Fields and Vector Calculus the Electric Potential Is a Scalar ﬁeld

Fields and vector calculus The electric potential is a scalar ﬁeld.

The velocity of the air flow at any given point is a vector. These vectors will be different at different points, so they are functions of position (and also of time). Thus, the air velocity is a vector field. Similarly, the pressure and temperature are scalar quantities that depend on position, or in other words, they are scalar fields. (b) The magnetic field inside an electrical machine is a vector that depends on position, or in other words a vector field.

For example: (a) Suppose we want to model the ﬂow of air around an aeroplane.

Although we will mainly be concerned with scalar and vector ﬁelds in three-dimensional space, we will sometimes use two-dimensional examples because they are easier to visualise.

Vector ﬁelds and scalar ﬁelds

Although we will mainly be concerned with scalar and vector ﬁelds in three-dimensional space, we will sometimes use two-dimensional examples because they are easier to visualise.

Vector ﬁelds and scalar ﬁelds

These vectors will be different at different points, so they are functions of position (and also of time). Thus, the air velocity is a vector field. Similarly, the pressure and temperature are scalar quantities that depend on position, or in other words, they are scalar fields. (b) The magnetic field inside an electrical machine is a vector that depends on position, or in other words a vector field.

Although we will mainly be concerned with scalar and vector ﬁelds in three-dimensional space, we will sometimes use two-dimensional examples because they are easier to visualise.

Vector ﬁelds and scalar ﬁelds

Thus, the air velocity is a vector field. Similarly, the pressure and temperature are scalar quantities that depend on position, or in other words, they are scalar fields. (b) The magnetic field inside an electrical machine is a vector that depends on position, or in other words a vector field.

Although we will mainly be concerned with scalar and vector ﬁelds in three-dimensional space, we will sometimes use two-dimensional examples because they are easier to visualise.

Vector ﬁelds and scalar ﬁelds

Similarly, the pressure and temperature are scalar quantities that depend on position, or in other words, they are scalar fields. (b) The magnetic field inside an electrical machine is a vector that depends on position, or in other words a vector field.

Although we will mainly be concerned with scalar and vector ﬁelds in three-dimensional space, we will sometimes use two-dimensional examples because they are easier to visualise.

Vector ﬁelds and scalar ﬁelds

(b) The magnetic ﬁeld inside an electrical machine is a vector that depends on position, or in other words a vector ﬁeld.

Although we will mainly be concerned with scalar and vector ﬁelds in three-dimensional space, we will sometimes use two-dimensional examples because they are easier to visualise.

Vector ﬁelds and scalar ﬁelds

In many applications, we do not consider individual vectors or scalars, but functions that give a vector or scalar at every point. Such functions are called vector fields or scalar fields. For example: (a) Suppose we want to model the flow of air around an aeroplane. The velocity of the air flow at any given point is a vector. These vectors will be different at different points, so they are functions of position (and also of time). Thus, the air velocity is a vector field. Similarly, the pressure and temperature are scalar quantities that depend on position, or in other words, they are scalar fields. The electric potential is a scalar field. Although we will mainly be concerned with scalar and vector fields in three-dimensional space, we will sometimes use two-dimensional examples because they are easier to visualise.

Vector ﬁelds and scalar ﬁelds

In many applications, we do not consider individual vectors or scalars, but functions that give a vector or scalar at every point. Such functions are called vector fields or scalar fields. For example: (a) Suppose we want to model the flow of air around an aeroplane. The velocity of the air flow at any given point is a vector. These vectors will be different at different points, so they are functions of position (and also of time). Thus, the air velocity is a vector field. Similarly, the pressure and temperature are scalar quantities that depend on position, or in other words, they are scalar fields. (b) The magnetic field inside an electrical machine is a vector that depends on position, or in other words a vector field. Although we will mainly be concerned with scalar and vector fields in three-dimensional space, we will sometimes use two-dimensional examples because they are easier to visualise.

Vector ﬁelds and scalar ﬁelds

In many applications, we do not consider individual vectors or scalars, but functions that give a vector or scalar at every point. Such functions are called vector fields or scalar fields. For example: (a) Suppose we want to model the flow of air around an aeroplane. The velocity of the air flow at any given point is a vector. These vectors will be different at different points, so they are functions of position (and also of time). Thus, the air velocity is a vector field. Similarly, the pressure and temperature are scalar quantities that depend on position, or in other words, they are scalar fields. (b) The magnetic field inside an electrical machine is a vector that depends on position, or in other words a vector field. The electric potential is a scalar field. Vector fields and scalar fields

In many applications, we do not consider individual vectors or scalars, but functions that give a vector or scalar at every point. Such functions are called vector fields or scalar fields. For example: (a) Suppose we want to model the flow of air around an aeroplane. The velocity of the air flow at any given point is a vector. These vectors will be different at different points, so they are functions of position (and also of time). Thus, the air velocity is a vector field. Similarly, the pressure and temperature are scalar quantities that depend on position, or in other words, they are scalar fields. (b) The magnetic field inside an electrical machine is a vector that depends on position, or in other words a vector field. The electric potential is a scalar field. Although we will mainly be concerned with scalar and vector fields in three-dimensional space, we will sometimes use two-dimensional examples because they are easier to visualise. , and similarly for any other value of x.

