# How can I find the curves in the problem?

It's the first exercise in this book (It has a drawing).

Let $\rho = f(\theta)$ be the equation of a plane curve in polar coordinates. Through the pole $O$ draw a line perpendicular to the radius vector $OM$, and let $T$ and $N$ be the points where this line cuts the tangent and the normal. Find expressions for the distances $OT$, $ON$, $MN$, and $MT$ in terms of $f(\theta)$ and $f'(\theta)$.

Find the curves for which each of these distances, in turn, is constant.

I tried to find the curve with $OT$ being constant.

\begin{aligned} M (\rho\cos \theta, \rho\sin\theta)\\ T (k\sin\theta, k\cos\theta)\\ \frac{y_T-y_M}{x_T-x_M} = \frac{dy_M}{dx_M} \\ (k\cos\theta-\rho\sin\theta)(\rho'\cos\theta-\rho\sin\theta)=\\(k\sin\theta-\rho\cos\theta)(\rho'\sin\theta+\rho\cos\theta) \end{aligned}

What should I do next?

-
You need to find the point $T$ (and likewise N, M) in terms of $f(\theta), f'(\theta)$ first. –  Calvin Lin Jan 6 '13 at 16:04

We'll solve just the case for which $OT$ is constant. The missing cases you can easily complete.

Let $\vec {r}$ the position vector of point M, and $\vec{i}$ and $\vec{j}$ the versors of axes $x$ and $y$ respectively. Let the set of vectors $\{\vec{u},\vec{v}\}$ an orthonormal basis and $\rho =f(\theta)$.

See the figure below:

So $$\vec {r} = f(\theta) \vec {u} \quad (I)$$ $$\vec {u}= \cos(\theta)\vec {i} +\sin(\theta) \vec {j}$$ $$\vec {v}= \dfrac{d\vec{u}}{d\theta} = -\sin(\theta)\vec {i} +\cos(\theta) \vec {j}$$ Let $\vec {t}$ a tangent vector at point M. The vector $\vec {t}$ can be defined as: $$\vec {t} = \dfrac{d\vec{r}}{d\theta}.$$

So $$\vec {t} = f'(\theta) \vec {u} + f(\theta) \vec{v}.$$ Let's calculate $OT$ now.

We know that $$\overrightarrow{OT}=\vec {r} + \overrightarrow{MT}.\quad (1)$$ But $$\overrightarrow{MT}= \lambda \vec{t} \quad (2)$$ and $$\overrightarrow{OT}= \mu \vec{v}. \quad(3)$$

So using equations $(I)$, $(1)$, $(2)$ and $(3)$ we get: $$f(\theta) \vec{u} + \lambda[f'(\theta) \vec{u} + f(\theta) \vec{v}]= \mu \vec{v} \Rightarrow$$ $$(f(\theta)+\lambda f'(\theta)) \vec{u} + \lambda f(\theta) \vec{v}= \mu \vec{v}. \quad (4)$$ It follows that $$\lambda = - \frac{f(\theta)}{f'(\theta)} \quad (5)$$ and $$\mu = - \frac{f^2(\theta)}{f'(\theta)}. \quad (6)$$ Therefore $$OT= \frac{f^2(\theta)}{f'(\theta)}. \quad (7)$$ As $\triangle NMT$ is a right triangle we can easily calculate $ON$, $MN$, and $MT$. For example:

We know that $$ON \cdot OT = OM^2, \quad (8)$$ then using $(7)$ and $(8)$ we get: $$ON = f'(\theta). \quad (9)$$ If we want a curve for which $OT$ is constant ($k$) then $$\frac{f^2(\theta)}{f'(\theta)}=k \Rightarrow$$ $$\Rightarrow k \dfrac{df}{d\theta}=f^2(\theta) \Rightarrow$$ $$\Rightarrow k \frac{d\rho}{\rho^2}= d\theta \Rightarrow$$ $$\Rightarrow -k \rho^{-1}=\theta + C$$ and we are done.

-
Thanks! Can you explain a bit why $\vec {v}= \dfrac{d\vec{u}}{d\theta}$ but $\vec {t} = \dfrac{d\vec{r}}{d\theta}$? And what tools did you use to draw the beautiful figure? –  zjk Jan 11 '13 at 13:03
@zjk. You're welcome. If you take the derivative of a versor you'll get another versor which is orthogonal to the first one. $$\vec {u} \cdot \vec {u} =1 \Rightarrow \dfrac{d (\vec {u} \cdot \vec {u})}{d \theta} \Rightarrow$$ $$2 \vec {u} \cdot \dfrac{d \vec {u}}{d \theta}=0 \Rightarrow \vec {u} \perp \dfrac{d \vec {u}}{d \theta}$$ –  RicardoCruz Jan 11 '13 at 14:42
@zjk. As regards to $\dfrac{d \vec {r}}{d \theta}$, if you define two points $M$ and $M'$ as $M = O + \vec {r}(\theta)$ and $M' = O + \vec{r}(\theta + \Delta \theta)$ you will get a straight line defined by $MM'$ which is a secant line of the curve $\rho = f(\theta)$. When $\Delta \theta \to 0$, then $M' \to M$ and the line defined by $MM'$ approaches to a tangent line. So if you calculate $\dfrac {d \vec{r}}{d \theta} = \lim_{\Delta \theta \to 0}\frac{\vec {r}(\theta + \Delta \theta)-\vec {r}(\theta)}{\Delta \theta}$, you will ge a tangent vector to the curve at point $M$. –  RicardoCruz Jan 11 '13 at 14:49
@zjk. As regards to the figure, it was drawn using Geogebra. –  RicardoCruz Jan 11 '13 at 14:50