Comprimentos de uma curva paramétrica

Recall that a parametric curve is given by the equations $x=f(t)$ and $y=g(t)$, for $a\le t\le b$. To find the arc length of a parametric curve, we have to assume two facts:

(1) as $t$ goes from $a$ to $b$, we trace the curve exactly once;

(2) as $t$ increases, $x$ also increases. (This way, we prevent our parametrization from "reversing" directions at any point.)

Given these assumptions, the arc length is equal to

$$L=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\, dt.$$

For example, let's calculate the arc length of the curve defined by $x=t^2$ and $y=t^3$, as $t$ ranges from $0$ to $1$. The graph of this curve is shown below:

Since $\frac{dx}{dt}=2t$ and $\frac{dy}{dt}=3t^2$, we have

$$L=\int_0^1 \sqrt{(2t)^2+(3t^2)^2}\, dt=\int_0^1 \sqrt{4t^2+9t^4}\, dt$$ $$=\int_0^1 t\sqrt{4+9t^2}\, dt.$$

Now, we perform a $u$-substitution by letting $u=4+9t^2$. Then $du=18t\, dt$, and so the integral becomes

$$L=\frac{1}{18}\int_4^{13} \sqrt{u}\, du=\frac{1}{27}u^{1.5}\bigg|_4^{13}=\frac{1}{27}\left(13^{1.5}-4^{1.5}\right)$$ $$=\frac{1}{27}\left(13\sqrt{13}-8\right).$$

This is our final arc length.

Let's start with the formula derived in the previous section (Arc Length Between Two Points)

$$L=\int_{f(a)}^{f(b)} \sqrt{1+\left(\frac{dy}{dx}\right)^2}\, dx,$$

where $f(a)$ is the starting $x$-value, and $f(b)$ is the ending $x$-value. We now observe that since $x=f(t)$,

$$dx=f'(t)\, dt=\frac{dx}{dt}\, dt.$$

So, since $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}},$$ It follows that

$$L=\int_a^b \sqrt{1+\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)^2}\cdot \frac{dx}{dt}\, dt$$ $$=\int_a^b \sqrt{\left[1+\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)\right]\cdot \left(\frac{dx}{dt}\right)^2}\, dt$$ $$=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\, dt.$$

This is our final formula for the arc length of a parametric curve.