Tangents to Polar Curves

calculuspolar coordinates

I am given the polar curve $r=3\cos(θ)$. I am to list all of the points (three in total) $(r,θ)$ where the tangent line is horizontal.

From $r=3\cos(θ)$, I was able to derive that

$$x = r\cos(\theta) = 3\cos(\theta)cos(\theta) = 3\cos^2(\theta)$$

$$y = r\sin(\theta) = 3\cos(\theta)\sin(\theta)$$

Therefore
$$\frac{dy}{d\theta} = 3\cos(\theta)(\cos(\theta)) + \sin(\theta)(-3\sin\theta) = 3\cos^2(\theta)-3\sin^2(\theta)$$

The question asks for where the tangent line is horizontal. I know that a tangent line is horizontal when $\frac{dy}{d\theta} = 0$, so I plugged that into my equation.

$$3\cos^2(\theta)-3\sin^2(\theta) = 0$$

$$cos^2(\theta) – sin^2(\theta) = 0$$

For the difference between $\cos^2(x)$ and $\sin^2(x)$ to the zero, then $\cos^2(x) = \sin^2(x)$. Such a phenomenon only occurs every $\frac{1}{8}$th of the unit circle.

$$:. \theta = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4 \quad etc$$

However, most of these are apparently wrong.

Plugging in $\theta = \pi/4$ to come up with $(r,\theta)$ = $(3\frac{\sqrt2}{2},\theta)$ is correct, but $3\pi/4$ and $5\pi/4 $ are not correct.

I'm confused as to what else the answers could possible be. Please help in any way you can.

Best Answer

As I mention in my comment, the horizontal line actually occurs where $\frac{dy}{dx}$ becomes $0$. Thus, since this derivative is $\frac{dy}{d\theta}$ divided by $\frac{dx}{d\theta}$, to be fully correct, you have to ensure that the denominator, i.e., $\frac{dx}{d\theta}$, is not $0$. However, that's not the case here as $\frac{dx}{d\theta} = -6\cos(\theta)\sin(\theta)$.

The actual issue is that $r = 3\cos(\theta)$ must be non-negative. This is because $r$ is the distance from the origin in the Cartesian coordinate plane and distances are always non-negative. Thus, some of your proposed values of $\theta$ are not allowed if $\cos(\theta)$ is negative. In particular, this limits the range of $\theta$ to being in the first & fourth quadrants, e.g., $-\frac{\pi}{2} \lt \theta \lt \frac{\pi}{2}$ or $\frac{3\pi}{2} \lt \theta \lt \frac{5\pi}{2}$, so values like $\frac{3\pi}{4}$ and $\frac{5\pi}{4}$ are not included as they're in the second & third quadrants.

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[Math] Polar to Parametric Equation

You can rewrite $x=r \cos \theta$ as $r=\frac{x}{\cos\theta}$ and plug that in. You immediately get $$x=\sin\theta\cos\theta+\cos^2\theta$$ Doing the same trick for $r=\frac{y}{\sin\theta}$ gives you $$y=\sin^2\theta+\sin\theta\cos\theta$$

From here on it's not hard - the slope of the tangent is $\frac{dy/d\theta}{dx/d\theta}$

[Math] Horizontal and vertical asymptotes of polar curve $r = \theta/(\pi – \theta) \, , \, \in[0,\pi]$

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))

enter image description here

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))

enter image description here

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.

Best Answer

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[Math] Polar to Parametric Equation

[Math] Horizontal and vertical asymptotes of polar curve $r = \theta/(\pi – \theta) \, , \, \in[0,\pi]$

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