Differential geometry acceleration as the sum of two vectors

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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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?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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