Angle between the radius vector and tangent

Angle between the radius vector drawn to a point on a curve and the tangent to the curve at that point. Let the equation of the curve in polar coordinates be $\rho = f(\theta)$.

Let P be any fixed point $(\rho,\theta)$ on the curve. If $\theta$, which we assume as the independent variable, takes on an increment $\Delta \theta$, then $\rho$ will take on a corresponding increment $\Delta \rho$.

Figure 6.8: Angle between the radius vector drawn to a point on a curve and the tangent to the curve at that point.

Denote by Q the point $(\rho + \Delta \rho,\theta + \Delta \theta)$. Draw PR perpendicular to OQ. Then $OQ = \rho + \Delta \rho$, $PR = \rho \sin\Delta \theta$, and $OR = \rho \cos\Delta \theta$. Also,

$$\tan PQR = \frac{PR}{RQ} = \frac{PR}{OQ - OR} = \frac{\rho \sin \Delta \theta}{\rho + \Delta \rho - \rho \cos \Delta \theta}.$$

Denote by $\psi$ the angle between the radius vector OP and the tangent PT. If we now let $\Delta \theta$ approach the limit zero, then

(a)

the point Q will approach indefinitely near P;

(b)

the secant PQ will approach the tangent PT as a limiting position; and

(c)

the angle PQR will approach $\psi$ as a limit.

Hence

$$\tan\ \psi =\ \lim_{\Delta \theta \to 0} \frac{\rho \Delta \thet... ...\frac{\rho \Delta \theta}{2 \rho \sin^2 \frac{\Delta \theta}{2} + \Delta \rho}$$

(since, from 39, §1.1, $\rho - \rho \cos \Delta \theta = \rho (1 - \cos \Delta \theta)= 2 \rho \sin^2 \frac{\Delta \theta}{2}$). Dividing both numerator and denominator by $\Delta \theta$, this is

$$=\ \lim_{\Delta \theta \to 0} \frac{\frac{\rho \sin \Delta \theta... ...lta \theta}{2}}{\frac{\Delta \theta}{2}} + \frac{\Delta \rho}{\Delta \theta}}.$$

Since

$$\lim_{\Delta \theta \to 0} \left ( \frac{\Delta \rho}{\Delta \the... ...\lim_{\Delta \theta \to 0} \left ( \sin \frac{\Delta \theta}{2} \right ) = 0,$$

also

$$\lim_{\Delta \theta \to 0} \left ( \frac{\sin \Delta \theta}{\Delta \theta} \right ) = 1$$

and

$$\lim_{\Delta \theta \to 0} \frac{\sin \frac{\Delta \theta}{2}}{\frac{\Delta \theta}{2}} = 1$$

by §3.10, we have

$$\tan\ \psi = \frac{\rho}{\frac{d\rho}{d\theta}}$$ (6.12)

From the triangle OPT we get

$$\tau = \theta + \psi.$$ (6.13)

Having found $\tau$, we may then find $\tan\,\tau$, the slope of the tangent to the curve at P. Or since, from (6.13),

$$\tan \tau = \tan (\theta + \psi) = \frac{\tan \theta + \tan \psi}{1 - \tan \theta \tan \psi}$$

we may calculate $\tan\, \psi$ from (6.12) and substitute in the formula

$${\rm slope\ of\ tangent}\ = \tan \tau = \frac{\tan \theta + \tan \psi}{1 - \tan \theta \tan \psi}.$$ (6.14)

Example 6.7.1   Find $\psi$ and $\tau$ in the cardioid $\psi = a(1 -\cos\,\theta)$. Also find the slope at $\theta = \frac{\pi}{6}$.

Solution. $\frac{d\psi}{d\theta} = a \sin \theta$. Substituting in (6.12) gives

$$\tan \psi = \frac{\rho}{\frac{d\rho}{d\theta}} = \frac{a(1 - \c... ...{2}}{2a \sin \frac{\theta}{2} \cos \frac{\theta}{2}} = \tan \frac{\theta}{2},$$

by items 39 and 37, §1.1. Since $\tan \psi = \tan \frac{\theta}{2}$, we have $\psi = \frac{\theta}{2}$.

Substituting in (6.13), $\tau = \theta + \frac{\theta}{2} = \frac{3\theta}{2}$. so

$$\tan\, \tau = \tan \frac{\pi}{4} = 1.$$

To find the angle of intersection $\phi$ of two curves $C$ and $C'$ whose equations are given in polar cooordinates, we may proceed as follows:

Figure 6.9: The angle between two curves.

angle TPT$'$ = angle OPT$'$ - angle OPT,

or, $\phi = \psi' - \psi$. Hence

$$\ \tan\, \phi = \frac{\tan\, \psi' - \tan\, \psi}{1 + \tan\, \psi' \tan\, \psi},$$ (6.15)

where $\tan\, \psi'$ and $\tan\, \psi$ are calculated by (6.12) from the two curves and evaluated for the point of intersection.

Example 6.7.2   Find the angle of of intersection of the curves $\rho = a\sin\, 2\theta$, $\rho = a\cos\, 2\theta$.

Solution. Solving the two equations simultaneously, we get at the point of intersection

$$\tan\, 2\theta = 1, \,\ \ \ \ 2\theta = 45^o=\pi/4,\ \ \ \theta = \frac{45}{2}^o=\pi/8.$$

From the first curve, using (6.12),

$$\tan \psi' = \frac{1}{2} \tan 2 \theta = \frac{1}{2},$$

for $\theta = \frac{45}{2}^o=\pi/8$. From the second curve,

$$\tan \psi = -\frac{1}{2} \cot 2 \theta = -\frac{1}{2},$$

for $\theta = \frac{45}{2}^o=\pi/8$.

Substituting in ((6.15),

$$\tan \psi = \frac{ \frac{1}{2} + \frac{1}{2} }{ 1 - \frac{1}{4} } = \frac{4}{3}.$$

therefore $\psi = \arctan \frac{4}{3}$.