Differential geometry acceleration as the sum of two vectors

  • Thread starter reb659
  • Start date
  • #1
reb659
64
0

Homework Statement



a(t)=<1+t^2,4/t,8*(2-t)^(1/2)>

Express the acceleration vector
a''(1) as the sum of a vector parallel to a'(1) and a vector orthogonal to a'(1)

Homework Equations





The Attempt at a Solution



I took the first two derivatives and calculated a'(t)=<2t, -4t^2, -4/(2-t)^(1/2)> and a''(t)=<2, 8/t^3, -2/(2-x)^(3/2)>. I figured the tangent vector field a'(1)/|a'(1)| will give me the vector parallel to a'(1). But how do I get the orthogonal vector? I was thinking the cross product between a'(1)/|a'(1)| and a''(1) would give me the orthogonal vector to a'(1), but the vectors ended up being linearly independent so I couldn't represent a''(1) as a sum of the other two.

I then tried calculating the principal normal vector field but I would need to take the derivative of the tangent vector field a'(t)/|a'(t)| and that ended up being incredibly messy and I'm sure that isn't the right way to do this.
 
Last edited:

Answers and Replies

  • #2
chiro
Science Advisor
4,815
134

Homework Statement



a(t)=<1+t^2,4/t,8*(2-t)^(1/2)>

Express the acceleration vector
a''(1) as the sum of a vector parallel to a'(1) and a vector orthogonal to a'(1)

Homework Equations





The Attempt at a Solution



I took the first two derivatives and calculated a'(t)=<2t, -4t^2, -4/(2-t)^(1/2)> and a''(t)=<2, 8/t^3, -2/(2-x)^(3/2)>. I figured the tangent vector field a'(1)/|a'(1)| will give me the vector parallel to a'(1). But how do I get the orthogonal vector? I was thinking the cross product between a'(1)/|a'(1)| and a''(1) would give me the orthogonal vector to a'(1), but the vectors ended up being linearly independent so I couldn't represent a''(1) as a sum of the other two.

I then tried calculating the principal normal vector field but I would need to take the derivative of the tangent vector field a'(t)/|a'(t)| and that ended up being incredibly messy and I'm sure that isn't the right way to do this.

You have the right idea.

The normal is going to be T x N where T is the tangent and N is what is known as the binormal vector.

The normal should be "normalized" (ie length of 1).

Heres a page with the ideas:

http://mathworld.wolfram.com/BinormalVector.html
 
  • #3
reb659
64
0
So I should able to write a(1) as a linear combination of T(1) and N(1), correct?

But how do I get N(t)? Taking derivatives of T(t) is very messy and the professor said it didn't involve any long calculations. Furthermore we haven't learned about the binormal and normal in this section yet, thats not until later but that seems like the only way to do this and to explicitly get N(t) doesn't seem feasible with those equations. Is there any way to do this without it? I am so confused...
 
Last edited:
  • #4
reb659
64
0
Basically the main question I have is if the only way to do this would be to explicitly calculate the expression for N(t).
 
  • #5
chiro
Science Advisor
4,815
134
Basically the main question I have is if the only way to do this would be to explicitly calculate the expression for N(t).

You can use the second derivative.
 
  • #6
reb659
64
0
How so?
 
Last edited:

Suggested for: Differential geometry acceleration as the sum of two vectors

Replies
2
Views
937
Replies
3
Views
4K
Replies
14
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
18
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
Top