Home page for accesible maths 3 Functions of two or more variables 3.1 Some examples 3.3 Expressing

f

\nabla\phi

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3.2 The gradient vector

You’ve studied partial differentiation in MATH102. We’ll use the same notation for partial derivatives $f_{x}=\frac{\partial f}{\partial x}$ , $f_{y}=\frac{\partial f}{\partial y}$ as used in MATH102.

Example. $f(x,y)=x^{2}+xy+y^{2}$ . Then

$\quad$

For a function $f$ from $\mathbb{R}^{3}$ to $\mathbb{R}$ , the vector-valued function $(f_{x}\;\;f_{y}\;\;f_{z})$ is called the gradient vector of $f$ , denoted $\nabla f$ or grad $f$ .

Example. Let $f(x,y,z)=x^{2}y^{2}z$ . Then

One partial derivative zero. If $f=f(x,y)$ and $f_{y}=0$ throughout the plane (or some region such as a rectangle or a circle), then, of course, $f$ does not vary with $y$ , so it is constant along horizontal lines and is actually a function of $x$ only. In other words, $f(x,y)=g(x)$ for some function $g$ of one variable. Similarly, if $f$ is a function of $x,\,y,\,z$ and $f_{x}=0$ throughout a region, then it equals $g(y,z)$ for some function $g$ of two variables.

First order approximation to the variation of $f$ .

If $f$ is a function of $x$ only, then, by the very meaning of differentiation, $f(x+\delta x)-f(x)\approx f^{\prime}(x)\,\delta x$ for small $\delta x$ , with error small compared to $\delta x$ . Suppose now that $f$ is a function of $x$ and $y$ , and that $f_{x}$ and $f_{y}$ exist and vary continuously. By the principle just stated, applied twice, we have

$\displaystyle f(x+\delta x,y)-f(x,y)$	$\displaystyle\approx$	$\displaystyle f_{x}(x,y)\>\delta x,$
$\displaystyle f(x+\delta x,y+\delta y)-f(x+\delta x,y)$	$\displaystyle\approx$	$\displaystyle f_{y}(x+\delta x,y)\>\delta y$
	$\displaystyle\approx$	$\displaystyle f_{y}(x,y)\>\delta y\qquad\mbox{since $f_{y}$ varies continuously}.$

By adding these two statements, we obtain the following approximation to the change in $f$ , $\delta f$ :

\delta f=f(x+\delta x,y+\delta y)-f(x,y)\approx f_{x}\,\delta x+f_{y}\,\delta y% \qquad(*),

in which $f_{x}$ and $f_{y}$ are evaluated at $(x,y)$ , and the error is small compared to $\delta x$ and $\delta y$ .

Similarly, for a function of three variables,

\delta f\approx f_{x}\,\delta x+f_{y}\,\delta y+f_{z}\,\delta z=(\nabla f).\,% \delta\mbox{\boldmath$r$}\qquad(**).

Curves $f(x,y)=c$

Suppose that $(x,y)$ and $(x+\delta x,y+\delta y)$ both lie on the curve $f(x,y)=c$ . Then $\delta f=0$ , so by (*), $f_{x}\delta x+f_{y}\delta y\approx 0$ for small $\delta x$ and $\delta y$ . But this is exactly the same thing as saying $\nabla f\cdot(\delta x\;\;\delta y)=0$ . We therefore have the following theorem:

Theorem 3.3

The vector $\nabla f(x_{0},y_{0})$ is normal to the curve $f(x,y)=c$ at the point $(x_{0},y_{0})$ .

In other words, if $\nabla f(x_{0},y_{0})\neq(0\;\;0)$ then there exists a tangent line to the curve at $(x_{0},y_{0})$ , and $\nabla f(x_{0},y_{0})$ is perpendicular to it.

We can (almost always) use this to determine an equation for the tangent to the curve at $(x_{0},y_{0})$ .

Example 3.4

Determine the equation of the tangent line to the curve $x^{2}+y^{3}=2$ at the point $(1,1)$ .

We calculate $\nabla f$ : $f_{x}=2x$ , $f_{y}=3y^{2}$ so $\nabla f(1,1)=(2\;\;3)$ . Thus $(2\;\;3)$ is normal to the curve at $(1,1)$ . In particular, the equation of the tangent line is:

(2\;\;3)\cdot(x\;\;y)=(2\;\;3)\cdot(1\;\;1)

so $2x+3y=5$ .

Remark. If $\nabla f(x_{0},y_{0})=(0\;\;0)$ then we can’t define the tangent line to the curve at $(x_{0},y_{0})$ . This can happen for a variety of reasons, none of which will be discussed in this course.

Example 3.5. The ellipse $3x^{2}+4y^{2}=12$ .

