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If a curve has unit speed, is the magnitude of its tangent and normal vectors equal to $1$? I am having trouble seeing this.

if r is the curve, then the tangent is $r'$. Also, normal vector is $r''/|r''|$

My professor wrote $K$(curvature) = $|r''| = |-torsion*normal| = |torsion|$. Where did the normal and the negative go?

$torsion = [(r' x r'') *r''']/|r' x r''|^2$

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  • $\begingroup$ That depends on how you compute such vectors. For any unit tangent vector, we could always multiply it by some scalar and get a new vector which is tangent to the curve with a magnitude different from $1$. You need to explain more carefully what you mean by "tangent' and "normal" vector for your question to make sense. $\endgroup$ – Spencer May 1 at 3:50
  • $\begingroup$ For instance can you provide the formula you use to compute these vectors? $\endgroup$ – Spencer May 1 at 3:50
  • $\begingroup$ if r is the curve, then the tangent is r'. Also, normal vector is r''/|r''| $\endgroup$ – blo May 1 at 3:51
  • $\begingroup$ That helps. Please edit your question to include those formulas. $\endgroup$ – Spencer May 1 at 3:53
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The speed of a curve by definition is $\|\vec{r}'\|$. If you define your "tangent vector" to be $\vec{r}'$ then having unit speed means that this tangent vector has a magnitude equal to $q$.

Your normal vector, because of the way it is defines always has a magnitude equal to $1$. This is because the vector is defined as $\vec{r}''$ divided by its own magnitude. When you divide a vector by it's magnitude the resulting vector has a magnitude equal to $1$.

The norm of a vector times a scalar obeys the following rule $ \| \alpha \vec{v} \| = | \alpha | \| \vec{v} \| $. The torsion that your professor gave is a scalar, so we can apply this rule.

$$\| -torsion * \vec{normal} \| = | -torsion| \| \vec{normal} \| $$ $$ = |-torsion| * 1 $$ $$ = | - torsion | = |torsion| $$

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  • $\begingroup$ My professor wrote K(curvature) = |r''| = |-torque*normal| = |torque|. Where did the normal and the negative go? $\endgroup$ – blo May 1 at 3:56
  • $\begingroup$ Yes I see that you just added that bit. You need to define "torque" now in the question statement. $\endgroup$ – Spencer May 1 at 3:57
  • $\begingroup$ torque = [(r' x r'') *r''']/|r' x r''|^2 $\endgroup$ – blo May 1 at 3:59
  • $\begingroup$ Please edit your question to include that formula. $\endgroup$ – Spencer May 1 at 3:59
  • $\begingroup$ Are you sure that your professor has the torque defined as a scalar and not a vector? $\endgroup$ – Spencer May 1 at 4:04

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