Slopes of polar curves

Three quick questions from earlier lessons.

On the circle $r = 2$ , $\dfrac{dr}{d\theta} = 0$ everywhere. Does that make every tangent horizontal?

Turn the polar curve into a parametric one, with $\theta$ as the parameter.

x = r\cos\theta

,

y = r\sin\theta

, so

\dfrac{dy}{dx} = \dfrac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}

, where

r' = \dfrac{dr}{d\theta}

.

$r = 1 + \sin\theta$ at $\theta = 0$ .

r = 1

,

r' = 1

r = 1, r′ = cos 0 = 1.

\dfrac{dy}{dx} = \dfrac{1}{1}

Top: 1·0 + 1·1; bottom: 1·1 − 1·0.

= 1

The slope.

Your turn: $r = 3$ at $\theta = \dfrac{\pi}{2}$ (top of the circle).

slope

0

r′ = 0: (0 + 3·0)/(0 − 3·1).

Find the slope $\dfrac{dy}{dx}$ of the polar curve $r = 8 + \cos\theta$ at $\theta = \dfrac{\pi}{2}$ .
Find the slope $\dfrac{dy}{dx}$ of the polar curve $r = 5 + \cos\theta$ at $\theta = \dfrac{\pi}{2}$ .

Where does the polar curve $r = f(\theta)$ , for instance $r = 1 + \sin\theta$ , have a horizontal tangent?
1. Where $\dfrac{dy}{d\theta} = 0$ and $\dfrac{dx}{d\theta} \ne 0$
2. Where $\dfrac{dr}{d\theta} = 0$
3. Where $\dfrac{dx}{d\theta} = 0$

Three questions to check you have got it.

Find the slope $\dfrac{dy}{dx}$ of the polar curve $r = 8 + \cos\theta$ at $\theta = \dfrac{\pi}{2}$ .
Find the slope $\dfrac{dy}{dx}$ of the polar curve $r = 5 + \cos\theta$ at $\theta = \dfrac{\pi}{2}$ .
Find the slope $\dfrac{dy}{dx}$ of the polar curve $r = 2 + \cos\theta$ at $\theta = \dfrac{\pi}{2}$ .

Polar slopes

$\dfrac{dy}{dx} = \dfrac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$ .

Watch out: $dr/d\theta$ is not the slope.

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