The gradient of a scalar field

1 The gradient of a scalar field

Consider the height $ϕ$ above sea level at various points on a hill. Some contours for such a hill are shown in Figure 14.

Figure 14:

{Contour map of a hill}

We are interested in how $ϕ$ changes from one point to another. Starting from $A$ and making a displacement $\underset{̲}{d}$ the change in height ( $ϕ$ ) depends on the direction of the displacement. The magnitude of each $\underset{̲}{d}$ is the same.

Displacement	Change in $ϕ$
$A B$	$40 - 30 = 10$
$A C$	$40 - 30 = 10$
$A D$	$30 - 30 = 0$
$A E$	$20 - 30 = - 10$

The change in $ϕ$ clearly depends on the direction of the displacement. For the paths shown $ϕ$ increases most rapidly along $A B$ , does not increase at all along $A D$ (as $A$ and $D$ are both on the same contour and so are both at the same height) and decreases along $A E$ .

The direction in which $ϕ$ changes fastest is along the line of greatest slope which is orthogonal (i.e. perpendicular) to the contours. Hence, at each point of a scalar field we can define a vector field giving the magnitude and direction of the greatest rate of change of $ϕ$ locally.

A vector field, called the gradient, written grad $ϕ$ , can be associated with a scalar field $ϕ$ so that at every point the direction of the vector field is orthogonal to the scalar field contour which is the direction of the maximum rate of change of $ϕ$ .

For a second example consider a metal plate heated at one corner and cooled by an ice bag at the opposite corner. All edges and surfaces are insulated. After a while a steady state situation exists in which the temperature $ϕ$ at any point remains the same. Some temperature contours are shown in Figure 15.

Figure 15:

{ Temperature contours and heat flow lines for a metal plate}

The direction of the heat flow is along the flow lines which are orthogonal to the contours (see the dashed lines in Figure 15(b)); this heat flow is proportional to $\underset{̲}{F}$ = grad $ϕ$ .

Definition
The gradient of the scalar field $ϕ = f (x, y, z)$ is $grad ϕ = \underset{̲}{\nabla} ϕ = \frac{\partial ϕ}{\partial x} \underset{̲}{i} + \frac{\partial ϕ}{\partial y} \underset{̲}{j} + \frac{\partial ϕ}{\partial z} \underset{̲}{k}$

Often, instead of grad $ϕ$ , the notation $\underset{̲}{\nabla} ϕ$ is used. ( $\underset{̲}{\nabla}$ is a vector differential operator called ‘del’ or ‘nabla’ defined by $\frac{\partial}{\partial x} \underset{̲}{i} + \frac{\partial}{\partial y} \underset{̲}{j} + \frac{\partial}{\partial z} \underset{̲}{k}$ . As a vector differential operator, it retains the characteristics of a vector while also carrying out differentiation.)

The vector grad $ϕ$ gives the magnitude and direction of the greatest rate of change of $ϕ$ at any point, and is always orthogonal to the contours of $ϕ$ . For example, in Figure 14, grad $ϕ$ points in the direction of $A B$ while the contour line is parallel to $A D$ i.e. perpendicular to $A B$ . Similarly, in Figure 15(b), the various intersections of the contours with the lines representing grad $ϕ$ occur at right-angles.

For the hill considered earlier the direction and magnitude of grad $ϕ$ are shown at various points in Figure 16. Note that the magnitude of grad $ϕ$ is greatest (as indicated by the length of the arrow) when the hill is at its steepest (as indicated by the closeness of the contours).

Figure 16:

{ Grad phi and the steepest ascent direction for a hill}

Key Point 3

$ϕ$ is a scalar field but grad $ϕ$ is a vector field.

Example 9

Find grad $ϕ$ for

$ϕ = x^{2} - 3 y$
$ϕ = x y^{2} z^{3}$

Solution

grad $ϕ = \frac{\partial}{\partial x} (x^{2} - 3 y) \underset{̲}{i} + \frac{\partial}{\partial y} (x^{2} - 3 y) \underset{̲}{j} + \frac{\partial}{\partial z} (x^{2} - 3 y) \underset{̲}{k} = 2 x \underset{̲}{i} + (- 3) \underset{̲}{j} + 0 \underset{̲}{k} = 2 x \underset{̲}{i} - 3 \underset{̲}{j}$
grad $ϕ = \frac{\partial}{\partial x} (x y^{2} z^{3}) \underset{̲}{i} + \frac{\partial}{\partial y} (x y^{2} z^{3}) \underset{̲}{j} + \frac{\partial}{\partial z} (x y^{2} z^{3}) \underset{̲}{k} = y^{2} z^{3} \underset{̲}{i} + 2 x y z^{3} \underset{̲}{j} + 3 x y^{2} z^{2} \underset{̲}{k}$

Example 10

For $f = x^{2} + y^{2}$ find grad $f$ at the point $A (1, 2)$ . Show that the direction of grad $f$ is orthogonal to the contour at this point.

