polar coordinates
As a sort of exercise, I tried to derive the formula for arc length in polar coordinates, using the following logic:
$$dS = r(\theta)d\theta\\
\implies S=\int r(\theta)d\theta$$
However, it turns out the formula is
$$S = \int \sqrt {r^2+\left(\frac {dr}{d\theta}\right)^2}d\theta$$
I could follow the derivation for the correct formula, but why is mine wrong?
Thanks
Use $$\sin^2+\cos^2=1\implies \sqrt {1-\cos x}=\dfrac {\sin x}{\sqrt {1+\cos x}} $$ Let $ u=1-\cos x $, so $ du =\sin xdx $. Then, we get $$\int \dfrac {1}{\sqrt {1-u}}\cdot \dfrac {\sin x}{\sin x} du$$ Now, cancel out and do one more $u$ substitution, integrate and then plug back in to get $$2a\sqrt {2 (1-\cos x)}|^{2\pi}_{0}$$
$\int_0^{2\pi} r\ d\theta$ will give you the average value of $r\cdot 2\pi$
Arc Length in polar coordinates
In the figure, I have taken a curve in Polar coordinates which I have broken up into pieces. If we calculate $\int_0^{2\pi} r\ d\theta$ we will get the sum of the dark lines. It will give you the average value of $r,$ times the range of $\theta$ but it will underestimate the arc length
If you want the arc length, you are going to need the diagonals. in that figure.
What is the length of the diagonal. As $d\theta$ becomes small, they are aproximately right triangles with one leg $r\ d\theta$ and the other leg, $dr$
And hypotenuse $\sqrt {(r\ d\theta)^2 + (dr)^2)}$
$\int \sqrt {(r^2 + (\frac{dr}{d\theta})^2} \ d\theta$ will give you the arc length
Best Answer
Your first formula is incorrect. For instance, consider the line $y=x$ and you want to get the length of the line segment from $x=0$ to $x=1$. The length is $\sqrt2$, and the equation in polar coordinates is $\theta=\dfrac{\pi}4$. If we use the first formula, we get the length to be $0$.
In terms of $x$, $y$, we have $$S = \int \sqrt{(dx)^2+(dy)^2} = \int \sqrt{1+y'^2} dx$$ Setting $x=r \cos(t)$ and $y=r\sin(t)$, we get $$dx = dr \cos(t) - r \sin(t)dt$$ and $$dy = dr \sin(t) + r \cos(t)dt$$ Hence, $$(dx)^2 + (dy)^2 = (dr)^2 + (rdt)^2$$