How do I find the arc length of a curve?

Question

http://tutorial.math.lamar.edu/Classes/CalcII/PolarArcLength.aspx

I see there is a formula for the arc length on that web page, and I think I understand it ...

What I don't understand is: how do I use that formula if I have a third variable $t$ as a parameter, and I know the functions $r(t)$ and $\theta(t)$?

The $\theta(t)$ I am working with is $$\theta(t) = \theta_i(1-t)+\theta_ft$$ where $\theta_i$ and $\theta_f$ are the (constant) initial and final angles, so I think this theta just increases linearly, not sure if that helps simplify things.

The $r(t)$ is not as simple as the theta but it is log-linear increase between the (constant) initial and final radius: $$r(t) = r_i^{1-t}\,r_f^t$$

I know that $t$ is going from $0$ to $1$, but I want to get rid of the $t$ and write $r$ as a function of $\theta$ somehow...

How do I find what $ds$ is from the parametrized versions of $r$ and $\theta$?

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Sorry if it's kinda noob question, I'm not good at this stuff. Thanks for any help :)

Answer 1

If we are given $t\mapsto r(t)$ and $t\mapsto\theta(t)$ we can set up the parametric representation $t\mapsto\bigl(x(t),y(t)\bigr)$ with $$x(t)=r(t)\cos\theta(t),\quad y(t)=r(t)\sin\theta(t)\ .$$ It follows that $$x'=r'\cos\theta-r\sin\theta\>\theta',\quad y'=r'\sin\theta+r\cos\theta\>\theta'\ ,$$ where the $'$ denotes differentiation with respect to $t$. This implies $$x'^2+y'^2=r'^2+r^2\>\theta'^2\ ,$$ so that $$ds=\sqrt{x'^2(t)+y'^2(t)}\>dt=\sqrt{r'^2(t)+r^2(t)\,\theta'^2(t)}\>dt\ .$$

Answer 2

Actually, for the purpose of integration, it is more convenient to use the common variable $t$.

Express the integrand in terms of $t$,

$$\frac{dr}{d\theta} =\frac{dr}{dt}\frac{dt}{d\theta} = \left(\frac{r_f}{r_i}\right)^t\left[1+\left(\frac{r_i}{r_f}-1\right)t\right] \frac{1}{\theta_f-\theta_i}$$

Thus, the arc integral becomes

$$L=\int_A \sqrt{r^2+\left( \frac{dr}{d\theta}\right)^2} d\theta$$

$$=(\theta_f-\theta_i) \int_0^1 \left(\frac{r_f}{r_i}\right)^t \sqrt{ r_i^2+\frac{1}{(\theta_f-\theta_i)^2}\left[1+\left(\frac{r_i}{r_f}-1\right)t\right]^2} dt$$

or,

$$L= \int_0^1 \left(\frac{r_f}{r_i}\right)^t \sqrt{\left[(\theta_f-\theta_i) r_i\right]^2+\left[1+\left(\frac{r_i}{r_f}-1\right)t\right]^2} dt$$