Differential geometry acceleration as the sum of two vectors

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SUMMARY

The discussion centers on expressing the acceleration vector a''(1) of the function a(t)=<1+t^2,4/t,8*(2-t)^(1/2)> as the sum of a vector parallel to the tangent vector a'(1) and a vector orthogonal to it. The derivatives calculated are a'(t)=<2t, -4t^2, -4/(2-t)^(1/2)> and a''(t)=<2, 8/t^3, -2/(2-x)^(3/2)>. The participants explore using the tangent vector field a'(1)/|a'(1)| for the parallel component and consider the cross product for the orthogonal component, but face challenges due to linear independence. The discussion suggests that utilizing the second derivative may provide a solution without complex calculations.

PREREQUISITES
  • Understanding of vector calculus, specifically acceleration and tangent vectors.
  • Familiarity with derivatives and their applications in physics.
  • Knowledge of vector operations, including cross products and normalization.
  • Basic concepts of differential geometry, particularly tangent and normal vectors.
NEXT STEPS
  • Study the properties of tangent and normal vectors in differential geometry.
  • Learn about the binormal vector and its role in vector calculus.
  • Explore the application of the second derivative in vector analysis.
  • Review examples of expressing vectors as linear combinations in vector spaces.
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Students and professionals in mathematics, physics, and engineering who are dealing with vector calculus and differential geometry, particularly those needing to understand vector decomposition in acceleration analysis.

reb659
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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:
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reb659 said:

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
 
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, that's 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...
 
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Basically the main question I have is if the only way to do this would be to explicitly calculate the expression for N(t).
 
reb659 said:
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.
 
How so?
 
Last edited:

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