2 I When (x, y) = (0, 0.4) we have u = ((1 − 0.4 )/5, 0) = (0.168, 0)

I When (x, y) = (1.2, 0.4) we again have u = (0.168, 0)

2 I When y = 0.8 we have u = ((1 − 0.8 )/5, 0) = (0.072, 0) 2 I When y = −0.8 we have u = ((1 − (−0.8) )/5, 0) = (0.072, 0)

The vector ﬁeld u = ((1 − y 2)/5, 0)

y=1

y=−1 , and similarly for any other value of x.

I When (x, y) = (1.2, 0.4) we again have u = (0.168, 0)

2 I When y = 0.8 we have u = ((1 − 0.8 )/5, 0) = (0.072, 0) 2 I When y = −0.8 we have u = ((1 − (−0.8) )/5, 0) = (0.072, 0)

The vector ﬁeld u = ((1 − y 2)/5, 0)

y=1

y=−1

2 I When (x, y) = (0, 0.4) we have u = ((1 − 0.4 )/5, 0) = (0.168, 0) , and similarly for any other value of x. 2 I When y = 0.8 we have u = ((1 − 0.8 )/5, 0) = (0.072, 0) 2 I When y = −0.8 we have u = ((1 − (−0.8) )/5, 0) = (0.072, 0)

The vector ﬁeld u = ((1 − y 2)/5, 0)

y=1

y=−1

2 I When (x, y) = (0, 0.4) we have u = ((1 − 0.4 )/5, 0) = (0.168, 0)

I When (x, y) = (1.2, 0.4) we again have u = (0.168, 0) 2 I When y = 0.8 we have u = ((1 − 0.8 )/5, 0) = (0.072, 0) 2 I When y = −0.8 we have u = ((1 − (−0.8) )/5, 0) = (0.072, 0)

The vector ﬁeld u = ((1 − y 2)/5, 0)

y=1

y=−1

2 I When (x, y) = (0, 0.4) we have u = ((1 − 0.4 )/5, 0) = (0.168, 0)

I When (x, y) = (1.2, 0.4) we again have u = (0.168, 0), and similarly for any other value of x. 2 I When y = −0.8 we have u = ((1 − (−0.8) )/5, 0) = (0.072, 0)

The vector ﬁeld u = ((1 − y 2)/5, 0)

y=1

y=−1

2 I When (x, y) = (0, 0.4) we have u = ((1 − 0.4 )/5, 0) = (0.168, 0)

I When (x, y) = (1.2, 0.4) we again have u = (0.168, 0), and similarly for any other value of x. 2 I When y = 0.8 we have u = ((1 − 0.8 )/5, 0) = (0.072, 0) The vector ﬁeld u = ((1 − y 2)/5, 0)

y=1

y=−1

2 I When (x, y) = (0, 0.4) we have u = ((1 − 0.4 )/5, 0) = (0.168, 0)

y=1

y=−1

2 I When (x, y) = (0, 0.4) we have u = ((1 − 0.4 )/5, 0) = (0.168, 0)

I When (x, y) = (1.2, 0.4) we again have u = (0.168, 0), and similarly for any other value of x. 2 I When y = 0.8 we have u = ((1 − 0.8 )/5, 0) = (0.072, 0) 2 I When y = −0.8 we have u = ((1 − (−0.8) )/5, 0) = (0.072, 0) when (x, y) = (0, 1) we have u = (−0.25, 0); when (x, y) = (−1, 0) we have u = (0, −0.25); when (x, y) = (0, −1) we have u = (0.25, 0). when (x, y) = (0, 1.2) we have u = (−0.3, 0); when (x, y) = (−1.2, 0) we have u = (0, −0.3); when (x, y) = (0, −1.2) we have u = (0.3, 0).

I When (x, y) = (1, 0) we have u = (0, 0.25);

I When (x, y) = (1.2, 0) we have u = (0, 0.3);

The vector ﬁeld u = (−y/4, x/4) when (x, y) = (0, 1.2) we have u = (−0.3, 0); when (x, y) = (−1.2, 0) we have u = (0, −0.3); when (x, y) = (0, −1.2) we have u = (0.3, 0).

when (x, y) = (0, 1) we have u = (−0.25, 0); when (x, y) = (−1, 0) we have u = (0, −0.25); when (x, y) = (0, −1) we have u = (0.25, 0).

I When (x, y) = (1.2, 0) we have u = (0, 0.3);

The vector ﬁeld u = (−y/4, x/4)

I When (x, y) = (1, 0) we have u = (0, 0.25); when (x, y) = (0, 1.2) we have u = (−0.3, 0); when (x, y) = (−1.2, 0) we have u = (0, −0.3); when (x, y) = (0, −1.2) we have u = (0.3, 0).

when (x, y) = (−1, 0) we have u = (0, −0.25); when (x, y) = (0, −1) we have u = (0.25, 0).

