How Do I Find Tangent and Normal Lines for Polar Equations?

Question

Given the equation $r=1+\sin(\theta)$, I am trying to compute:

(a) $\frac{dy}{dx}$

(b) the equation of the tangent and normal lines to the curve at the indicated θ–value (which is $\frac\pi6$ in this case).

Computing $\frac{dy}{dx}$ is easy, since it involves simply plugging everything into the equation $\frac{(f'(\theta)\sin(\theta) + f(\theta)\cos(\theta)) }{ (f'(\theta)\cos(\theta) - f(\theta)\sin(\theta)}$.

When I used this equation, I got a $\frac{dy}{dx}$ of $\frac{\cos(\theta)(2\sin(\theta)+1) }{ \cos^2(\theta)-\sin(\theta)-\sin^2(\theta)}$.

At this point, I am stuck. I can't figure out how to compute the tangent and normal lines from here and I am getting an indeterminate form $\frac{\sqrt3} 0$ whenever I plug in the value of $\frac\pi6$.

Could someone please assist me in finding these equations? Thanks in advance!

Answer 1

$$\frac{dy}{dx}=\frac{\frac{dr}{d\theta} \sin \theta + r \cos \theta} {\frac{dr}{d\theta} \cos \theta - r \sin \theta}$$

For $r = 1+\sin \theta, \quad \dfrac{dr}{d\theta} = \cos \theta$. So

\begin{align} \frac{dy}{dx} &=\frac{\cos \theta \sin \theta + r \cos \theta} {\cos^2 \theta - r \sin \theta} \\ &=\frac{\sin \theta + r} {\cos \theta - r \tan \theta} \\ \end{align}

At $\theta_0 = \frac \pi 6, \quad \sin \theta_0 = \frac 12, \quad \cos \theta_0 = \frac{\sqrt{3}}{2}, \quad \quad r_0 = \frac 32$ $\quad x_0 = r_0 \cos \theta_0 = \frac{3\sqrt 3}{4}, \quad y_0 = r_0 \sin \theta_0 = \frac 34$.

So $\displaystyle m =\frac{dy}{dx} =\frac{\frac 12 + \frac 32} {\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}} = undefined$

In other words, the tangent line is a vertical line.

So the tangent line at $\theta_0 = \frac \pi 6$ would be \begin{align} x &= x_0 \\ r \cos \theta &= \frac{3\sqrt 3}{4} \\ r &= \frac{3\sqrt 3}{4 \cos \theta} \end{align}

The equation of the normal line would be computed using $y = y_0$

Answer 2

With certain exceptions, there are two ways to write an equation of a given line by using a derivative at a point $(x_1,y_1)$. You're probably thinking of one of them already, but the other is just as valid.

\begin{align} y - y_1 = \frac{dy}{dx}(x - x_1), \tag1 \\ x - x_1 = \frac{dx}{dy}(y - y_1). \tag2 \end{align}

The exceptions are that Equation $(1)$ does not work for lines parallel to the $y$ axis and Equation $(2)$ does not work for lines parallel to the $x$ axis. But you only need one equation that works, so find one you can use and try using it.