The gradient at $(x,y)$ is

Hence $\nabla f$ at $(1,3/2)$ is $(6\;\;12)$ , and the tangent at $(1,3/2)$ is

Alternatively, we can describe the ellipse via the parametrized curve $\gamma:[0,2\pi)\rightarrow{\mathbb{R}}^{2}$ , $\gamma(t)=(2\cos t\;\;\sqrt{3}\sin t)$ . Then $(1\;\;3/2)=\gamma(\pi/3)$ . Since $\gamma^{\prime}(t)=(-2\sin t\;\;\sqrt{3}\cos t)$ , we have: $\gamma^{\prime}(\pi/3)=\hskip 284.527559pt$

Thus $(-2\;\;1)$ is a tangent vector to the curve at $(1,3/2)$ . In particular, another way of writing the tangent line is: $(x\;\;y)=(1\;\;3/2)+\lambda(-2\;\;1)$ .

It is straightforward to check that this really does give the same line as the equation $x+2y=4$ .

Surfaces $f(x,y,z)=c$

Suppose that $\mbox{\boldmath$r$}_{0}=(x_{0}\;\;y_{0}\;\;z_{0})$ and $\mbox{\boldmath$r$}_{0}+\delta\mbox{\boldmath$r$}=(x_{0}+\delta x\;\;y_{0}+% \delta y\;\;z_{0}+\delta z)$ lie on the surface $f(x,y,z)=c$ , so that $\delta\mbox{\boldmath$r$}$ is a displacement within the surface. By (**),

f_{x}\delta x+f_{y}\delta y+f_{z}\delta z=(\nabla f).\delta\mbox{\boldmath$r$}% \approx 0.

Hence we have the analogous statement to Thm. 3.3:

Theorem 3.6

Let $\mbox{\boldmath$r$}_{0}$ be a point on the surface $f(x,y,z)=c$ . Then $\nabla f(\mbox{\boldmath$r$}_{0})$ is normal to the surface at $\mbox{\boldmath$r$}_{0}$ . Thus if $\nabla f(\mbox{\boldmath$r$}_{0})\neq(0\;\;0\;\;0)$ then there exist a normal line and a tangent plane to the surface $f(x,y,z)=c$ at $\mbox{\boldmath$r$}_{0}$ ; the normal line is given by $(x\;\;y\;\;z)=\mbox{\boldmath$r$}_{0}+\lambda\nabla f(\mbox{\boldmath$r$}_{0})$ , and the tangent plane is given by the equation:

\nabla f(\mbox{\boldmath$r$}_{0})\cdot\mbox{\boldmath$r$}=\nabla f(\mbox{% \boldmath$r$}_{0})\cdot\mbox{\boldmath$r$}_{0}

Note once again that whereas lines in ${\mathbb{R}}^{2}$ can be given by an equation of the form $ax+by=c$ , an equation in ${\mathbb{R}}^{3}$ of the form $ax+by+cz=d$ describes a plane; a normal vector to this plane is $(a\;\;b\;\;c)$ . If we are given a normal vector ${\bf n}$ and a vector ${\bf u}$ in the plane, then the equation of the plane is: ${\bf n}\cdot\mbox{\boldmath$r$}={\bf n}\cdot{\bf u}$ .

Example 3.7. Find the normal line and tangent plane to the surface $xy=z^{2}$ at $(4,1,2)$ .

The surface is $f(x,y,z)=0$ , where $f(x,y,z)=xy-z^{2}$ .

$\quad$

Example: the gradient of $r$ and functions of $r$

Let $\mbox{\boldmath$r$}=(x\;\;y\;\;z)$ and

$r=|\mbox{\boldmath$r$}|=(x^{2}+y^{2}+z^{2})^{1/2}$ , the distance to the origin,

$\displaystyle{\hat{\mbox{\boldmath$r$}}=\frac{1}{r}\mbox{\boldmath$r$}=\left(% \frac{x}{r}\;\;\frac{y}{r}\;\;\frac{z}{r}\right)}$ , the unit vector in the direction of $𝒓$ .

Proposition: We have $\nabla r=\hat{\mbox{\boldmath$r$}}$ . Also, $\nabla f(r)=f^{\prime}(r)\hat{\mbox{\boldmath$r$}}$ .

Proof. By the (one-dimensional) chain rule,

$\displaystyle{\frac{\partial r}{\partial x}=}$

and similarly for $y$ and $z$ . Also,

$\displaystyle{\frac{\partial}{\partial x}f(r)=}$

An important special case is:

\nabla\left(\frac{GM}{r}\right)=-\frac{GM}{r^{2}}\,\hat{\mbox{\boldmath$r$}}

which is the gravitational force exerted at $P$ by a mass $M$ at the origin (see Examples 3.1 and 3.2).