Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/216

Illustrative Example 2. Find the partial derivatives of $u=\sin(ax+by+cz).$

Solution.	${\tfrac {\partial u}{\partial x}}=a\cos(ax+by+cz)$, treating $y$ and $z$ as constants,
	${\tfrac {\partial u}{\partial y}}=b\cos(ax+by+cz)$, treating $x$ and $z$ as constants,
	${\tfrac {\partial u}{\partial z}}=c\cos(ax+by+cz)$, treating $y$ and $x$ as constants.

Again turning to the function

$z=f(x,\ y),$

we have, by (A), §123, defined ${\tfrac {\partial z}{\partial x}}$ as the limit of the ratio of the increment of the function ($y$ being constant) to the increment of $x$, as the increment of $x$ approaches the limit zero. Similarly, (B), §123, has defined ${\tfrac {\partial z}{\partial y}}$. It is evident, however, that if we look upon these partial derivatives from the point of view of § 94, then

${\frac {\partial z}{\partial x}}$

may be considered as the ratio of the time rates of change of $z$ and $x$ when $y$ is constant, and

${\frac {\partial z}{\partial y}}$

as the ratio of the time rates of change of $z$ and $y$ when $x$ is constant.

124. Partial derivatives interpreted geometrically. Let the equation of the surface shown in the figure be

Plane through point

$z=f(x,\ y).$

Pass a plane EFGH through the point P (where $x=a$ and $y=b$) on the surface parallel to the XOZ-plane. Since the equation of this plane is

$y=\ b,$

the equation of the section JPK cut out of the surface is

$z=f(x,\ b),$

if we consider EF as the axis of $Z$ and EH as the axis of $X$: In this plane ${\tfrac {\partial z}{\partial x}}$ means the same as ${\tfrac {dz}{dx}}$ and we have

${\tfrac {\partial z}{\partial x}}$	$=\tan MTP$
	$=$ slope of section JK at P.