Radius of curvature of $r=2 \cdot e^{3 \phi}$

Question

I have to find the curvature of $r=2 \cdot e^{3 \phi}$

$\dfrac{\left[1+\left(\dfrac{d y}{d x}\right)^2\right]^{\frac 3 2}}{\dfrac{d^2 y}{d x^2}}$

$x=r\cdot cos(\phi)=2 \cdot e^{3 \phi}\cdot cos(\phi)$

$y=r\cdot sin(\phi)=2 \cdot e^{3 \phi}\cdot sin(\phi)$

$\dfrac{d x}{d \phi}=-2 e^{3 \phi}\cdot (sin(\phi) - 3 cos(\phi))$

$\dfrac{d y}{d \phi}=2 e^{3 \phi}\cdot (3 sin(\phi) + cos(\phi))$

$\dfrac{d y}{d x}=\dfrac{2 e^{3 \phi}\cdot (3 sin(\phi) + cos(\phi))}{-2 e^{3 \phi}\cdot (sin(\phi) - 3 cos(\phi))} =\dfrac{-3 sin(\phi) - cos(\phi)}{sin(\phi) - 3 cos(\phi)}$

$\dfrac{d^2 y}{d \phi^2}=4 e^{3 \phi}\cdot (4 sin(\phi) + 3 cos(\phi))$

$\dfrac{d^2 y}{d ^2 x}=\dfrac{\dfrac{2 e^{3 \phi}\cdot (4 sin(\phi) - 3 cos(\phi))}{50}}{\dfrac{2 e^{3 \phi}\cdot (sin(\phi) + 3 cos(\phi))}{10}}=\dfrac{4 sin(\phi) - 3 cos(\phi)}{5\cdot (sin(\phi) + 3 cos(\phi))}$

now using the radius of curvature formula i find an incredibly complex formula:

$\rho=\dfrac{\left[1+\left(\dfrac{d y}{d x}\right)^2\right]^{\frac 3 2}}{\dfrac{d^2 y}{d x^2}}=\dfrac{\left[1+\left(\dfrac{3 sin(\phi) - cos(\phi)}{sin(\phi) + 3 cos(\phi)}\right)^2\right]^{\frac 3 2}}{\dfrac{4 sin(\phi) - 3 cos(\phi)}{5\cdot (sin(\phi) + 3 cos(\phi))}}$

I can't find a way to write this in a more human way.

Answer 1

It is easier to use the formula for curvature in terms of the function $r(\theta)$:

If a curve is defined in polar coordinates as $r(\theta)$, then its curvature is $$ \kappa(\theta) = \frac{|r^2 + 2r'^2 - rr''|}{(r^2 + r'^2)^{3/2}} $$ where here the prime refers to the differentiation with respect to $\theta$.

The result is $e^{-3\theta}/(2\sqrt{10})$.

Answer 2

The curve $\gamma$ in question is a logarithmic spiral. From the selfsimilarity properties of these spirals it immediately follows that $\rho(\phi)=c r(\phi)$ for some constant $c$. The following (human) computations will confirm this fact.

I shall use the parametric representation $$\gamma:\quad\phi\mapsto{\bf z}(\phi):=2e^{3\phi}(\cos\phi,\sin\phi)\ .$$One computes $$\eqalign{{\bf z}'(\phi)&=2e^{3\phi}(-\sin\phi+3\cos\phi, \cos\phi+3\sin\phi)\cr &=2\sqrt{10}e^{3\phi}\bigl(\cos(\phi+\alpha),\sin(\phi+\alpha)\bigr)\ ,\cr}$$ whereby $\alpha:=\arctan{1\over3}$. It follows that the arc length $\phi\mapsto s(\phi)$ along $\gamma$ satisfies $$s'(\phi)=\bigl|{\bf z}'(\phi)\bigr|=2\sqrt{10}e^{3\phi}\ ,$$ whereas the forward tangent direction is given by $$\theta(\phi)=\arg\bigl({\bf z}'(\phi)\bigr)=\phi+\alpha\ .$$ Now the curvature $\kappa$ is the turning speed of the tangent with respect to arc length, hence $$\kappa(\phi)={d\theta\over ds}={\theta'(\phi)\over s'(\phi)}={1\over2\sqrt{10}}e^{-3\phi}\ ,$$ so that we finally obtain $$\rho(\phi)={1\over\kappa(\phi)}=2\sqrt{10}e^{3\phi}\ .$$