# curvature function $\frac{|r' \times r''|}{|r'|^3}$

I am learning curvature now and found that $$\frac{|T'|}{|r'|}$$ is $$\frac{|r' \times r''|}{|r'|^3}$$

I understand what $\frac{|T'|}{|r'|}$ means, it is change of direction as time changes divided by change of arc length as time changes because that is $\frac{\frac{|dT|}{|dt|}}{\frac{|ds|}{|dt|}}$.

But I cannot understand what $\frac{|r' \times r''|}{|r'|^3}$ really means. Can someone explain verbally?

--edited--

I just watched a video on it and it explained that as $$|r'\times r''|$$ gets bigger the curve becomes steeper and I understand it. But what if the change of $r'$ is steeper than $90^{\circ}$? $T'$ must be bigger but the area of parallelogram $|r' \times r''|$ gets smaller. Does it not?

• Please use a superscript before the prime. To denote the derivative of $T$, say, write this: $T^{'}$ - $T$, then the caret symbol ^ - then the prime ' . Commented Aug 30, 2018 at 7:49
• In LaTeX, you don't want r\prime or r\prime\prime. Instead, just do r' and r''. The apostrophe is a special macro that effectively makes r^{\prime} and r^{\prime\prime}. Commented Aug 30, 2018 at 8:17
• just edited and done
– 강승태
Commented Aug 30, 2018 at 8:22
• @강승태 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– user
Commented Sep 17, 2018 at 20:28

Using the quotient rule and $|u|'=\frac{u\cdot u'}{|u|}$, we get $$T=\frac{r'}{|r'|}\implies T'=\frac{r''r'\cdot r'-r'r'\cdot r''}{|r'|^3}\tag1$$ Noting that $T'\cdot r'=0$ (simply compute the dot product), we have \begin{align} |T'|^2 &=\frac{r''r'\cdot r'-r'r'\cdot r''}{|r'|^3}\cdot\frac{r''}{|r'|}\\ &=\frac{|r''|^2|r'|^2-(r'\cdot r'')^2}{|r'|^4}\\ &=\frac{|r''|^2|r'|^2\left(1-\cos^2(\theta)\right)}{|r'|^4}\\ &=\frac{|r''|^2|r'|^2\sin^2(\theta)}{|r'|^4}\\ &=\left|\frac{r''\times r'}{|r'|^2}\right|^2\tag2 \end{align} Therefore, $$\frac{|T'|}{|r'|}=\frac{|r''\times r'|}{|r'|^3}\tag3$$

In Other Words

$\frac{|r'\times r''|}{|r'|}$ measures the component of $r''$ perpendicular to $r'$. If the speed is doubled, the component of $r''$ perpendicular to $r'$ is quadrupled; this is because $r'$ is doubled and the rate of change of direction of $r'$ is doubled. Thus, to account for speed, we have $\frac{|r'\times r''|}{|r'|^3}$.

• please, tell me why $|u|'=\frac{u\cdot u'}{|u|}$? From where did you derive this formula? thanks!
– Iuli
Commented May 16, 2020 at 10:37
• Differentiate $|u|^2=u\cdot u$ and divide by $2|u|$.
– robjohn
Commented May 16, 2020 at 12:42

For the derivation of the formula take a look to Deriving curvature formula.

For an intuitive explanation let consider

$$\frac{|r\prime \times r\prime\prime|}{|r\prime|^3}=\left|\frac{r\prime}{|r\prime|} \times \frac{r\prime\prime}{|r\prime|^2}\right|$$

we have that $\frac{r\prime\prime}{|r\prime|^2}$ in an acceleration term and since $\frac{r\prime}{|r\prime|}$ is tangent to the curve the cross product $\left|\frac{r\prime}{|r\prime|} \times \frac{r\prime\prime}{|r\prime|^2}\right|$ is greater when then perpendicular component of the acceleration is greater that is when the curvature is greater.

• It is easier when compared to velocity and acceleration . thanks. But I still wonder, hypothetically, if a car suddenly turns more than $90^{\circ}$ like, $180^{\circ}$ for example, Would not the $|r' \times r''|$ smaller than when $90^{\circ}$? Because cross product is $|a||b|\sin\theta$? and $\sin\theta$ is greatest at $90^{\circ}$ and decrease thereafter?
– 강승태
Commented Aug 30, 2018 at 8:47
• The angle to be considered when we take the cross product is between the tangential direction to the curve and the direction of the acceleration. If the acceleration is purely tangential the product is zero since the cross product only act on the perpendicular component which is related to the curvature.
– user
Commented Aug 30, 2018 at 8:55

My personal history with this is that $\frac{|T'|}{|r'|}$ is something relatively easy to understand conceptually. But it's terrible to compute with. There are too many complications as you go from $r$ to $r'$ to $T$ to $T'$ to a magnitude.

Then there is $\frac{|r'\times r''|}{|r'|^3}$, which is in practice much easier to compute. And it can be proven that this is equal to the other expression. Its value is that it is in practice less work to compute $r'$, then $r''$, then a cross product, and then a magnitude.

But you can still try to make sense of it. $r'\times r''$ is a vector parallel to the binormal vector but scaled by some amount. Its magnitude is proportional to the speed and the magnitude of the acceleration and the sine of the angle between those things. Dividing by $|r'|^3$ leaves you with $\frac{|r''|}{|r'|^2}\sin(\theta)$. You can think about why $\kappa$ would be proportional to $|r''|$ and $\sin(\theta)$, while being inversely proportional to $|r'|^2$.

We have $$\frac{|r' \times r''|}{|r'|^3} =\frac{\left|\frac{r’}{|r’|}\times r’’\right|}{|r’|^2}.$$ Now the numerator is the normal component of the acceleration as it’s the height of a parallelogram with one side equals one. The bigger this component is the bigger is the curvature.

On the other hand curvature decreases quadratically when the velocity increases. So you may read $$\frac{\left|\frac{r’}{|r’|}\times r’’\right|}{|r’|^2}=\frac{\text{normal component of the acceleration}}{\text{square of the velocity}},$$ which corresponds to our physical experience.