Subject: Applied Mathematics 4

Topic: Calculus of Variation

Difficulty: Medium

By Green's theorem, Area enclosed by a simple chord curve is given by: $ A = \frac{1}{2} \int_c x \,\, dy - y \,\, dx = \frac{1}{2} \int_c (x \frac{dy}{dx} - y) \,\, dx = \frac{1}{2} \int (xy' - y) \,\, dx \hspace{0.25cm} ...(1) $

Also, length of arc is given by, $ S = \int_c \sqrt{1 + (\frac{dy}{dx})^2} \,\, dx = \int_c \sqrt{1 + (y')^2} \,\, dx \hspace{0.25cm} ...(2) $

To use Lagrange's multiplier, we multiply (2) by $ \lambda $ and add it to (1)

$ F = \int _c [\frac{1}{2} (xy' - y) + \lambda \sqrt{1 + (y')^2}] \,\, dx $

For maxima and minima conditions,

$ \frac{\partial F_1}{\partial y} - \frac{d}{dx} (\frac{\partial F_1}{\partial y'}) = 0 $

where, $ F_1 = \frac{1}{2} (xy' - y) + \lambda \sqrt{1 + (y')^2} $

$ \frac{\partial F_1}{\partial y} = - \frac{1}{2} \,\, $&$ \,\, \frac{\partial F_1}{\partial y'} = \frac{1}{2}x + \frac{\lambda}{2 \sqrt{1 + (y')^2 }} \times 2y' $

$ \frac{\partial F_1}{\partial y} = - \frac{1}{2} \,\, $&$ \,\, \frac{\partial F_1}{\partial y'} = \frac{1}{2}x + [\frac{\lambda y'}{\sqrt{1 + (y')^2 }}] \\ $

Now,

$ \frac{\partial F_1}{\partial y} - \frac{d}{dx} (\frac{\partial F_1}{\partial y'}) = 0 \\ - \frac{1}{2} - \frac{d}{dx}[\frac{1}{2} x + \frac{\lambda y'}{\sqrt{1 + (y')^2 }}] = 0 \\ - \frac{1}{2} - \frac{1}{2} - \frac{d}{dx} [\frac{\lambda y'}{\sqrt{1 + (y')^2 }}] = 0 \\ -1 - \frac{d}{dx} [\frac{\lambda y'}{\sqrt{1 + (y')^2 }}] = 0 \\ \frac{d}{dx} [\frac{\lambda y'}{\sqrt{1 + (y')^2 }}] = -1 \\ $

Integrating,

$ \frac{\lambda y'}{\sqrt{1 + (y')^2 }} = - x + c_1 \\ - \frac{\lambda y'}{\sqrt{1 + (y')^2 }} = x - c_1 \\ (\frac{\lambda y'}{\sqrt{1 + (y')^2 }})^2 = (x - c_1)^2 \hspace{0.5cm} [Squaring] \\ \frac{\lambda^2 (y')^2}{1 + (y')^2} = (x - c_1)^2 \\ \lambda^2 (y')^2 = (x - c_1)^2(1 + (y')^2) \\ \lambda^2 (y')^2 = (x - c_1)^2 + (x - c_1)^2 (y')^2 \\ [\lambda^2 - (x - c_1)^2](y')^2 = (x - c_1)^2 \\ (y')^2 = \frac{(x - c_1)^2}{\lambda^2 - (x - c_1)^2} \\ \frac{dy}{dx} = \frac{(x - c_1)}{\sqrt{\lambda^2 - (x - c_1)^2}} \\ dy = \frac{(x - c_1)}{\sqrt{\lambda^2 - (x - c_1)^2}} \,\, dx $

Integrating, we get,

$ y = - \sqrt{\lambda^2 - (x - c_1)^2} + c_2 \\ y - c_2 = - \sqrt{\lambda^2 - (x - c_1)^2} \\ (y - c_2)^2 = \lambda^2 - (x - c_1)^2 \hspace{0.5cm} [Squaring] \\ (x - c_1)^2 + (y - c_2)^2 = \lambda^2 $

Which is a circle. Therefore, a circle of given length l encloses maximum area.