01 - NA

Maxima are Stationary Points

Let $f(x)$ be a nice function then

Maxima and Minima of $f$ occur when $f^{\prime }(x)=0$

Note that a nice function refers to a continuously differentiable function

Shortest Path between Two Points

Let $(x_1,y_1)$ and $(x_2,y_2)$ be points then

Straight line minimises distance

Proof

Rotate plane (so distances unchanged) so that $y_1=y_2$

Let path be denoted by nice (continuously differentiable) function $y(x)$ then

Length of path is given by

$$I[y]={\int }_{x_1}^{x_2}\sqrt{1+y^{\prime }(x{)}^2}dx$$

where $I[y]$ is a functional that assigns real number to each $y$

As $y^{\prime }(x{)}^2\ge 0$ then integral is minimised when

$$y^{\prime }(x)=0$$

with constraint $y(x_1)=y(x_2)=y_1$

Hence straight line path minimises the distance

Biatholon Problem

Consider person traveling from point $A$ to point $B$
where first crossing field and then river at speed $c_1$ and $c_2$ respectively

Then

Fastest Route satisfies

$$\frac{\sin {\psi }_1}{c_1}=\frac{\sin {\psi }_2}{c_2}$$

where ${\psi }_1$ and ${\psi }_2$ are from

Note that this is identical to Snell’s Law

Proof

For any $P$ along border from field to river then the fastest way is a straight line

Let

$$A=(x_A,y_A),B=(x_B,y_B),P=(x_P,y_P)$$

Then need $y_P$ minimises total time so at optimal value of $y_P$ then

$$\frac{d}{d{y}_P}\left(\frac{\sqrt{(x_A-x_P{)}^2+(y_A-y_P{)}^2}}{c_1}+\frac{\sqrt{(x_B-x_P{)}^2+(y_B-y_P{)}^2}}{c_2}\right)=0$$

Hence

$$\frac{(y_A-y_P)}{c_1\sqrt{(x_A-x_P{)}^2+(y_A-y_P{)}^2}}=\frac{(y_P-y_B)}{c_2\sqrt{(x_B-x_P{)}^2+(y_B-y_P{)}^2}}$$

Thus optimum position of $P$ is such that angles ${\psi }_1,{\psi }_2$ satisfy

$$\frac{\sin {\psi }_1}{c_1}=\frac{\sin {\psi }_2}{c_2}$$

Tangent Angle

Let angle $\varphi =\varphi (x)$ be of path to $x$-axis then

$$\tan \psi =\frac{dy}{dx}=y^{\prime }(x)$$

Arclength Relations

Let angle $\varphi =\varphi (x)$ be of path to $x$-axis then
Let $s$ be the arclength defined by

$$d{s}^2=d{x}^2+d{y}^2$$

Then

$$\sin \psi =\frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}=\frac{dy}{ds}\text{ and }\cos \psi =\frac{1}{\sqrt{1+y^{\prime 2}}}=\frac{dx}{ds}$$

Curvature of Path

Let angle $\varphi =\varphi (x)$ be of path to $x$-axis then
Let $s$ be the arclength defined by

$$d{s}^2=d{x}^2+d{y}^2$$

Then

$$\kappa =\frac{d\psi }{ds}=\frac{y^{\prime \prime }}{(1+y^{\prime 2}{)}^{3/2}}$$

where $\kappa$ is the curvature of the path (invariant under rotation of axes)