Solution

$grad f = \frac{\partial f}{\partial x} \underset{̲}{i} + \frac{\partial f}{\partial y} \underset{̲}{j} + \frac{\partial f}{\partial z} \underset{̲}{k} = 2 x \underset{̲}{i} + 2 y \underset{̲}{j} + 0 \underset{̲}{k} = 2 x \underset{̲}{i} + 2 y \underset{̲}{j}$

and at $A (1, 2)$ this equals $2 \times 1 \underset{̲}{i} + 2 \times 2 \underset{̲}{j} = 2 \underset{̲}{i} + 4 \underset{̲}{j}$ .

Since $f = x^{2} + y^{2}$ then the contours are defined by $x^{2} + y^{2} =$ constant, so the contours are circles centred at the origin. The vector grad $f$ at $A (1, 2)$ points directly away from the origin and hence grad $f$ and the contour are orthogonal; see Figure 17. Note that $\underset{̲}{r} (A) = \underset{̲}{i} + 2 \underset{̲}{j} = \frac{1}{2} grad f$ .

Figure 17:

{ Grad f is perpendicular to the contour lines}

The change in a function $ϕ$ in a given direction (specified as a unit vector $\underset{̲}{a}$ ) is determined from the scalar product $(grad ϕ) \cdot \underset{̲}{a}$ . This scalar quantity is called the directional derivative.

Note:

$\underset{̲}{a}$ along a contour implies $\underset{̲}{a}$ is perpendicular to $grad ϕ$ which implies $\underset{̲}{a} \cdot grad ϕ = 0$ .
$\underset{̲}{a}$ perpendicular to a contour implies $\underset{̲}{a} \cdot grad ϕ$ is a maximum.

Task!

Given $ϕ = x^{2} y^{2} z^{2}$ , find

grad $ϕ$
grad $ϕ$ at $(- 1, 1, 1)$ and a unit vector in this direction.
the derivative of ϕ at ( 2 , 1 , − 1 ) in the direction of
1. $\underset{̲}{i}$
2. $\underset{̲}{d} = \frac{3}{5} \underset{̲}{i} + \frac{4}{5} \underset{̲}{k} .$

$grad ϕ = \frac{\partial ϕ}{\partial x} \underset{̲}{i} + \frac{\partial ϕ}{\partial y} \underset{̲}{j} + \frac{\partial ϕ}{\partial z} \underset{̲}{k} = 2 x y^{2} z^{2} \underset{̲}{i} + 2 x^{2} y z^{2} \underset{̲}{j} + 2 x^{2} y^{2} z \underset{̲}{k}$
At $(- 1, 1, 1)$ , grad $ϕ = - 2 \underset{̲}{i} + 2 \underset{̲}{j} + 2 \underset{̲}{k}$
A unit vector in this direction is
$\frac{grad ϕ}{|grad ϕ|} = \frac{- 2 \underset{̲}{i} + 2 \underset{̲}{j} + 2 \underset{̲}{k}}{\sqrt{{(- 2)}^{2} + 2^{2} + 2^{2}}} = \frac{1}{2 \sqrt{3}} (- 2 \underset{̲}{i} + 2 \underset{̲}{j} + 2 \underset{̲}{k}) = - \frac{1}{\sqrt{3}} \underset{̲}{i} + \frac{1}{\sqrt{3}} \underset{̲}{j} + \frac{1}{\sqrt{3}} \underset{̲}{k}$
At ( 2 , 1 , − 1 ) , grad ϕ = 4 i ̲ + 8 j ̲ − 8 k ̲
1. To find the derivative of $ϕ$ in the direction of $\underset{̲}{i}$ take the scalar product
  $(4 \underset{̲}{i} + 8 \underset{̲}{j} - 8 \underset{̲}{k}) \cdot \underset{̲}{i} = 4 \times 1 + 0 + 0 = 4$ . So the derivative in the direction of $\underset{̲}{d}$ is 4.
2. To find the derivative of $ϕ$ in the direction of $\underset{̲}{d} = \frac{3}{5} \underset{̲}{i} + \frac{4}{5} \underset{̲}{k}$ take the scalar product
  $(4 \underset{̲}{i} + 8 \underset{̲}{j} - 8 \underset{̲}{k}) \cdot (\frac{3}{5} \underset{̲}{i} + \frac{4}{5} \underset{̲}{k}) = 4 \times \frac{3}{5} + 0 + (- 8) \times \frac{4}{5} = \frac{12}{5} - \frac{32}{5} = - 4$ .
  So the derivative in the direction of $\underset{̲}{d}$ is $- 4$ .

