Find two points on $r=\theta$ such that the tangent lines to the graph at the points are parallel

Question

I was asked by my instructor to find two different points on the curve $\gamma(t)$ given by:

$$ \gamma(t) = (t\cos(t),t\sin(t)) \\ t\in(-\infty,\infty)$$

Such that the tangent lines to $\gamma(t)$ at those points are parallel. I noticed this is an Archimedean spiral ($r=\theta$), and managed to construct the following formula for the slope of a tangent line to $\gamma(t)$ at $t=t_0$:

$$m=\frac{\sin(t_0)+t_0\cos(t_0)}{\cos(t_0)-t_0\sin(t_0)}$$

I tried to substitute $\sin(t)=\cos(t)$ in order to get an explicit solution, but it won't work. If for example I try to find two points where the tangent lines are parallel to the $x$ axis, I am left with the equation $\tan(t)=-t$, which of course I cannot solve (only Taylor can help but I prefer to avoid using it, if possible).

My instructor emphasized that I have to find 2 points explicitly, and that 2 of them are enough (meaning I don't need to find all of the points with respect to some sort of a pattern).

Thanks!

Answer 1

The curve is bilaterally symmetric: $x(t)=-x(-t)$ and $y(t)=y(-t)$.

So if tangent line is vertical or horizontal for some t, the symmetric point -t will have parallel tangent line as well.

From the tangent all points such that $\sin(t_0) + t_0 \cos(t_0)=0$ correspond to horizontal tangents and $\cos(t_0) - t_0 \sin(t_0)=0$ correspond to vertical tangents.

Answer 2

COMMENT.- You just need equality of derivatives. Since $r=\theta$ we have $\dfrac{dy}{dx}=\dfrac{\sin(\theta)+\theta\cos(\theta)}{ \cos(\theta)-\theta\sin(\theta)}$.

It follows that you need two distinct values $\theta_1$, $\theta_2$ such that $\dfrac{\sin(\theta_1)+\theta_1\cos(\theta_1)}{ \cos(\theta_1)-\theta_1\sin(\theta_1)}=\dfrac{\sin(\theta_2)+\theta_2\cos(\theta_2)}{ \cos(\theta_2)-\theta_2\sin(\theta_2)}$

Unfortunately, you have a transcendental equation $f(x)=\dfrac{\sin(x)+x\cos(x)}{\cos(x)-x\sin(x)}=a$ where $a$ can be an arbitrary real number. You cannot find exact solutions, just approximative ones. For example, for $a=1$ you can find $\theta_1\approx0.4026$ and $\theta_2\approx2.7097$. This way you have

$$f(\theta_1)\approx0.999895\\f(\theta_2)\approx0.999932$$