5. Tangents

Tangent parallel to the initial line $\dfrac{dy}{d\theta} = 0$

Tangent perpendicular to the initial line $\dfrac{dx}{d\theta} = 0$

Example

Find the equation of the tangent to the curve $r = a\sin 2\theta$ parallel to the initial line

where $0 \leq \theta \leq \dfrac{\pi}{2}$

Answer

when $\dfrac{dy}{d\theta} = 0$

$$r\sin \theta = y$$

so

$$y = a\sin \theta \sin 2\theta$$

using double angle formula

$$y = 2a\sin^2 \theta \cos \theta = 2a(\cos \theta - \cos^3 \theta)$$

differentiate

$$\dfrac{dy}{d\theta} = 2a(-\sin \theta + 3\sin \theta \cos^2 \theta)$$

pull out $\sin \theta$

$$\dfrac{dy}{d\theta} = 2a\sin\theta(3\cos^2 \theta - 1)$$

for $\dfrac{dy}{d\theta} = 0$ both

$$\sin \theta = 0, \quad \quad \cos^2 \theta = \dfrac{1}{3}$$

looking at $\sin \theta = 0$

$$\theta = 0, \quad \quad r = 0$$

this isn't a solution since $r > 0$

looking at $\cos^2 \theta = \dfrac{1}{3}$

$$\cos \theta = \dfrac{1}{\sqrt{3 }}, \quad \quad \sin \theta = \dfrac{\sqrt{2}}{\sqrt{3}}$$

using double angle formula for $r = a \sin 2\theta$

$$r = 2a\sin \theta \cos \theta$$

subbing in $\sin \theta$ and $\cos \theta$

$$r = 2a \cdot \dfrac{\sqrt{2}}{\sqrt{3}} \cdot \dfrac{1}{\sqrt{3}} = \dfrac{2a \sqrt{2}}{3}$$

if you plot this you will get a circle which intersects $r = a\sin \theta$ at the tangent points

now since $y = r\sin \theta$

$$y = \dfrac{2a \sqrt{2}}{3} \cdot \dfrac{\sqrt{2}}{\sqrt{3}} = \dfrac{4a}{3\sqrt{3}}$$

now to get the equation of the tangent we sub $y$ into $r = \dfrac{y}{sin\theta}$

$$r = \dfrac{4a}{3\sqrt{3}} \cosec \theta$$

on a graph this looks like