The Tractrix is a curve that passes through the point (1, 0) of the horizontal axis and has the property that the length of the segment of the tangent line, at any point on the curve, between the point of tangency and the vertical axis is constant.
To find a parametrization of the tractrix in the previous figure, it is suggested to use as a parameter at the angle t, and you must use derivatives and differential equations to get:
$α (\theta) = (sin (\theta), cos (\theta) + ln (tan (\theta / 2)))$,
where $\theta \in [0, π / 2]$.
$$l: m=\dfrac{y-y_t}{x}=y'(x)$$ $$\Rightarrow y'(x)=\dfrac{y-y_t}{x} \hspace{2cm} (1) $$
We know that:
$x=a\cdot sin(\theta) \hspace{2cm} (i)$
and
$$a^2=x^2+(y_t-y)^2$$ $$\Rightarrow a^2-x^2=(y_t-y)^2$$ $$\Rightarrow \sqrt{a^2-x^2}=y_t-y$$ $$\Rightarrow \sqrt{a^2-x^2}+y=y_t \hspace{2cm} (2)$$ substituting (2) in (1) $$y'(x)=\dfrac{-\sqrt{a^2-x^2}}{x}$$ $$\Rightarrow \dfrac{dy}{dx} \dfrac{-\sqrt{a^2-x^2}}{x}$$ $$\Rightarrow dy = \dfrac{-\sqrt{a^2-x^2}}{x}dx$$ $$\Rightarrow \int dy = \int \dfrac{-\sqrt{a^2-x^2}}{x}dx$$ $$\Rightarrow y=a \cdot \ln\left(\dfrac{a+\sqrt{a^2-x^2}}{x}\right)-\sqrt{a^2-x^2}$$ $$\Rightarrow y=a \cdot \ln\left(\dfrac{a}{x}\left(1+\sqrt{1-\dfrac{x^2}{a^2}}\right)\right)-a \cdot \sqrt{1-\dfrac{x^2}{a^2}}$$
$$\Rightarrow y=a \cdot \ln\left(\dfrac{1}{\sin(\theta)}\left(1+\sqrt{1-\sin^2(\theta)}\right)\right)-a \cdot \sqrt{1-\sin^2(\theta)} \hspace{2cm} by (i)$$ $$\Rightarrow y=a \cdot \ln\left(\dfrac{1}{\sin(\theta)}\left(1+\cos(\theta)\right)\right)-a \cdot \cos(\theta)$$ $$\Rightarrow y=a \cdot \ln\left(\cot\left(\dfrac{\theta}{2}\right)\right)-a \cdot \cos(\theta)$$
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