Why does $\sin\phi=r\frac{d\theta}{ds}$ and $\cos\phi=\frac{dr}{ds}?$

Question

The relation between $p$ and $r$ where $p$ is the length of the perpendicular from the fixed point $O$ on the tangent to the curve at any point $P$ is called pedal equation of the curve.

I want to find the radius of curvature of the pedal curves.

Let $p=f(r)$ be the pedal equation.

$\psi=\theta+\phi........(1)$

and $p=r\sin\phi..........(2)$

Differentiating $(1)$ with respect to $s$,

$\frac{d\psi}{ds}=\frac{d\theta}{ds}+\frac{d\phi}{ds}.............(3)$

Differentiating $(2)$ with respect to $r$,

$\frac{dp}{dr}=\sin\phi+r\cos\phi\frac{d\phi}{dr}$

$\frac{dp}{dr}=\sin\phi+r\cos\phi\frac{d\phi}{ds}\frac{ds}{dr}$

$\frac{dp}{dr}=r\frac{d\theta}{ds}+r\frac{dr}{ds}\frac{ds}{dr}\frac{d\phi}{ds}$

$\because \sin\phi=r\frac{d\theta}{ds}$ and $\cos\phi=\frac{dr}{ds}$

$\frac{dp}{dr}=r(\frac{d\theta}{ds}+\frac{d\phi}{ds})=r\frac{d\psi}{ds}$....from $(3)$

$\frac{dp}{dr}=r\frac{1}{\rho}$

We know that radius of curvature$=\rho=\frac{ds}{d\phi}$

$\therefore \rho=r\frac{dr}{dp}$

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In this derivation i do not understand why does $\sin\phi=r\frac{d\theta}{ds}$ and $\cos\phi=\frac{dr}{ds}?$.Please help me.

Answer 1

Move along the curve from $P$ to a nearby point $P'$ with polar coordinates $(r+\delta r, \theta +\delta\theta)$. We denote the change in arc length $\delta s$. For $P$ and $P'$ sufficiently close together, we can assume that the curve between them approximates to a straight line.

We therefore have a triangle $OPP'$ with $OP=r,\; PP'=\delta s,\;OP'=r+\delta r,\; \angle POP'=\delta \theta\; \text{and}\; \angle OP'P=\phi.$

(Apologies for the lack of diagram — you may want to draw one!)

Now using the cosine rule, $$ \begin {align} r^2&=(r+\delta r)^2+\delta s^2-2(r+\delta r)\delta s \cos \phi \\ 2(r+\delta r)\delta s\cos\phi&=2r\delta r+\delta r^2+\delta s^2 \\ 2r\delta s \cos \phi&\approx2r\delta r \\ \cos\phi&\approx\frac{\delta r}{\delta s} \end {align} $$ And the sine rule, $$ \begin {align} \frac {\sin\phi}{r}&=\frac{\sin\delta\theta}{\delta s} \\ \sin\phi&=r\frac{\sin\delta\theta}{\delta \theta}\frac{\delta \theta}{\delta s} \\ \sin\phi&\approx r\frac{\delta \theta}{\delta s} \end {align} $$

Answer 2

It is worth going back from differential to finite difference segments. The relations you ask are seen from trig directly.

Also curvature is defined as rate of change of tangential rotation with respect to arc $s$.

$$ \frac{ 1 } { \rho } = \frac{ d \psi } {ds } $$

So the definitive relation you gave is incorrect.

BTW, rate of change of $p$ on arc is curvature multiplied by $\sqrt{r^2-p^2}. $