Exercises

Find grad ϕ for the following scalar fields
1. $ϕ = y - x$ .
2. $ϕ = y - x^{2}$ ,
3. $ϕ = x^{2} + y^{2} + z^{2}$ .
Find grad ϕ for each of the following two-dimensional scalar fields given that r ̲ = x i ̲ + y j ̲ and r = x 2 + y 2 (you should express your answer in terms of r ̲ ).
1. $ϕ = r$ ,
2. $ϕ = ln r$ ,
3. $ϕ = \frac{1}{r}$ ,
4. $ϕ = r^{n}$ .
If ϕ = x 3 y 2 z , find,
1. $\underset{̲}{\nabla} ϕ$
2. a unit vector normal to the contour at the point $(1, 1, 1)$ .
3. the rate of change of $ϕ$ at $(1, 1, 1)$ in the direction of $\underset{̲}{i}$ .
4. the rate of change of $ϕ$ at $(1, 1, 1)$ in the direction of the unit vector $\underset{̲}{n} = \frac{1}{\sqrt{3}} (\underset{̲}{i} + \underset{̲}{j} + \underset{̲}{k})$ .
Find a unit vector which is normal to the sphere $x^{2} + {(y - 1)}^{2} + {(z + 1)}^{2} = 2$ at the point $(0, 0, 0)$ .
Find unit vectors normal to $ϕ_{1} = y - x^{2}$ and $ϕ_{2} = x + y - 2$ . Hence find the angle between the curves $y = x^{2}$ and $y = 2 - x$ at their point of intersection in the first quadrant.

1. $\frac{\partial}{\partial x} (y - x) \underset{̲}{i} + \frac{\partial}{\partial y} (y - x) \underset{̲}{j} = - \underset{̲}{i} + \underset{̲}{j}$
2. $- 2 x \underset{̲}{i} + \underset{̲}{j}$
3. $[\frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2})] \underset{̲}{i} + [\frac{\partial}{\partial y} (x^{2} + y^{2} + z^{2})] \underset{̲}{j} + [\frac{\partial}{\partial z} (x^{2} + y^{2} + z^{2})] \underset{̲}{k} = 2 x \underset{̲}{i} + 2 y \underset{̲}{j} + 2 z \underset{̲}{k}$
1. $\frac{\underset{̲}{r}}{r}$ ,
2. $\frac{\underset{̲}{r}}{r^{2}}$ ,
3. $- \frac{\underset{̲}{r}}{r^{3}}$ ,
4. $n r^{n - 2} \underset{̲}{r}$
1. $3 x^{2} y^{2} z \underset{̲}{i} + 2 x^{3} y z \underset{̲}{j} + x^{3} y^{2} \underset{̲}{k}$ ,
2. $\frac{1}{\sqrt{14}} (3 \underset{̲}{i} + 2 \underset{̲}{j} + \underset{̲}{k})$ ,
3. $3$ ,
4. $2 \sqrt{3}$
The vector field $\underset{̲}{\nabla} ϕ$ where $ϕ = x^{2} + {(y - 1)}^{2} + {(z + 1)}^{2}$ is $2 x \underset{̲}{i} + 2 (y - 1) \underset{̲}{j} + 2 (z + 1) \underset{̲}{k}$
The value that this vector field takes at the point $(0, 0, 0)$ is $- 2 \underset{̲}{j} + 2 \underset{̲}{k}$ which is a vector normal to the sphere.
Dividing this vector by its magnitude forms a unit vector: $\frac{1}{\sqrt{2}} (- \underset{̲}{j} + \underset{̲}{k})$
$10 8^{\circ}$ or $7 2^{\circ}$ (intersect at $(1, 1)$ ) [At intersection, grad $ϕ_{1} = - 2 \underset{̲}{i} + \underset{̲}{j}$ and grad $ϕ_{2} = \underset{̲}{i} + \underset{̲}{j} .$ ]

1 The gradient of a scalar field

Key Point 3

Example 9

Solution

Example 10

Solution

Task!

Answer

Exercises

Answer