Question

Prove that the evolute of the tractrix $x = a(\cos t + \log \tan \dfrac{t}{2})$, $y = a\sin t$ is the catenary $y = a\cosh \left( {\dfrac{x}{a}} \right)$.

Answer 1

We will use the fact that evolute of a curve is the envelope of the normals of that curve. So, we will first find the slope of normal of the curve and then the equation of normal to the curve. After that, we will keep in LHS the factor containing $x$ and in RHS, the factors containing $t$ and find x using two such equations and finally use L’ Hopital’s theorem.

Complete step-by-step answer:
Since, we know that the evolute of a curve is the envelope of the normals of that curve.
So, let us first find the slope of normal that will be $- \dfrac{{dx}}{{dy}}$.
We can write $- \dfrac{{dx}}{{dy}} = - \dfrac{{dx}}{{dt}} \times \dfrac{{dt}}{{dy}}$ ………(1)
So, we will first find $\dfrac{{dx}}{{dt}}$:
$\dfrac{{dx}}{{dt}} = a\dfrac{d}{{dt}}\left( {\cos t + \log \tan \dfrac{t}{2}} \right)$
Since, we know that $\dfrac{d}{{dx}}\cos x = - \sin x$ and $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
Hence, \dfrac{{dx}}{{dt}} = a\left\\{ { - \sin t + \dfrac{1}{{\tan \dfrac{t}{2}}} \times \dfrac{d}{{dt}}\left( {\tan \dfrac{t}{2}} \right)} \right\\} (Chain Rule)
Now, since we know that $\dfrac{d}{{dt}}\tan x = {\sec ^2}\left( x \right)$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$
\therefore \dfrac{{dx}}{{dt}} = a\left\\{ { - \sin t + \dfrac{{\cos \dfrac{t}{2}}}{{\sin \dfrac{t}{2}}} \times \left( {{{\sec }^2}\dfrac{t}{2}} \right) \times \dfrac{d}{{dt}}\left( {\dfrac{t}{2}} \right)} \right\\} (Chain Rule)
Since, we know that $\sec x = \dfrac{1}{{\cos x}}$.
\therefore \dfrac{{dx}}{{dt}} = a\left\\{ { - \sin t + \dfrac{{\cos \dfrac{t}{2}}}{{\sin \dfrac{t}{2}}} \times \dfrac{1}{{{{\cos }^2}\dfrac{t}{2}}} \times \dfrac{1}{2}} \right\\} = - a\left( {\sin t + \dfrac{1}{{2.\sin \dfrac{t}{2}.\cos \dfrac{t}{2}}}} \right)
We also know that $2\sin x\cos x = \sin 2x$.
\therefore \dfrac{{dx}}{{dt}} = a\left\\{ { - \sin t + \dfrac{1}{{\sin t}}} \right\\} = a\left( {\dfrac{{1 - {{\sin }^2}t}}{{\sin t}}} \right).
We will now use that ${\sin ^2}x + {\cos ^2}x = 1$
$\therefore \dfrac{{dx}}{{dt}} = a\left( {\dfrac{{{{\cos }^2}t}}{{\sin t}}} \right)$ ……..(2)
Now, let us find $\dfrac{{dy}}{{dt}}$:
$\dfrac{{dy}}{{dt}} = a\dfrac{d}{{dt}}\left( {\sin t} \right)$
Since, we know that $\dfrac{d}{{dx}}\sin x = \cos x$
So, $\dfrac{{dy}}{{dt}} = a\cos t$ …….(3)
Putting (2) and (3) in (1), we will get:-
$- \dfrac{{dx}}{{dy}} = - \dfrac{{dx}}{{dt}} \times \dfrac{{dt}}{{dy}} = - \dfrac{{a{{\cos }^2}t}}{{\sin t}} \times \dfrac{1}{{a\cos t}} = - \dfrac{{\cos t}}{{\sin t}} = - \cot t$ ……(4)
Now, we have to find the equation of normal which is given by:
$y - {y_1} = \dfrac{{ - dx}}{{dy}}(x - {x_1})$.
Now, we are given that $x = a(\cos t + \log \tan \dfrac{t}{2})$, $y = a\sin t$.
So, equation of normal will be given by:
$y - a\sin t = \dfrac{{ - dx}}{{dy}}\left( {x - a\left( {\cos t + \log \tan \dfrac{t}{2}} \right)} \right)$
Now putting (4) in this, we will have:
$y - a\sin t = - \cot t\left( {x - a\left( {\cos t + \log \tan \dfrac{t}{2}} \right)} \right)$
We can rewrite it as:
$\tan t\left( {y - a\sin t} \right) = - \left( {x - a\left( {\cos t + \log \tan \left( {\dfrac{t}{2}} \right)} \right)} \right)$
On simplifying the terms on the right, we will get:
$\Rightarrow \tan t\left( {y - a\sin t} \right) = - x + a\cos t + a\log \tan \left( {\dfrac{t}{2}} \right)$
Simplifying it by bringing all terms on LHS, we get:-
