# Negative Arc Length Problem (Cardioid Example)

I had a question to calculate the arc length of the cardiod $r=1+cos(\theta)$.

My calculation went as follows: $\displaystyle\int_0^{2\pi} \sqrt{((1+\cos(\theta))^2+\sin^2(\theta)}\mathrm d\theta = \int_0^{2\pi} \sqrt{4\cos^2\bigg(\frac{\theta}{2}\bigg)}\mathrm d\theta= \int_0^{2\pi} 2\cos\bigg(\frac{\theta}{2}\bigg)\mathrm d\theta = 4\sin\bigg(\frac{\theta}{2}\bigg)\bigg|_0^{2\pi}=0$

The step where I have gone wrong is that: $$\sqrt{4\cos^2\bigg(\frac{\theta}{2}\bigg)}$$ should be $$\bigg|2\cos\bigg(\frac{\theta}{2}\bigg)\bigg|$$ which leads to an arc length of $8$ but I don't see why this should happen. It is obvious to me that arc length cannot be zero in this case but why is the modulus sign needed here?

Also, I know about the argument saying that we could integrate from $0$ to $\pi$ and double the arc length since the curve is symmetric but could those arguments be avoided here since that is not the issue.

Can we have negative arc length? Does arc length become negative when the curve dips below the x-axis? Please could this be explained in more detail.

"can we have negative arc-length ?".

Yes, you can define absolute or signed arclength.

The first is the standard acception and is used when calculating the total length of a curve, and goes as per the comments and answers already given.

The second is used especially in physics.

Take just a circle (constant $r$) and $\alpha = \alpha(t)$. Then the integral of $r|da|$ will give the total length of the arc, while the integral of $r\,d\alpha$ will give the (signed) final arc position , cancelling any back and forth (CCW -CW).

The modulus is there because for every $x\in\mathbb R$, the value of $\sqrt{x^2}$ is equal to $|x|$, not $x$. Remember, $\sqrt{a}$ is defined as the non-negative solution to the equation $x^2=a$. The function $x\mapsto\sqrt{x}$ is a positive function on $[0,\infty)$.

For example, if $x=-2$, then $\sqrt{(-2)^2}=2\neq -2$.

So, since $4\cos^2\left(\frac\theta2\right) = \left(2\cos\left(\frac\theta2\right)\right)^2$ applying the rule $\sqrt{x^2}=|x|$ to the expression $$\sqrt{4\cos^2\left(\frac\theta2\right)}$$

yields $$\left|2\cos\left(\frac\theta2\right)\right|.$$

Can we have negative arclength?

No, the arclength is an integral of a positive function, so it is positive.

Does arclength become negative when the curve dips below the x axis?

No.