My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree2Thanks
  • 1 Post By skipjack
  • 1 Post By zylo
Reply
 
LinkBack Thread Tools Display Modes
September 14th, 2018, 08:59 PM   #1
Member
 
Joined: Apr 2017
From: New York

Posts: 77
Thanks: 6

UTV and UNV

Hello everyone,
My question is about Unit tangent vector and Unit normal vectors.

I attached the question and part of my solution.

I know the formulas of both UTV and UNV.

What I did on page 1 is to calculate UTV given the vector r(t) and to continue I know that I have to find the derivative of T(t) and magnitude of T(t) and I know that UNV= T'(t) / ||T'(t)||.

My main question is how will I apply the given point (1, 1, 0)? Which component should be plugged in or all? Or what is the next step after finding general UNV equation?

Second question is : Is there a shorter or easier way to answer this question, since the derivative of T(t) is getting nastier and messier?

Thanks in advance.
Attached Images
File Type: jpg UTV.jpg (23.4 KB, 21 views)

Last edited by skipjack; September 15th, 2018 at 01:56 AM.
Leonardox is offline  
 
September 15th, 2018, 01:53 AM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 19,510
Thanks: 1741

The attached question doesn't ask for the unit normal vector.

For the given point, t = 4. You differentiated correctly, so $\dot{\bf{r}}$(4) = (-1, 0, 8). Divide that by √65.
skipjack is online now  
September 15th, 2018, 09:46 AM   #3
Member
 
Joined: Apr 2017
From: New York

Posts: 77
Thanks: 6

Unit vector

This is what I found by plugging t=4

Questions :
How did we find t=4?
Sqrt65 is magnitude of what?
What formula did we use?

I’m very new at this topic and trying to understand for geometrically and algebraicly.
Thanks for the help.
If you live in NYC I already owe you a cup of coffee
Attached Images
File Type: jpg 01188997-3B92-4356-87D3-CD0333BD1254.jpg (20.7 KB, 4 views)
Leonardox is offline  
September 15th, 2018, 11:30 AM   #4
Member
 
Joined: Apr 2017
From: New York

Posts: 77
Thanks: 6

OK, after some readings and search this is what level I came to:

unit vector = v / magnitude of vector

here our vector is tangent vector which can be found by the derivative of given curve.

then derivative of given line divided by magnitude of same vector ( derivative of given vector's magnitude) will give me the unit vector.

but I still couldn't find how it became sqrt65 and how t=4
Leonardox is offline  
September 15th, 2018, 11:40 AM   #5
Global Moderator
 
Joined: Dec 2006

Posts: 19,510
Thanks: 1741

For e^(4-t) to equal 1, t must be 4.

The magnitude of (-1, 0, 8) is √((-1)² + (0)² + (8)²), which is √65.
skipjack is online now  
September 15th, 2018, 11:47 AM   #6
Member
 
Joined: Apr 2017
From: New York

Posts: 77
Thanks: 6

how did we find (-1, 0 , 8 )
ok magnitude of this is sqrt65 true

and why e^4-t=1

or you mean = -1?

where did (-1, 0, 8 ) come from ?
Leonardox is offline  
September 15th, 2018, 12:53 PM   #7
Global Moderator
 
Joined: Dec 2006

Posts: 19,510
Thanks: 1741

The question specifies the point (1, 1, 0). For the first coordinate to be 1, t must be 4.

The vector (-1, 0, 8) comes from substituting t = 4 into the tangent vector $\dot{\bf{r}}$ obtained by differentiation.
Thanks from Leonardox
skipjack is online now  
September 15th, 2018, 01:57 PM   #8
Member
 
Joined: Apr 2017
From: New York

Posts: 77
Thanks: 6

Thanks everything is perfectly clear now.
I learned it.

As a completion of this question how do we write the unit vector:
-1/sqrt65+8/sqrt65= 7/sqrt65

Or -1/sqrt65 , 8/sqrt65
Attached Images
File Type: jpg 76D6CA5B-EF57-4E04-A80A-1487D395241E.jpg (20.8 KB, 1 views)
File Type: jpg 13F622D0-F418-4BA7-AE13-2D54C89B94C5.jpg (20.2 KB, 1 views)
Leonardox is offline  
September 15th, 2018, 02:39 PM   #9
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,431
Thanks: 105

Let r(t) be vector from origin to point on curve.

r=x(t)i+y(t)j+z(t)k

Then dr is clearly tangent to curve and dr/dt is a tangent vector.
dr/dt=dx/dti+dy/dtj+dz/dtk

The unit tangent vector is then
T=(dr/dt)/|dr/dt|


since T.T=1
d(T.T)/dt=2(dT/dt).T=0 $\displaystyle \rightarrow$ dT/dt is normal to T. Then
N=(dT/dt)/|dT/dt|.

That's the principle behind what you are rying to do. If you know how to differentiate and find length of a vector, the rest is trivial, boring, and uninteresting algebra.
Thanks from Leonardox
zylo is offline  
September 17th, 2018, 07:08 AM   #10
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,431
Thanks: 105

Actally, OP notation is easier for typing summary and calculation layout.

R=(x,y,z)

R'=(x',y',z')

T=R'/|R'|, T.T=1 -> T.T'=0

N=T'/|T'|

B=TXN, Binormal

k=|T'|/|R'|, definition of curvature, or geometric definition.*


*Draw unit tangent T at two neighboring points dR. r is radius of curvature. k=1/r
|ds|=|dR|=|R'|dt
|ds|=r|d$\displaystyle \theta$|=r|dT|
|R'|/r=|T'|
zylo is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
unv, utv



Thread Tools
Display Modes






Copyright © 2018 My Math Forum. All rights reserved.