Differential geometry acceleration as the sum of two vectors

Is there some way to get N(1) from this?In summary, the problem asks to express the acceleration vector a''(1) as the sum of a vector parallel to the tangent vector a'(1) and a vector orthogonal to a'(1). The attempt at a solution involves finding the normal vector N(t) using the cross product of a'(t) and a''(t), but the calculations can become messy. The use of the second derivative may also be useful in finding the normal vector.
  • #1
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.
 
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  • #2
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
 
  • #3
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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  • #4
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
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.
 
  • #6
How so?
 
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1. What is differential geometry acceleration?

Differential geometry acceleration is a mathematical concept that describes the change in velocity of an object in a curved space. It takes into account the curvature of the space and the object's position and velocity at a given point.

2. How is differential geometry acceleration calculated?

Differential geometry acceleration is calculated as the sum of two vectors: the tangential acceleration, which accounts for the change in the magnitude of velocity, and the centripetal acceleration, which accounts for the change in direction of velocity due to the curvature of space.

3. What is the difference between tangential and centripetal acceleration?

Tangential acceleration is the component of acceleration that changes the magnitude of velocity, while centripetal acceleration is the component that changes the direction of velocity. Together, they make up the total differential geometry acceleration.

4. How is differential geometry acceleration used in real-world applications?

Differential geometry acceleration is used in a wide range of fields, including physics, engineering, and computer graphics. It is particularly useful in understanding the motion of objects in curved spaces, such as satellites orbiting around a planet or cars navigating a race track.

5. Are there any limitations or assumptions when using differential geometry acceleration?

One limitation of differential geometry acceleration is that it assumes a smooth, continuous space. It may not accurately describe the motion of objects in highly irregular or discontinuous spaces. Additionally, it assumes that the object's mass remains constant, which may not be the case in certain scenarios.

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