$\Rightarrow x + y\tan t - a\dfrac{{{{\sin }^2}t}}{{\cos t}} - a\cos t - a\log \tan \left( {\dfrac{t}{2}} \right) = 0$
Rearranging the terms to get the following expression:
$\Rightarrow x + y\tan t - a\left( {\dfrac{{{{\sin }^2}t + {{\cos }^2}t}}{{\cos t}}} \right) - a\log \tan \left( {\dfrac{t}{2}} \right) = 0$
We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
Hence, we get the following expression:
$\Rightarrow x + y\tan t - a\dfrac{1}{{\cos t}} - a\log \tan \left( {\dfrac{t}{2}} \right) = 0$
We know that $\cos x = \dfrac{1}{{\sec x}}$
$\Rightarrow x + y\tan t - a\sec t - a\log \tan \left( {\dfrac{t}{2}} \right) = 0$ ……….…… (5)
Now this equation gives us the family of normals of the given tractrix with the parameter $t$. The envelope of the family of the equation (5) is the evolute of the given tractrix.
Now, let us partially differentiate it with respect to t, then we will get:-
$\Rightarrow y{\sec ^2}t - a\sec t\tan t - \dfrac{1}{2}a\left( {\dfrac{{{{\sec }^2}\dfrac{t}{2}}}{{\tan \dfrac{t}{2}}}} \right) = 0$
On simplifying it, we will get:-
$\Rightarrow y{\sec ^2}t - a\sec t\tan t - a\cos ect = 0$
Now, writing the function as inverses, we will get:-
$\Rightarrow \dfrac{y}{{{{\cos }^2}t}} - \dfrac{{a\sin t}}{{{{\cos }^2}t}} - \dfrac{a}{{\sin t}} = 0$
Taking a common and taking LCM to get:
$\Rightarrow \dfrac{{y\sin t - a\left( {{{\sin }^2}t + {{\cos }^2}t} \right)}}{{{{\cos }^2}t\sin t}} = 0$
We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
Hence, we get: $y\sin t - a = 0$
We can write it as: $y = \dfrac{a}{{\sin t}}$. ………….(6)
Now, putting this in (5), we will get:-
$\Rightarrow x + \left( {\dfrac{a}{{\sin t}}} \right)\tan t - a\sec t - a\log \tan \left( {\dfrac{t}{2}} \right) = 0$
Simplifying it to get:
$\Rightarrow x + a\sec t - a\sec t - a\log \tan \left( {\dfrac{t}{2}} \right) = 0$
This can be written as:
$\Rightarrow x - a\log \tan \left( {\dfrac{t}{2}} \right) = 0$
Taking the expression without x on the RHS, we will get:-
$\Rightarrow x = a\log \tan \left( {\dfrac{t}{2}} \right)$
We can rewrite it as:
$\Rightarrow \log \tan \left( {\dfrac{t}{2}} \right) = \dfrac{x}{a}$
Now, we know that $\log a = b \Rightarrow a = {e^b}$
Hence, we get: $\tan \left( {\dfrac{t}{2}} \right) = {e^{\left( {\dfrac{x}{a}} \right)}}$ ……………..(7)
Now, consider (6): $y = \dfrac{a}{{\sin t}}$.
We know that $\sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$.
Hence, we have: $y = a\left( {\dfrac{{1 + {{\tan }^2}\left( {\dfrac{t}{2}} \right)}}{{2\tan \left( {\dfrac{t}{2}} \right)}}} \right)$
Using (7) in this, we will get:-
$\Rightarrow y = a\left( {\dfrac{{1 + {e^{2\left( {\dfrac{x}{a}} \right)}}}}{{2{e^{\left( {\dfrac{x}{a}} \right)}}}}} \right)$
Taking ${e^{\left( {\dfrac{x}{a}} \right)}}$ common from both numerator and denominator, we will get:-
$\Rightarrow y = a\left( {\dfrac{{{e^{\left( {\dfrac{x}{a}} \right)}} + {e^{ - \left( {\dfrac{x}{a}} \right)}}}}{2}} \right)$
We will now use the formula: $\cosh \left( x \right) = \dfrac{{{e^x} + {e^{ - x}}}}{2}$.
Hence, we have:-
$y = a\cosh \left( {\dfrac{x}{a}} \right)$
Hence, proved that envelope of the family of normals, the evolute of the given tractrix is the catenary $y = a\cosh \left( {\dfrac{x}{a}} \right)$.

Note: Evolute is the locus of the Centre of curvature. But here it is defined as the envelope of normals of that curve. Therefore the normals of the curve touch the evolute.
Additional Information: Tractrix is sometimes used to describe the solid formed by rotation of a curve about the x-axis.