Find all angles such that $r(\theta)$ has a horizontal tangent

Question

Find all angles $\theta$ on the interval $[0, 2\pi)$ where the curve given by $$r(\theta) = 10+9\sin(\theta)$$ has a horizontal tangent.

My approach:

$r'(\theta) = 9\cos(\theta)$ $$\frac{dy}{dx} = \frac{r'(\theta)\sin(\theta)+r(\theta)\cos(\theta)}{r'(\theta)\cos(\theta)-r(\theta)\sin(\theta)}$$

For horizontal tangent, the numerator must be $0$, so we get

$9\cos(\theta)\sin(\theta)+(10+9\sin(\theta))\cos(\theta) = 0$

$9\sin(\theta)\cos(\theta)+5\cos(\theta) = 0$

and from this I obtain that $\theta=n\pi-\frac{\pi}{2}$ and my solutions $\theta=\frac{\pi}{2}$ and $\theta = \frac{3\pi}{2}$

I don't have the solution, and I'm fairly new to polar curves and tangent lines. Is my approach correct?

Answer 1

As I pointed out in my comment, you're missing some solutions. Your last equation

$$(5+9\sin\theta)\cos\theta=0$$

factorizes as

$$\cos\theta=0 \; \; \text{ or } \; \; \sin\theta=\frac{-5}{9}\; .$$

You only gave the solutions to the first part. But there are two solutions to the last part as well. I've marked the solutions in the figure below:

In cartesian coordinates they are

$$(0,19) \;\; (0,-1) \;\; \left(\frac{5\sqrt{56}}{9},-\frac{25}{9}\right) \;\; \left(-\frac{5\sqrt{56}}{9},-\frac{25}{9}\right) \; .$$

Answer 2

Defining a curve $\gamma$ in the $(x,y,)$-plane by its polar representation $\theta\mapsto r(\theta)>0$ is tantamount to the parametric representation $$\gamma:\quad\theta\mapsto\bigl(x(\theta),y(\theta)\bigr):=\bigl(r(\theta)\cos\theta, \>r(\theta)\sin\theta\bigr)\ .$$ The tangent to $\gamma$ is horizontal when $$y'(\theta)=r'(\theta)\sin\theta+r(\theta)\cos\theta=0\ .$$ In your example this means that $$0=9\cos\theta\sin\theta+(10+9\sin\theta)\cos\theta=(10+18\sin\theta)\cos\theta\ .$$ Now determine the $\theta$-values mod $2\pi$ that satisfy this equation.