Question

find the slope for r=6/theta, theta=pi/3

Answer 1

\[\large\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\]
where \(\dfrac{dy}{dx}\) denotes the slope of the tangent line.

Recall that when converting from rectangular to polar coordinates, you have the relations
\[\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\]
Keep in mind that \(r\) is a function of \(\theta\). Denote this by \(r=f(\theta)\).
\[\begin{cases}x=f(\theta)\cos\theta\\y=f(\theta)\sin\theta\end{cases}\]
Then \(f'(\theta)=\dfrac{dr}{d\theta}\).

Differentiate both equations with respect to \(\theta\):
\[\begin{cases}
\dfrac{dx}{d\theta}=f'(\theta)\cos\theta-f(\theta)\sin\theta=\dfrac{dr}{d\theta}\cos\theta-r\sin\theta\\\\
\dfrac{dy}{d\theta}=f'(\theta)\sin\theta+f(\theta)\cos\theta=\dfrac{dr}{d\theta}\sin\theta+r\cos\theta
\end{cases}\]
which gives
\[\large\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\cos\theta-r\sin\theta}{\frac{dr}{d\theta}\sin\theta+r\cos\theta}\]