Page:Calculus Made Easy.pdf/198

independent variables, $t$ and $x$. This being understood, let us take the expression

${\begin{aligned}y&=F(x+at)+f(x-at),\\{\text{or,}}\;\quad y&=F(w)+f(v);\\{\text{where}}\;\quad w&=x+at,\quad {\text{and}}\quad v=x-at.\\{\text{Then}}\;\quad {\frac {\partial y}{\partial x}}&={\frac {\partial F(w)}{\partial w}}\cdot {\frac {\partial w}{\partial x}}+{\frac {\partial f(v)}{\partial v}}\cdot {\frac {\partial v}{\partial x}}\\&=F'(w)\cdot 1+f'(v)\cdot 1\end{aligned}}$

(where the figure $1$ is simply the coefficient of $x$ in $w$ and $v$);

${\begin{aligned}{\text{and}}\;\ {\frac {\partial ^{2}y}{\partial x^{2}}}&=F''(w)+f''(v).&&\\{\text{Also}}\;\ {\frac {\partial y}{\partial t}}&={\frac {\partial F(w)}{\partial w}}\cdot {\frac {\partial w}{\partial t}}+{\frac {\partial f(v)}{\partial v}}\cdot {\frac {\partial v}{\partial t}}\\&=F'(w)\cdot a-f'(v)a;\\{\text{and}}\;\ {\frac {\partial ^{2}y}{\partial t^{2}}}&=F''(w)a^{2}+f''(v)a^{2};\\{\text{whence}}\;\ {\frac {\partial ^{2}y}{\partial t^{2}}}&=a^{2}\,{\frac {\partial ^{2}y}{\partial x^{2}}}.\end{aligned}}$

This differential equation is of immense importance in mathematical physics.

 

Maxima and Minima of Functions of two Independent Variables.

 

Example (5). Let us take up again Exercise IX. p. 110, No. 4.

Let $x$ and $y$ be the length of two of the portions of the string. The third is $30-(x+y)$, and the area of the