Unit tangent and normal vectors

  • Thread starter cdotter
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Homework Statement


r(t)=ti+t^2j
Find the velocity, speed, acceleration, unit tangent, and unit normal vectors.


Homework Equations


Velocity=r'(t)
Speed=magnitude of r'(t)
Acceleration=r''(t)
Unit tangent=r'(t)/magnitude of r'(t)
Unit normal=d/dt[unit tangent]/magnitude of d/dt[unit tangent]


The Attempt at a Solution



Velocity=i+2tj
Speed=[itex]\sqrt{1^2+(2t)^2} = \sqrt{1+4t^2}[/itex]
Acceleration=2j
Unit tangent=[tex]\frac{i+2tj}{\sqrt{1+4t^2}}[/tex]

I'm pretty sure that's all right so far. I get mixed up in the algebra at the unit normal.

For d/dt[unit tangent] I have [tex]\frac{2j\sqrt{1+4t^2}-\frac{1}{2}(1+4t^2)^{-1/2}(8t)}{1+4t^2}[/tex].

Is that correct? How do I take the magnitude of that mess? I can't really see a way to simplify it.
 
Last edited:

Answers and Replies

  • #2
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your unit normal still has to have a direction doesn't it? if you find the i and j components, then you can find the magnitude the same way you found the magnitude of r'(t)
 
  • #3
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That's the problem, I have no idea how to simplify it to something where I can square the components, add them, and take the square root. :cry:
 
  • #4
tiny-tim
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hi cdotter! :wink:
Unit normal=d/dt[unit tangent]/magnitude of d/dt[unit tangent]

nooo!

the unit normal is simply the unit vector perpendicular to the unit tangent! :smile:
 
  • #5
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hi cdotter! :wink:


nooo!

the unit normal is simply the unit vector perpendicular to the unit tangent! :smile:

Maybe I called it the wrong thing...the principal unit normal vector?

http://www.ltcconline.net/greenl/courses/202/vectorFunctions/tannorm.htm

I just found this trick to simplify it so maybe that will work:

"Since the unit vector in the direction of a given vector will be the same after multiplying the vector by a positive scalar, we can simplify by multiplying by the factor"
 
  • #6
1,033
1
Maybe I called it the wrong thing...the principal unit normal vector?

http://www.ltcconline.net/greenl/courses/202/vectorFunctions/tannorm.htm

I just found this trick to simplify it so maybe that will work:

"Since the unit vector in the direction of a given vector will be the same after multiplying the vector by a positive scalar, we can simplify by multiplying by the factor"

haha funny thing is I was literally just reading that link. I wasn't going to send it to you because it is the exact question that you are given.
 
  • #7
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haha funny thing is I was literally just reading that link. I wasn't going to send it to you because it is the exact question that you are given.

:rofl: I didn't realize the exact same problem was there until you said it, I just saw the trick/technique.
 
  • #8
tiny-tim
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Maybe I called it the wrong thing...the principal unit normal vector?

ah!

as you can see, my definition also works (and is much quicker) …

the only difference is that my method comes up with two unit normal vectors (oppsotie each other),

the one you need is "is the unique vector that points into the curve", ie the one towards the centre of curvature (the concave side)

for a 3D curve, you do need the book's method, to decide which one is the principal vector

but for a 2D curve like this, just choosing the concave side is enough! :wink:
 

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