Showing the derivative of a vector is orthogonal to the vector

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Homework Help Overview

The discussion revolves around understanding the relationship between the derivative of a vector function and its position vector, particularly in the context of curves on a sphere. The original poster is trying to clarify the implications of the condition |r(t)|=c, where r(t) represents the position vector, and how it relates to the orthogonality of the derivative vector.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to reconcile the statement that a curve lies on a sphere with the condition |r(t)|=c, questioning when |r(t)| might not be constant. Some participants suggest considering the definition of |r(t)| and its implications for orthogonality.

Discussion Status

Participants are exploring the definitions and implications of the condition |r(t)|=c, with some providing examples to illustrate when the position vector and its derivative may not be orthogonal. There is a productive exchange of ideas, with some clarification emerging regarding the relationship between the position vector and the concept of a constant radius.

Contextual Notes

There is an ongoing discussion about the assumptions related to the continuity of the curve and the implications of varying distances from the origin for general curves.

davidp92
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Homework Statement


http://i.imgur.com/6j8W6.jpg
I'm trying to understand that example in the text. I can imagine a curve on a sphere having the derivative vector being orthogonal to the position vector. What I don't understand is, how does "if a curve lies on a sphere with center the origin" mean the same thing as "if |r(t)|=c (constant)"?

Homework Equations


The problem is I don't understand why the statement says "Show that if |r(t)|=c ..."
Doesn't the example mean that every derivative vector of a curve is orthogonal to the position vector since |r(t)|=c for all t as long as the curve is continuous? (thinking it in terms of sqrt(x^2+y^2+z^2))
When is |r(t)| not equal to a constant?
 
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If |r(t)| is a constant, then you are essentially talking about a radius. Think about what the definition of |r(t)| is (it looks like you have an idea in the relevant equations section). If it is not equal to a constant, then the position vector and the derivative are not necessarily orthogonal. It is easy to find an example of this (for example, r(t) = <t,t>).
 
I don't quite get your question. |r(t)|=c means the position r(t) is at a constant distance of c from the origin at all times. If it's always at a distance of c from the origin, then it lies on a sphere of radius c around the origin. For a general curve, the distance from the origin will vary with time.
 
lineintegral1 said:
If |r(t)| is a constant, then you are essentially talking about a radius. Think about what the definition of |r(t)| is (it looks like you have an idea in the relevant equations section). If it is not equal to a constant, then the position vector and the derivative are not necessarily orthogonal. It is easy to find an example of this (for example, r(t) = <t,t>).

How is |r(t)| not equal to a constant for r(t)=<t,t>?

Thanks for replying!
 
Dick said:
I don't quite get your question. |r(t)|=c means the position r(t) is at a constant distance of c from the origin at all times. If it's always at a distance of c from the origin, then it lies on a sphere of radius c around the origin. For a general curve, the distance from the origin will vary with time.

Ohh, I get it now. I wasn't thinking of it the way you have it in the bold.

Thanks!
 

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