Proving derivatives of a parametrized line are parallel

V0ODO0CH1LD
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Homework Statement



Show that if σ(t) for (t in I) is a parametrization of a line, then σ''(t) is parallel to σ'(t).

Homework Equations





The Attempt at a Solution



I thought that if σ(t) is a parametrization of a line then it could be expressed as σ(t) = vt + a, but then σ'(t) = v and σ''(t) = 0. Can two vectors be parallel? Is it because the distance between them is constant? Or did I make a mistake earlier on?
 
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V0ODO0CH1LD said:

Homework Statement



Show that if σ(t) for (t in I) is a parametrization of a line, then σ''(t) is parallel to σ'(t).

Homework Equations





The Attempt at a Solution



I thought that if σ(t) is a parametrization of a line then it could be expressed as σ(t) = vt + a, but then σ'(t) = v and σ''(t) = 0. Can two vectors be parallel? Is it because the distance between them is constant? Or did I make a mistake earlier on?

That would be the equation of a line where the point ##\sigma(t)## moves with constant velocity. But what about something of the form ##\sigma(t) = f(t)\vec D + \vec a##? Wouldn't that give a straight line too?
 
Ah, thank you! But both ##\vec{D}## and ##\vec{a}## are still just regular vectors, right?
 
V0ODO0CH1LD said:
Ah, thank you! But both ##\vec{D}## and ##\vec{a}## are still just regular vectors, right?

Yes. ##\vec D## is a direction vector and ##\vec a## is a position vector to a point on the line. Both vectors are constants.
 
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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