Todai Entrance Exam: Math 2016 – Problem 2

(1)

The surface area of S around (x, y) is determined by the product of the circumference 2\pi y(x) and the curve length AB \sqrt{dx^{2} + dy^{2}}, so:

    \[ S = \int_{A}^{B} 2\pi y(x)\sqrt{dx^{2} + dy^{2}}  \]

Remember that y depends on x.

    \[ S = 2\pi \int_{-1}^{1} y(x)\sqrt{1 + \frac{dy^{2}}{dx^{2}}}dx = 2\pi \int_{-1}^{1} F(y, y')dx \]

So the statement follows.

(2)

    \[ F - y'\frac{\partial F}{\partial y'} = c \Leftrightarrow \frac{\mathrm{d} F}{\mathrm{d} x} - \frac{\mathrm{d} y'\frac{\partial F}{\partial y'}}{\mathrm{d} x} = 0 \Leftrightarrow \left (\frac{\partial F}{\partial y}\frac{\mathrm{d} y}{\mathrm{d} x} + \frac{\partial F}{\partial y'}\frac{\mathrm{d} y'}{\mathrm{d} x}  \right ) - \left (\frac{\partial F}{\partial y'}\frac{\mathrm{d} y'}{\mathrm{d} x} + y'\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial F}{\partial y'}  \right ) = 0 \]

    \[ \Leftrightarrow \frac{\partial F}{\partial y}\frac{\mathrm{d} y}{\mathrm{d} x}  + y'\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial F}{\partial y'} = 0 \Leftrightarrow \frac{\partial F}{\partial y} + \frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial F}{\partial y'} = 0 \]

(3)

    \[ F - y'\frac{\partial F}{\partial y'} = c \Leftrightarrow y\sqrt{1 + (y')^2} - \frac{y(y')^{2}}{\sqrt{1 + (y')^2}} = c \Leftrightarrow \frac{y}{\sqrt{1 + (y')^2}} = c \Leftrightarrow c^{2}(y')^2 - y^{2} = -c^{2} \]

(4)

    \[ \frac{\mathrm{d} y}{\mathrm{d} x} = \pm \sqrt{\frac{y^{2} - c^{2}}{c^{2}}} \]

This is a separable non-linear first-order differential equation. So the solution is:

    \[ x = \pm \text{arcosh}\left ( \frac{y}{a} \right ) \Leftrightarrow y = \text{cosh}\left ( \pm\frac{x}{a} \right )  \]


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