Computing Slopes of Tangent Lines

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To compute slopes of tangent lines to a polar curve $r=f(\theta)$ , we treat it as a parametrized curve with $\theta=t$ and $r=f(t)$ . (Equivalently, we can use $\theta$ as our parameter). This means that

$x= r \cos(\theta) = f(t) \cos(t); \qquad y = r \sin(\theta) = f(t) \sin(t).$ Taking derivatives we get

$dx/dt = -f(t) \sin(t) + f'(t) \cos(t); \qquad dy/dt = f(t)\cos(t) + f'(t) \sin(t).$ That's enough information for us to compute

$dy/dx$ :

$\begin{eqnarray*}\frac{dy}{dx} &=& \frac{dy/dt}{dx/dt} \cr &=& \frac{f(t)\cos(t) + f'(t) \sin(t)}{-f(t) \sin(t) + f'(t) \cos(t)} \cr &=& \frac{r \cos(\theta) + r' \sin(\theta)}{-r\sin(\theta) + r' \cos(\theta)}, \end{eqnarray*}$ where

$r'$ means

$dr/d\theta$ .

In the following video, we apply this method to compute $dy/dx$ as a function of $\theta$ for a circle $r=2$ and for a cardioid $r=1+\sin(\theta)$ .