calculuspitangent linetrigonometry
I am having problems determining the horizontal and vertical tangents for $r=1-\sin(\theta)$.
I thought the tangent lines occurred at $\frac{5\pi}6$, $\frac\pi6$, $\frac{3\pi}2$ while the vertical tangent lines occurred at $\frac{11\pi}6$ and $\frac{7\pi}6$.
I solved for $x$ to get $x=\cos(\theta)-\sin(\theta)\cdot\cos(\theta)$ which produced $-\frac{\sqrt3}4$ for $\frac{5\pi}6$, $\frac{\sqrt3}4$ for $\frac\pi6$, $0$, and $\frac{3\pi}2$.
An asymptote is going to mean that $r\rightarrow\infty$, which happens when the denominator approaches zero. Vertical or horizontal means that $\theta$ is a multiple of $\pi/2$ (vertical for even, horizontal for odd multiples). The only asymptote would therefore be when $\theta\rightarrow\pi$, which would be a horizontal one above (and parallel to) the negative $x$ axis.
Now as $\theta\rightarrow\pi$, $$ y = r\sin\theta = \frac{\theta\sin\theta}{\pi-\theta} = \frac{\theta\sin(\pi-\theta)}{\pi-\theta} \rightarrow \theta \rightarrow \pi $$ so the horizontal asymptote is $y=\pi$.
For visualization, we can reason that as $\theta$ increases from $0$ to $\pi$, $r$ increases from $0$ to $\infty$, crossing $1$ at the positive $y$ axis. Here are some plots for $\theta\in[0,.6\pi]$ and $[0,.99\pi]$ made with sage (online):
x_coords = [t/(pi-t)*cos(t) for t in srange(0, 0.6*pi, 0.02)]
y_coords = [t/(pi-t)*sin(t) for t in srange(0, 0.6*pi, 0.02)]
list_plot(zip(x_coords, y_coords))
x_coords = [t/(pi-t)*cos(t) for t in srange(0, 0.99*pi, 0.02)]
y_coords = [t/(pi-t)*sin(t) for t in srange(0, 0.99*pi, 0.02)]
list_plot(zip(x_coords, y_coords))
Horizontal and vertical tangents can be found by finding the roots of $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$ respectively (assuming that they are never simultaneously zero, in which case we'd have to resort to higher order derivatives). A simple way to find these roots is to observe that $$r = \left(\frac{\pi}{\theta}-1\right)^{-1} \implies \frac{dr}{d\theta} = -\left(\frac{\pi}{\theta}-1\right)^{-2} \cdot \left(-\pi\theta^{-2}\right) = \frac{\pi}{(\theta-\pi)^2} $$ so that $$ 0 = \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta-r\sin\theta \iff \tan\theta = \frac{1}{r} \frac{dr}{d\theta} = \frac{\pi}{\theta(\pi-\theta)} $$ and $$ 0 = \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta+r\cos\theta \iff -\cot\theta = \frac{1}{r} \frac{dr}{d\theta} = \frac{\pi}{\theta(\pi-\theta)} $$ which show us that they never vanish simultaneously. From the shape of $\frac{\pi}{\theta(\pi-\theta)}$, which is symmetric and positive on $(0,\pi)$ with global minimum at $\frac{\pi}{2}$ and vertical asymptotes at the endpoints, we see that it will meet $\tan\theta$ at exactly one point $\theta_0\in(0,\frac{\pi}{2})$; thus, there is one unique vertical tangent to our curve. From the graph above, we might estimate $$ \tan\theta_0\approx\frac{.4}{.25}=1.6 \implies \theta_0 \approx \frac{\pi}{2}-\sqrt{\frac{\pi}{2}\left(\frac{\pi}{2}-\frac{5}{4}\right)} \approx .8609 $$ where we used the quadratic equation to deduce $\theta$ from $\tan\theta$. However, the actual solution seems to be closer to $0.97803904765198235$:
t=var('t'); find_root(t*(pi-t)*tan(t)==pi, 0, pi/2)
For horizontal tangents, the same function $\frac{\pi}{\theta(\pi-\theta)}$ must now meet $-\cot\theta$, which is equivalent to the equation $$\tan\theta=\theta\left(1-\frac{\pi}{\theta}\right) $$ which on $[0,\pi]$ has only solutions at the two endpoints, i.e., on our curve, at the origin and at the horizontal asymptote found above.
$$\dfrac{dy}{d\theta}\cdot \frac{d\theta}{dx} = \dfrac{dy}{dx}$$ and we want to know when $\dfrac{dy}{dx} = 0$.
$$\frac{dx}{d\theta} = \frac{\cos3\theta}{r},\quad \frac{dy}{d\theta} = \frac{\sin 3\theta}{r} \implies \dfrac{dy}{dx} = \frac{\sin{3\theta}}{r}\cdot \dfrac{r}{\cos 3\theta}$$
Now solve for when $$\require{cancel} \dfrac{dy}{dx} = \frac{\sin{3\theta}}{\cancel{r}}\cdot \dfrac{\cancel{r}}{\cos 3\theta} = = \tan(3\theta)= 0$$
That's going to happen when $\sin 3\theta = 0$ (and note that when $\sin{3\theta} = 0 \iff 3\theta = k \pi \iff \theta = k\pi/3\;$ ($k$ any integer), we have that $\cos 3\theta = \pm 1\iff 3\theta = k\pi \iff \theta = k\pi/3$.)
The points at which the tangents are horizontal and/or vertical are given in polar coordinates, $(r, \theta)$.
Best Answer
Vertical tangents: Your values of $\theta$ are correct, so we can find the $x$ and $y$ coordinates of the intersections by just plugging into $x=\cos\theta(1-\sin\theta)$ and $y=\sin\theta(1-\sin\theta)$. That gives $$ (x,y) = (\pm \frac{3\sqrt{3}}{4},-\frac{3}{4}) $$ So the tangent lines are given by $x=\pm \frac{3\sqrt{3}}{4}$, and they intersect the curve at $y=-\frac34$.
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