Vectors: Find w such that dv/ds = w x v

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Homework Help Overview

The problem involves finding a vector w such that the derivative of a vector v, defined as the sum of the tangent (T), normal (N), and binormal (B) unit vectors of a curve in R3, can be expressed in the form dv/ds = w x v. The context is rooted in multivariate calculus and vector calculus concepts.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss splitting the vector v into components and using a matrix to find the determinant for the cross product. There is mention of mismatching components when attempting to derive the relationship. Some participants suggest alternative methods, such as considering the properties of the cross product and the implications of the orthogonal triad of unit vectors.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem. Some have offered hints and suggestions for alternative perspectives, but there is no explicit consensus on the next steps or a clear resolution yet.

Contextual Notes

Participants are working under the constraints of a multivariate calculus course, and there is an emphasis on understanding the relationships between the vectors involved without providing direct solutions.

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Homework Statement



T (tangent), B (binormal), N (normal) are an orthogonal triad of unit vectors of a curve in R3.Given:
  • dT/ds = kN
  • dN/ds = -kT + tB
  • dB/ds = -tN

Find vector w so that these equations may be written in the form:

dv/ds = w x v, where v = T + N + B

Homework Equations



Given above!

The Attempt at a Solution



I tried splitting v (T + N + B) into its components, as well as w, and putting them into a matrix to find the determinant (for the cross product).

The matrix consists of <i, j, k>, w = <w1, w2, w3>, and v = <(T1 + N1 + B1), (T2 + N2 + B2), (T3 + N3 + B3)>.

However, taking the determinant/cross product gives mismatching components (ie. j and k components supposedly add up to an i component). I'm not too sure where to go from here :(
Note: This is for a Multivariate Calculus course!
 
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YayMathYay said:

Homework Statement



T (tangent), B (binormal), N (normal) are an orthogonal triad of unit vectors of a curve in R3.Given:
  • dT/ds = kN
  • dN/ds = -kT + tB
  • dB/ds = -tN

Find vector w so that these equations may be written in the form:

dv/ds = w x v, where v = T + N + B

Homework Equations



Given above!

The Attempt at a Solution



I tried splitting v (T + N + B) into its components, as well as w, and putting them into a matrix to find the determinant (for the cross product).

The matrix consists of <i, j, k>, w = <w1, w2, w3>, and v = <(T1 + N1 + B1), (T2 + N2 + B2), (T3 + N3 + B3)>.

However, taking the determinant/cross product gives mismatching components (ie. j and k components supposedly add up to an i component). I'm not too sure where to go from here :(
Note: This is for a Multivariate Calculus course!

Hi YayMathYay, welcome to PF! :smile:

There's a much easier way to do this than by looking at the Cartesian components of the vectors.

Hint: What is \mathbf{v}\times ( \mathbf{w} \times \mathbf{v} )?
 
gabbagabbahey said:
Hi YayMathYay, welcome to PF! :smile:

There's a much easier way to do this than by looking at the Cartesian components of the vectors.

Hint: What is \mathbf{v}\times ( \mathbf{w} \times \mathbf{v} )?

First of all, thanks for the welcome! :)

I know the hint you gave me means \left(\mathbf{v} \bullet \mathbf{v}\right)\mathbf{w} - \left(\mathbf{v} \bullet \mathbf{w}\right)\mathbf{v}. I've tried this method before, but I wasn't too sure where to go from here..
 
YayMathYay said:
First of all, thanks for the welcome! :)

I know the hint you gave me means \left(\mathbf{v} \bullet \mathbf{v}\right)\mathbf{w} - \left(\mathbf{v} \bullet \mathbf{w}\right)\mathbf{v}. I've tried this method before, but I wasn't too sure where to go from here..

For starters, what is \mathbf{v} \cdot \mathbf{v} ? Next since \mathbf{T}, \mathbf{B} & \mathbf{N} are an orthogonal triad in \mathbb{R}^3, they span \mathbb{R}^3, and so any vector can be decomposed into a linear combination of those unit vectors. So, why not write \mathbf{w} = w_{T}\mathbf{T} + w_{N}\mathbf{N} + w_{B}\mathbf{B}... what does that make \mathbf{w} ( \mathbf{v} \cdot \mathbf{v} ) - \mathbf{v} ( \mathbf{w} \cdot \mathbf{v} )?
 

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