Prove that the evolute of the tractrix $x=a(\cos t+\log \tan\frac{t}{2}),y=a\sin t$ is the catenary $y=a\cosh (\frac{x}{a})$
Since evolute of a curve is the envelope of the normals of that curve.I first found the normal to the curve.
$x=a(\cos t+\log \tan\frac{t}{2}),y=a\sin t$
$\frac{dy}{dx}=\tan t$
So the equation of the normal is $y-a\sin t=-\cot t(x-a\cos t-a\log \tan \frac{t}{2})……….(1)$
Now taking $t$ as a parameter,i tried to find the envelope of the normal by partial differentiating eqn $(1)$ with respect to $t$,we get
$-a\cos t=x\csc^2t-a\cot t\sin t-a\cos t\csc^2t+a\frac{\cot t}{\sin t}-a\log \tan \frac{t}{2}\csc^2t………..(2)$
I am stuck here.It has got very complicated.How can i eliminate $t$ between $(1)$ and $(2).$Please help me.
Best Answer
I'm going to use this parametric equation for a tractrix: $x(t)=a(t-\tanh t)$, $y(t)=a\text{ sech } t$
Let us call the $x$-component of the parametric equation for the evolute $X$, and the $y$-component $Y$. If we call $\phi$ the tangential angle and $R$ the radius of curvature, then $X=x-R\sin(\phi)$ and $Y=y+R\cos(\phi)$. Note that $X$, $Y$, $R$, and $\phi$ are functions of $t$.
We note that $R=\displaystyle\frac{ds}{d\phi}=\frac{s'}{\phi'}$ where $s$ is the arc length.
We know that $s'=\sqrt{x'^2+y'^2}$, or equivalently $a\sqrt{\left(1-\text{sech}^2 \text{ }t\right)^2+ \text{sech}^2 \text{ }t\tanh^2t}$. This simplifies down to $a*|\tanh t \text{ }|$
We also know that $\displaystyle\frac{d}{dt}\tan(\phi)=\left(\frac{y'}{x'}\right)'=\frac{x'y''-x''y'}{x'^2}=\frac{\left(1-\text{ sech}^2 \text{ } t\right)\left(\text{sech } t\tanh^2t-\text{sech}^3 \text{ }t\right)+2\text{ sech}^3 \text{ }t\tanh^2 t}{\left(1-\text{sech}^2 \text{ }t\right)^2}=\frac{\text{sech }t \tanh^2t}{\left(1-\text{sech}^2 \text{ }t\right)^2}=\frac{d\phi}{dt}\sec^2\phi$
So $\displaystyle\frac{d\phi}{dt}=\frac{\text{sech }t \tanh^2t}{\left(1-\text{sech}^2 \text{ }t\right)^2}*\frac{\left(1-\text{sech}^2 \text{ }t\right)^2}{\tanh^2t}=\text{sech } t$, because $\displaystyle\sec^2\phi=\frac{x'^2+y'^2}{x'^2}$
Ergo $\displaystyle R=\frac{s'}{\phi'}=\frac{a*|\tanh t \text{ }|}{\text{sech }t}$
Knowing that $\displaystyle\sin\phi=\frac{y'}{\sqrt{x'^2+y'^2}}=-\frac{\text{sech }t\tanh t }{|\tanh t \text{ }|}$ and that $\displaystyle\cos\phi=\frac{x'}{\sqrt{x'^2+y'^2}}=\frac{1-\text{sech}^2\text{ }t}{|\tanh t \text{ }|}$, we can find $X$ and $Y$.
$\displaystyle X=x-R\sin\phi=a(t-\tanh t)+\frac{a*|\tanh t \text{ }|}{\text{sech }t}* \frac{\text{sech }t\tanh t }{|\tanh t \text{ }|}=a*t $
$\displaystyle Y=y+R\cos\phi=a\text{ sech } t+\frac{a*|\tanh t \text{ }|}{\text{sech }t}*\frac{1-\text{sech}^2 \text{ }t}{|\tanh t \text{ }|}= a*\cosh t$
Changing this parametric equation into implicit form, we get that the evolute of a tractrix is a catenary: $y=a*\cosh\left(\frac{x}{a}\right)$
P.S. there is a general form for an evolute: $$X=x-y'\frac{x'^2+y'^2}{x'y''-x''y'}$$ $$Y=y+x'\frac{x'^2+y'^2}{x'y''-x''y'}$$ You can derive this formula by generalizing the process I went through to find the evolute of a tractrix.
You can also read about it here:
http://mathworld.wolfram.com/Evolute.html
http://www.wikiwand.com/en/Evolute
P.P.S. My math might be a little off, I am only a high school freshman