Mumbai University > Electronics and Telecommunication > Sem 4 > Applied Maths 4

Marks: 6M

Year: May 2014

To find y = f(x) such that

$$\int\limits_{x_{1}}^{x_{2}} Fdx = \int\limits_{x_{1}}^{x_{2}} y\sqrt{1 + (y')^2}dx.........(1)$$

is minimum subject to the condition

$$\int\limits_{x_{1}}^{x_{2}} Gdx = \int\limits_{x_{1}}^{x_{2}} (\sqrt{1 + (y')^2})dx = l.........(2)$$

Lagrange's method is given as $\int\limits_{x_{1}}^{x_{2}} (F + \lambda G)dx$ where $\lambda$ is the Lagrange's multiplier

Hence, multiplying eq(2) by $\lambda$ and adding equation 1 we get,

$$\therefore H = F + \lambda G = \int\limits_{x_{1}}^{x_{2}} (y + \lambda) (\sqrt{1 + (y')^2})dx.........(3)$$

Since the integrand is free from x, we use $F - y' \frac{\partial F}{\partial y'} = c.........(4)$

Where $F = H = (y + \lambda)(\sqrt{1 + (y')^2})........(5)$

$$\therefore \frac{\partial F}{\partial y'} = (y + \lambda) \frac{y'}{\sqrt{1 + (y')^2}}$$

From equation (4) and (5) we get,

$$(y + \lambda)(\sqrt{1 + (y')^2}) - y' * (y + \lambda) * \frac{y'}{\sqrt{1 + (y')^2}} = c \\ \frac{(y + \lambda) (1 + (y')^2) - (y + \lambda)(y')^2}{\sqrt{1 + (y')^2}} = c \\ \therefore \frac{y + \lambda}{\sqrt{1 + (y')^2}} = c$$

Squaring both the sides we get,

$$(y + \lambda)^2 = c^2(1 + y')^2 \\ i.e. y'^2 = \frac{(y + \lambda)^2 - c^2}{c^2} \\ \therefore y' = \frac{\sqrt{(y + \lambda)^2 - c^2}}{c} \\ \therefore \frac{c}{\sqrt{(y + \lambda)^2 - c^2}}dy = dx$$

Integrating above equation we get,

$$c * cosh^{-1} \bigg(\frac{y + \lambda}{c}\bigg) = x + c' \\ \therefore cosh^{-1} \bigg(\frac{y + \lambda}{c}\bigg) = \frac{x + c'}{c} \\ \therefore y = c * cosh\bigg(\frac{x + c'}{c}\bigg) - \lambda$$

$\therefore y = c * cosh\bigg(\frac{x + c'}{c}\bigg) - \lambda$ is the required curve