Arc length of the cardioid

Question

Compute the length of the segment of the cardioid $(r, θ) = (1+ \cos(t), t) $ such that $ t \in [0, 2π].$

How do I find the arc length of the cardioid. I did $\mathbf{r}'=\langle -\sin(t),1\rangle$ so $|\mathbf{r}'|=\sqrt{\sin^2(t)+1}$ which you can't integrate. I think I may need to turn the polar coordinates into cartesian, and then do this process, but I don't know how to do that.

Answer 1

But the formula for the computation of the length of a curve in polar coordinates is\begin{align}\int_0^{2\pi}\sqrt{r^2(t)+(r')^2(t)}\,\mathrm dt&=\int_0^{2\pi}\sqrt{(1+\cos t)^2+(-\sin t)^2}\,\mathrm dt\\&=\int_0^{2\pi}\sqrt{2+2\cos t}\,\mathrm dt.\end{align}Can you take it from here?

Answer 2

I don't know whether turning the polar coordinates into Cartesian would work or not(???). Your integral can't be computed since it is a kind of Elliptic integral.

I rather stick to $$\text{length}=\int_0^{2\pi}{\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} d\theta}$$ for 2-D polar integration of arc-length. Notice the integrand is essentially $ds$.

Answer 3

\begin{align} ds &= \sqrt{r^2+\left( \frac{dr}{d\theta} \right)^2} \, d\theta \\ &= \sqrt{(1+\cos \theta)^2+(-\sin \theta)^2} \, d\theta \\ &= \sqrt{2+2\cos \theta} \, d\theta \\ &= \sqrt{4\cos^2 \frac{\theta}{2}} \, d\theta \\ &= 2\left| \cos \frac{\theta}{2} \right| \, d\theta \\ P &=2\int_{0}^{\pi} 2\cos \frac{\theta}{2} \, d\theta \\ &= 8\int_{0}^{\frac{\pi}{2}} \cos t \, d t \\ &= 8 \end{align}

Answer 4

I think I may need to turn the polar coordinates into cartesian, and then do this process, but I don't know how to do that.

While manageable, working in Cartesian coordinates requires a lot more effort and is not advisable.

The equation's Cartesian form is a fairly simple quartic polynomial equation,

$$r = 1 + \cos\theta \implies (x^2-x+y^2)^2 = x^2 + y^2$$

Use the quadratic formula to reveal the $4$ possible solutions for $y$ as a function of $x$,

$$y(x) = \pm \sqrt{\frac{1+2x-2x^2 \pm \sqrt{1+4x}}2}$$

and together their plots form the cardioid, color-coded in the figure below.

Because of the curve's symmetry, we need only $2$ of these solutions, e.g. the red and green solutions, to compute the arc length.

Implicitly differentiating the Cartesian equation and solving for $\dfrac{dy}{dx}$ gives

$$\frac{dy}{dx} = \frac{-2x^3+3x^2+(1-2x)y^2}{y(2x^2-2x-1+2y^2)}$$

Now plug in the corresponding $y$'s, then each $\dfrac{dy}{dx}$ into the arc length integral, $\displaystyle\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$. Finding the limits for either integral is straightforward; the cardioid passes through $y=0$ when $x\in\{0,2\}$, and solving $\text{red}=\text{green}$ shows the two solutions meet in the second quadrant at the point $\left(-\dfrac14,\dfrac{\sqrt3}4\right)$.

After some simplification, the total arc length of the curve is then $2$ times the integral sum

$$\int_{-\tfrac14}^0 \frac{2\,dx}{\sqrt{(1+4x)\left(3+\sqrt{1+4x}\right)}} + \int_{-\tfrac14}^2 \frac{2\,dx}{\sqrt{(1+4x)\left(3-\sqrt{1+4x}\right)}}$$

which reduce neatly upon substituting $u=\sqrt{1+4x}$ (no elliptic integrals necessary).

$$2 \left(\underbrace{\int_0^1 \frac{du}{\sqrt{3+u}}}_{(*)\,u\to-u} + \int_0^3 \frac{du}{\sqrt{3-u}}\right) \stackrel{(*)}= 2 \int_{-1}^3 \frac{du}{\sqrt{3-u}} \stackrel{u\to3-u}= 2 \int_0^4 \frac{du}{\sqrt u} = \boxed8$$