Differential length of a logarithmic spiral

Question

I am working on a problem that asks to find the magnetic field at the origin of a logarithmic spiral $r = e^\theta$ from $\theta = 0$ to $\theta = 2\pi$, where the angle $\theta$ is measured counterclockwise from the positive x axis.

I was able to do some research and found the following information about the curve.

I know that the tan() of the angle between a radial line and the tangent is 1 in this case because the coefficient on the exponent is 1.

My question is how the differential length $ds$ was determined? Also, I have no idea how $tan \beta = r d\theta / dr$. Can anyone help explain the geometry on this problem to me?

Thanks so much for your help.

Answer 1

$dr$ is the infinitesimal change in the radial direction while $r d\theta$ is the corresponding change in the perpendicular direction. The arc length is therefore: $$ ds = \sqrt{dr^2 + (rd\theta)^2} = \\ \sqrt{1 + \left(r\frac{d\theta}{dr}\right)^2}dr $$ Also, $tan\beta$ is the ratio of tangential to the radial displacement i.e. $$ tan\beta = \frac{rd\theta}{dr} $$

Answer 2

The solution starts with the Cartesian coordinates $$ x=r\cos\theta=a e^{b\theta}\cos\theta \, , \qquad y=r\sin\theta=a e^{b\theta}\sin\theta $$ for a spiral given by the polar equation $r(\theta)=ae^{b\theta}$.

Constructing the vector $\vec r= (a e^{b\theta}\cos\theta,a e^{b\theta}\sin\theta,0)$ we find the tangent vector $d\vec \ell$ by taking the derivative of $\vec r$ w/r to the parameter $\theta$: $$ d\vec \ell =\left(a b e^{b \theta } \cos (\theta )-a e^{b \theta } \sin (\theta ),a b e^{b \theta } \sin (\theta )+a e^{b \theta } \cos (\theta ),0\right)d\theta \, . $$ Then it follows that \begin{align} ds^2&=d\vec\ell\cdot d\vec \ell= a^2 \left(b^2+1\right) e^{2 b \theta }(d\theta)^2=r^2(\theta)(1+b^2) (d\theta)^2\, ,\\ &=r^2(\theta)\left(1+\frac{(dr/d\theta)^2}{r^2(\theta)}\right)(d\theta)^2\, . \end{align} Since $\frac{dr}{d\theta}=b r $ we also have $$ (r d\theta)^2= \left(\frac{dr}{b}\right)^2 $$ and thus $$ ds^2= \left(\frac{dr}{b}\right)^2+ (dr)^2= (dr)^2\left(1+\frac{1}{b^2}\right) $$ Finally, \begin{align} \vec r\cdot d\vec \ell&= r d\ell \cos\beta = a^2 b e^{2b\theta}d\theta\, ,\\ \vert \vec r\times d\vec \ell \vert &=r d\ell \sin\beta = a^2 e^{2b\theta} d\theta \end{align} so that $$ \tan(\beta)=\frac{1}{b}\, . $$

Answer 3

While considering infinitesmals or differential lengths, $d \theta \approx 0$ or in other words OS is parallel to OQ.

In the small right differential triangle SPQ the Pythagoras thm holds and can be drawn as given and is often helpful. ( You already labeled $dr,$ so continue) .. $\beta$ is the same alternate angle between parallels.

$$ \cos\beta=\frac {dr}{ ds}, \quad \tan \beta =\frac{PQ}{PS}=\frac{r \; d \theta}{dr} ,\quad \sin \beta =\frac{PQ}{QS}=\frac{r \; d \theta}{ds} ; $$

General log spiral has equation

$$ r = r_{ start } \cdot e ^ {(\cot \alpha \;\theta)} $$

In this particular equiangular case $\alpha = \pi/4 = \beta $