My Math Forum UTV and UNV

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September 14th, 2018, 09:59 PM   #1
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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?

Attached Images
 UTV.jpg (23.4 KB, 21 views)

Last edited by skipjack; September 15th, 2018 at 02:56 AM.

 September 15th, 2018, 02:53 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,879 Thanks: 1835 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.
September 15th, 2018, 10:46 AM   #3
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Joined: Apr 2017
From: New York

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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
 01188997-3B92-4356-87D3-CD0333BD1254.jpg (20.7 KB, 4 views)

 September 15th, 2018, 12:30 PM #4 Senior Member   Joined: Apr 2017 From: New York Posts: 118 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
 September 15th, 2018, 12:40 PM #5 Global Moderator   Joined: Dec 2006 Posts: 19,879 Thanks: 1835 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.
 September 15th, 2018, 12:47 PM #6 Senior Member   Joined: Apr 2017 From: New York Posts: 118 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 ?
 September 15th, 2018, 01:53 PM #7 Global Moderator   Joined: Dec 2006 Posts: 19,879 Thanks: 1835 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
September 15th, 2018, 02:57 PM   #8
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Joined: Apr 2017
From: New York

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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
 76D6CA5B-EF57-4E04-A80A-1487D395241E.jpg (20.8 KB, 1 views) 13F622D0-F418-4BA7-AE13-2D54C89B94C5.jpg (20.2 KB, 1 views)

 September 15th, 2018, 03:39 PM #9 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 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
 September 17th, 2018, 08:08 AM #10 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 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'|

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