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

Marks: 8M

Year: Dec 2015

Suppose we have a closed curve of fixed length $L = \int\sqrt{\dot{x}^2 + \dot{y}^2}dt$ and we swish to enclose the maximum amount of area Green's theorem allow us to express area as a line integral

$\int\frac{1}{2} (-y\dot{x} - x\dot{y})dt$ so that the variation problem becomes.

Maximize $J = \int\frac{1}{2} (-y\dot{x} - x\dot{y})dt - \lambda \sqrt{\dot{x}^2 + \dot{y}^2}dt$

The two variable Euless - Lagrange equations are

$\frac{\dot{y}}{2} - \frac{d}{dt} \bigg(-\frac{y}{2} - \frac{\lambda \dot{x}}{\sqrt{\dot{x}^2 + \dot{y}^2}}\bigg) = 0 \\ \frac{-\dot{x}}{2} - \frac{d}{dt} \bigg(\frac{y}{2} - \frac{\lambda \dot{y}}{\sqrt{\dot{x}^2 + \dot{y}^2}}\bigg) = 0$

Which may be written as

$\frac{d}{dt}\bigg(y + \frac{\lambda \dot{x}}{\sqrt{\dot{x}^2 + \dot{y}^2}}\bigg) = 0$

We then have

$y + \frac{\lambda \dot{x}}{\sqrt{\dot{x}^2 + \dot{y}^2}} = \cdot C$ and $-x + \frac{\lambda \dot{y}}{\sqrt{\dot{x}^2 + \dot{y}^2}} = \cdot D$

Simplifying both equations we get $(x - D)^2 + (y - C)^2 = \lambda^2$

Which is an equation of circle centered at (D, C) with radius $\lambda$.