Question

Find the length of normal at any point to the curve, $y = c\cosh \left( {\dfrac{x}{c}} \right)$ .
(A) Fixed
(B) $\dfrac{{{y^2}}}{{{c^2}}}$
(C) $\dfrac{{{y^2}}}{c}$
(D) $\dfrac{y}{{{c^2}}}$

Answer 1

In the given curve, $'c'$ is a constant and $\cosh$ is a hyperbolic function. Use the relation $y\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}}$ for finding the length of the normal to a curve. Now use the chain rule to evaluate the first derivative of the function $y$. You can use the identity of the hyperbolic function ${\cosh ^2}x - {\sinh ^2}x = 1$to simplify your derivative.

Complete step-by-step answer:
We are given an equation of curve $y = c\cosh \left( {\dfrac{x}{c}} \right)$ and we have to calculate the length of normal at any point to the curve.
Firstly, we need to understand the concept of normal of a curve. If you differentiate the equation of a curve, you will get a formula for the gradient of the curve. Before you learnt differentiation, you would have found the gradient of a curve by drawing a tangent and measuring the gradient of this. This is because the gradient of a curve at a point is equal to the gradient of the tangent at that point. The normal to the curve is the line perpendicular (at right angles) to the tangent to the curve at that point. Remember, if two lines are perpendicular, the product of their gradients is $- 1$ .
The curve consists of a hyperbolic function. The hyperbolic trigonometric functions extend the notion of the parametric equations for a unit circle $\left( {x = \cos t{\text{ and }}y = \sin t} \right)$ to the parametric equations for a hyperbola, which yield the following two fundamental hyperbolic equations:
$x = \cosh a = \dfrac{{{e^a} + {e^{ - a}}}}{2},y = \sinh a = \dfrac{{{e^a} - {e^{ - a}}}}{2}$
We know that, for a curve $y = f\left( x \right)$ , length of normal from the curve to the y-axis $\left( L \right) = y\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}}$
And we also know that $\dfrac{{d\left( {\cosh x} \right)}}{{dx}} = \sinh x$ and ${\cosh ^2}x - {\sinh ^2}x = 1 ………….(i)$
So, to find the length of the normal from the above formula, we will find $\dfrac{{dy}}{{dx}}$ first.
$y = f\left( x \right) \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {c\cosh \left( {\dfrac{x}{c}} \right)} \right)}}{{dx}}$
Now, by using the chain rule of differentiation, i.e. $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {c\cosh \left( {\dfrac{x}{c}} \right)} \right)}}{{d\left( {\dfrac{x}{c}} \right)}} \times \dfrac{{d\left( {\dfrac{x}{c}} \right)}}{{dx}}$
After solving this differentiation using (i), we get:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {c\cosh \left( {\dfrac{x}{c}} \right)} \right)}}{{d\left( {\dfrac{x}{c}} \right)}} \times \dfrac{{d\left( {\dfrac{x}{c}} \right)}}{{dx}} = c\sinh \dfrac{x}{c} \times \dfrac{1}{c} = \sinh \dfrac{x}{c}$
Now let’s put this value of $\dfrac{{dy}}{{dx}}$ in the formula of the length of normal
Therefore, $L = y\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} = y\sqrt {1 + {{\left( {\sinh \dfrac{x}{c}} \right)}^2}} = y\sqrt {1 + {{\sinh }^2}\dfrac{x}{c}}$
From relation (i), we can say that ${\cosh ^2}x = 1 + {\sinh ^2}x$. Now we can rewrite the above expression as:
$\Rightarrow L = y\sqrt {1 + {{\sinh }^2}\dfrac{x}{c}} = y\sqrt {{{\cosh }^2}\dfrac{x}{c}} = y \times \cosh \dfrac{x}{c}$
But according to the given information $y = c\cosh \left( {\dfrac{x}{c}} \right) \Rightarrow \dfrac{y}{c} = \cosh \left( {\dfrac{x}{c}} \right)$
Therefore, we get: $\Rightarrow L = y \times \cosh \dfrac{x}{c} = y \times \dfrac{y}{c} = \dfrac{{{y^2}}}{c}$

So, the correct answer is “Option C”.

Note: Understanding the concept of a hyperbolic function and normal of a curve is very important for solving this question. An alternative approach to this problem is to use the exponential form of hyperbolic function, i.e. $x = \cosh a = \dfrac{{{e^a} + {e^{ - a}}}}{2}{\text{ and }}y = \sinh a = \dfrac{{{e^a} - {e^{ - a}}}}{2}$ to change the given function $y = f\left( x \right)$ .