer 21

i know the formulas bt can u just help me with the answers

Answer 22

no, I'm afraid you have to participate, we can't just give out answers...
during this time you should have found \(\|\vec r'(t)\|\) and \(\vec r'(t)\)
if not, try that first; you need it in the formula we are writing

Answer 23

the answer for the unit tangent i got [(1-t^2) + 2t +( 1+t^2)] / [root of 2 (t^2 +1)^2

Answer 24

if I could find a pen I could check that :P

Answer 25

careful with the square root in the magnitude

Answer 26

\[\|\vec r(t)\|=\sqrt{(1-t^2)^2+(2t)^2+(1+t)^2}\]

Answer 27

what about the magnitude

Answer 28

\[\|\vec r(t)\|=\sqrt{1-2t^2+t^4+4t^2+1+2t^2+t^4}\]

Answer 29

that is the magnitude

Answer 30

yes i agree but we got same factors

Answer 31

\[\|\vec r(t)\|=\sqrt{2+4t^2+2t^4}=\sqrt{2(1+t^2)^2}=\sqrt2(1+t^2)\]I have to go teach a class I'm afraid, I hope someone else can continue helping you
good luck!

Answer 32

ok