Calc 3 questions concerning Normal and Tangent unit Vectors

In summary, the conversation discussed the differentiation of the tangent vector T(t), using the quotient rule and product and chain rules. There was also a discussion on whether to differentiate the un-normalized tangent vector and then normalize it, or to differentiate the normalized tangent vector directly.
  • #1
badtwistoffate
81
0
heres is one problem i did, i photo'd it so i wouldn't have to worry about it...
Am I doing it right? any problems can you see :-/ on like the 8th/9th line, I don't think I can do what I did...

http://img130.imageshack.us/img130/1332/test076ze.jpg [Broken]

Sooo
 
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  • #2
Since you have

[tex]\vec{T} (t) = \frac{1}{\sqrt{5t^2+1}} \left< 1,t,2t\right>= \left< \frac{1}{\sqrt{5t^2+1}},\frac{t}{\sqrt{5t^2+1}},\frac{2t}{\sqrt{5t^2+1}}\right> [/tex]

[tex]\vec{T} ^{\mbox{ }\prime} (t)[/tex] will require the quotient rule, the result is

[tex]\vec{T} ^{\mbox{ }\prime} (t) = \frac{1}{(5t^2+1)\sqrt{5t^2+1}}\left< -5t,1,2\right> = \frac{1}{(5t^2+1)^{\frac{3}{2}}}\left< -5t,1,2\right>[/tex]
 
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  • #3
So I do the quotient rule to each of the components of T(t)?
How did you get your answer for T'(t)?
 
  • #4
Yep, differentiate each component of T(t) according to the quotient rule and simplify.

How did I do it? Simple, I used Maple v10 :smile:
 
  • #5
Instead of the quotient rule, I think I would be inclined to write the components as
[tex](5t^2+1)^{-\frac{1}{2}}[/tex]
[tex]t(5t^2+1)^{-\frac{1}{2}}[/tex]
[tex]2t(5t^2+1)^{-\frac{1}{2}}[/tex]
and use the product and chain rules.
 
  • #6
Why not just differentiate the un-normalised tangent vector then normalise it afterwards? Seems computationally simpler to me.

Edit: Hm, doesn't work. I don't understand why differentiating a normalised tangent vector vs. a tangent vector should change the direction in which the resulting vector points. I also don't understand why the derivative of the normalised tangent vector will always be normal, seems to me that would be an acceleration and should therefore only be normal if the particle isn't picking up any kinetic energy.

Guess I'm going to have to break out my calculus book and do some reading :wink:
 
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1. What is a normal vector in Calculus 3?

A normal vector is a vector that is perpendicular to a given surface or curve at a specific point. In Calculus 3, normal vectors are often used to determine the direction of curvature of a surface, or the direction of motion of a particle on a curve.

2. How do you find the unit tangent vector in Calculus 3?

The unit tangent vector is a vector that is tangent to a curve at a specific point and has a length of 1. To find the unit tangent vector, you can take the derivative of the curve and then divide by the magnitude of the derivative.

3. What is the difference between a normal and a unit normal vector?

A normal vector is a vector that is perpendicular to a surface or curve, while a unit normal vector is a normal vector with a magnitude of 1. In other words, a unit normal vector is a normalized version of a normal vector.

4. How do you use normal and tangent vectors to find the rate of change of a function in Calculus 3?

In Calculus 3, normal and tangent vectors can be used to find the rate of change of a function through the use of the gradient vector. The gradient vector is a vector that points in the direction of the steepest increase of the function at a given point and has a magnitude equal to the rate of change at that point.

5. Can normal and tangent vectors be used in 3D space?

Yes, normal and tangent vectors can be used in 3D space. In Calculus 3, these vectors are often used to describe the direction of motion of a particle in 3D space or the direction of curvature of a 3D surface.

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