Finding the unit tagent vector, normal vec and curvature problem

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SUMMARY

The discussion focuses on calculating the unit tangent vector, normal vector, and curvature for the curve defined by the parametric equation r(t) = <1/3 * t^3, 1/2 * t^2, t>. The correct formula for the unit tangent vector T(t) is T(t) = r'(t)/|r'(t)|, while the unit normal vector N(t) is derived from N(t) = T'(t)/|T'(t)|. A common mistake highlighted is using r' instead of T when calculating N(t). The curvature can be determined once T(t) and N(t) are correctly established.

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mr_coffee
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Hello everyone, the problem says to:
For the curve gien by r(t) = <1/3* t^3, 1/2 * t^2, t>
find (a) The unit tagent vector;
(b) the unit normal vector;
(c) the curvature;

Well it seems easy enough! the formula's are just derivatives for instance:
The unit tagent vector says:
T(t) = r'(t)/|r'(t)| i got this one right, you can see my work on the image below:
but part (b) i missed..
The normal vector N is suppose to just be:
N(t) = T'(t)/|T'(t)|;
Here is my work and it does not match the back of the book.
http://show.imagehosting.us/show/800636/0/nouser_800/T0_-1_800636.jpg
 
Last edited by a moderator:
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In finding N, you did NOT find T'/|T'|. You used r' rather than T.

Yes, [tex]T= <t^2, t, 1>/\sqrt{t^4+ t^2+ 1}[/tex].
To find N(t) you have to differentiate THAT: differentiate
[tex]\left<\frac{t^2}{\sqrt{t^4+ t^2+ 1}},\frac{t}{\sqrt{t^4+ t^2+ 1}},\frac{1}{\sqrt{t^4+ t^2+ 1}}\right>[/tex].
 

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