Given: $$x = e^{-t} \sin t$$ $$y = e^{-t} \cos t$$ Find equation of tangent line when $x = 0$
That's what i tried to solve this problem:
$$\frac{dx}{dt} = e^{-t}(\cos t - \sin t)$$ $$\frac{dy}{dt} = e^{-t}(\cos t + \sin t)$$ $$\frac{dy}{dx} = \frac{\frac{dx}{dt}}{\frac{dy}{dt}} = \frac{\sin t - \cos t}{\cos t + \sin t} = \frac{\tan x - 1}{\tan x + 1}$$ $$y_{tan} = \frac{dy_0}{dx} (x - x_0) + y_0, \text{ with } x_0 = 0$$ $$y_{tan} = \frac{\tan 0 - 1}{\tan 0 + 1} (x) + y_0$$ $$y_{tan} = -x + y_0$$ And now I need to somehow find valid values of $y_0$ (looking at the graphic I expect two of them), so we have:
$$0 = e^{-t} * \sin t$$ $$y_0 = e^{-t} * \cos t$$ Let's solve for t. $$t = \pi k, k \in \mathbb{Z}$$ $$y_0 = e^{-t}, \text{ where } k \equiv 0 (mod 2)$$ or $$y_0 = -e^{-t}, \text{ where } k \equiv 1 (mod 2)$$
And that's the end of my reasoning. I can't find any valid value of $y_0$. How am I supposed to do that?
"when $x = 0$", which happens every time $t = k \pi$ for any integer $k$. In detail, when $\mathrm{e}^{-t} \sin(t) = 0$. Since the exponential is never zero, when $\sin(t) = 0$, which is $\sin^{-1}(0) + 2\pi k$ and $\pi - \sin^{-1}(0) + 2 \pi k$, for any integer $k$. Since $\sin^{-1}(0) = 0$, these are the set of even and odd integer multiples of $\pi$, respectively. Collectively, this is all the integer multiples of $\pi$. So we want the equation of the tangent line to the parametric curve whenever $t = k \pi$ for any integer $k$.
So we have the coordinates of the points of tangency, $(x,y) = (0, \mathrm{e}^{-k\pi} \cos(k \pi)) = (0, \mathrm{e}^{-k\pi} (-1)^k)$. You have calculated \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\tan(x) -1}{\tan(x) + 1} \\ &= \frac{\tan(k\pi)-1}{\tan(k\pi)+1} \\ &= \frac{0-1}{0+1} \\ &= -1 \text{.} \end{align*} Therefore, the (point-slope form) equations for the infinitely many lines tangent to the parametric curve when $x = 0$, indexed by the integer $k$, are $$ y - (-1)^k \mathrm{e}^{-k\pi} = (-1)(x-0) \text{.} $$