Showing the derivative of a vector is orthogonal to the vector

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!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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