Prove that v(t) is any vector that depends on time

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Homework Help Overview

The discussion revolves around proving properties of a time-dependent vector, specifically the velocity vector v(t) of a moving particle that maintains a constant magnitude. Participants explore the relationship between the derivative of this vector and its orthogonality to the vector itself, as well as the implications of these properties in different contexts, such as polar coordinates and charged particles in magnetic fields.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants suggest using the equation relating the dot product of the vector with itself to its magnitude and discuss taking derivatives of both sides. There are questions about the nature of orthogonality and how it relates to dot products. Some participants express uncertainty about treating the vector as a variable during differentiation.

Discussion Status

Several participants have provided insights and hints regarding the mathematical approach to the problem, including the use of derivatives and integrals. There is an ongoing exploration of the implications of orthogonality and the conditions under which the magnitude of the vector remains constant. No explicit consensus has been reached, but productive lines of reasoning are being developed.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available and the methods they can employ. There is a focus on understanding the mathematical principles behind vector differentiation and the implications of orthogonality in the context of the problem.

  • #31
Oblio said:
I need to work it to v^2?

Was the multiplying by two only by the knowledge of doing the first part?

yeah sort of... for the second part you somehow need to figure out that \frac{d(\vec{v}\cdot\vec{v})}{dt} = 2\vec{\frac{dv}{dt}}\cdot\vec{v}

we already proved this in the first part.

so yes, v.v = |v|^2

|v|^2 = C, so |v|=sqrt(C), which is a constant.
 
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  • #32
Alright, I wasn't sure if multiplying by two was also a reasonable 'step' without the preknowledge.
 

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