# If a curve has unit speed, is the magnitude of its tangent and normal vectors equal to $1$?

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?

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

• 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. – Spencer May 1 at 3:50
• For instance can you provide the formula you use to compute these vectors? – Spencer May 1 at 3:50
• if r is the curve, then the tangent is r'. Also, normal vector is r''/|r''| – blo May 1 at 3:51
• That helps. Please edit your question to include those formulas. – Spencer May 1 at 3:53

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.

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

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