Determine an expression for the length of the curve $r = f(\theta)$ between $\theta = a$ and $\theta = b$.
Is my proof correct?
We know that $$x=r\cos \theta=f(\theta)\cos \theta,$$ $$y=r\sin \theta=f(\theta)\sin \theta.$$Taking the derivative and using the product rule, we have $$\frac{dx}{d\theta} = f'(\theta)\cos \theta - f(\theta)\sin \theta,$$ $$\frac{dy}{d\theta} = f'(\theta)\sin \theta + f(\theta)\cos \theta.$$Using the arc length formula of a parametric curve$$L =\int ^b_a\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \\.$$$$=\int ^b_a\sqrt{\left(\dfrac{dx}{d\theta}\right)^2+\left(\dfrac{dy}{d\theta}\right)^2}\,d\theta \\.$$$$=\int ^b_a\sqrt{(f'(\theta)\cos \theta-f(\theta)\sin \theta)^2+(f'(\theta)\sin \theta+f(\theta)\cos \theta)^2}\,d\theta \\.$$$$=\int ^b_a\sqrt{(f'(\theta))^2(\cos^2 \theta+\sin^2 \theta)+(f(\theta))^2(\cos^2 \theta+\sin^2\theta)}\,d\theta \\.$$$$=\int ^b_a\sqrt{(f'(\theta))^2+(f(\theta))^2}\,d\theta \\.$$$$=\boxed{\int ^b_a\sqrt{r^2+\left(\dfrac{dr}{d\theta}\right)^2}\,d\theta}.$$
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$\begingroup$A remark leading to a shorter proof and, as well, a deeper understanding:
Your system:
$$\begin{cases}\frac{dx}{d\theta} &=& f'(\theta)\cos \theta - f(\theta)\sin \theta\\ \frac{dy}{d\theta} &=& f'(\theta)\sin \theta + f(\theta)\cos \theta\end{cases}$$
can be written under the form:
$$\begin{pmatrix}\frac{dx}{d\theta}\\\frac{dy}{d\theta}\end{pmatrix}= \underbrace{\begin{pmatrix}\cos \theta &- \sin \theta\\ \sin \theta &\cos \theta\end{pmatrix}}_{R_{\theta}}\begin{pmatrix}f'(\theta)\\f(\theta)\end{pmatrix}$$
The $\theta$ rotation matrix is very natural if you make a sketch, and preserves lengths. We have therefore:
$$\sqrt{\left(\dfrac{dx}{d\theta}\right)^2+\left(\dfrac{dy}{d\theta}\right)^2}=\sqrt{f'(\theta)^2+f(\theta)^2}$$
$\endgroup$ $\begingroup$An alternative view using complex variables: Clearly
$$z=re^{i\theta}$$
Now,
$$L=\int |\dot z|\ d\theta\\ \dot z=(ir+\dot r)e^{i\theta}\\ |\dot z|=\sqrt{r^2+\dot r^2}\\ L=\int \sqrt{r^2+\dot r^2}\ d\theta\\ $$
$\endgroup$ $\begingroup$Or directly from differential form of the right triangle in polar coordinates
$$ ds^2 = (r d \theta)^2+ dr^2 $$
$$s=\int_a^b \sqrt{r^2+ r'^2}\ d\theta $$
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