I When (x, y) = (1.2, 0) we have u = (0, 0.3);

The vector ﬁeld u = (−y/4, x/4)

I When (x, y) = (1, 0) we have u = (0, 0.25); when (x, y) = (0, 1) we have u = (−0.25, 0); when (x, y) = (0, 1.2) we have u = (−0.3, 0); when (x, y) = (−1.2, 0) we have u = (0, −0.3); when (x, y) = (0, −1.2) we have u = (0.3, 0).

when (x, y) = (0, −1) we have u = (0.25, 0).

I When (x, y) = (1.2, 0) we have u = (0, 0.3);

The vector ﬁeld u = (−y/4, x/4)

I When (x, y) = (1, 0) we have u = (0, 0.25); when (x, y) = (0, 1) we have u = (−0.25, 0); when (x, y) = (−1, 0) we have u = (0, −0.25); when (x, y) = (0, 1.2) we have u = (−0.3, 0); when (x, y) = (−1.2, 0) we have u = (0, −0.3); when (x, y) = (0, −1.2) we have u = (0.3, 0).

I When (x, y) = (1.2, 0) we have u = (0, 0.3);

The vector ﬁeld u = (−y/4, x/4)

I When (x, y) = (1, 0) we have u = (0, 0.25); when (x, y) = (0, 1) we have u = (−0.25, 0); when (x, y) = (−1, 0) we have u = (0, −0.25); when (x, y) = (0, −1) we have u = (0.25, 0). when (x, y) = (0, 1.2) we have u = (−0.3, 0); when (x, y) = (−1.2, 0) we have u = (0, −0.3); when (x, y) = (0, −1.2) we have u = (0.3, 0).

I When (x, y) = (1.2, 0) we have u = (0, 0.3);

The vector ﬁeld u = (−y/4, x/4)

I When (x, y) = (1, 0) we have u = (0, 0.25); when (x, y) = (0, 1) we have u = (−0.25, 0); when (x, y) = (−1, 0) we have u = (0, −0.25); when (x, y) = (0, −1) we have u = (0.25, 0).

I When (x, y) = (1.2, 0) we have u = (0, 0.3); when (x, y) = (−1.2, 0) we have u = (0, −0.3); when (x, y) = (0, −1.2) we have u = (0.3, 0).

The vector ﬁeld u = (−y/4, x/4)

I When (x, y) = (1, 0) we have u = (0, 0.25); when (x, y) = (0, 1) we have u = (−0.25, 0); when (x, y) = (−1, 0) we have u = (0, −0.25); when (x, y) = (0, −1) we have u = (0.25, 0).

I When (x, y) = (1.2, 0) we have u = (0, 0.3); when (x, y) = (0, 1.2) we have u = (−0.3, 0); when (x, y) = (0, −1.2) we have u = (0.3, 0).

The vector ﬁeld u = (−y/4, x/4)

I When (x, y) = (1, 0) we have u = (0, 0.25); when (x, y) = (0, 1) we have u = (−0.25, 0); when (x, y) = (−1, 0) we have u = (0, −0.25); when (x, y) = (0, −1) we have u = (0.25, 0).

I When (x, y) = (1.2, 0) we have u = (0, 0.3); when (x, y) = (0, 1.2) we have u = (−0.3, 0); when (x, y) = (−1.2, 0) we have u = (0, −0.3); The vector ﬁeld u = (−y/4, x/4)

I When (x, y) = (1, 0) we have u = (0, 0.25); when (x, y) = (0, 1) we have u = (−0.25, 0); when (x, y) = (−1, 0) we have u = (0, −0.25); when (x, y) = (0, −1) we have u = (0.25, 0).

I When (x, y) = (1.2, 0) we have u = (0, 0.3); when (x, y) = (0, 1.2) we have u = (−0.3, 0); when (x, y) = (−1.2, 0) we have u = (0, −0.3); when (x, y) = (0, −1.2) we have u = (0.3, 0). The vector ﬁeld u = (−y/4, x/4)

I When (x, y) = (1, 0) we have u = (0, 0.25); when (x, y) = (0, 1) we have u = (−0.25, 0); when (x, y) = (−1, 0) we have u = (0, −0.25); when (x, y) = (0, −1) we have u = (0.25, 0).

I When (x, y) = (1.2, 0) we have u = (0, 0.3); when (x, y) = (0, 1.2) we have u = (−0.3, 0); when (x, y) = (−1.2, 0) we have u = (0, −0.3); when (x, y) = (0, −1.2) we have u = (0.3, 0). I When (x, y) = (0.87, 0.50) we have u = (−0.22, −0.12);

I when (x, y) = (0.50, 0.87) we have u = (−0.12, −0.22);

I when (x, y) = (0.00, 1.00) we have u = (0.00, −0.25);

I when (x, y) = (−0.50, 0.87) we have u = (0.12, −0.22).

The vector ﬁeld u = (−x/4, −y/4) I when (x, y) = (0.50, 0.87) we have u = (−0.12, −0.22);

I when (x, y) = (0.00, 1.00) we have u = (0.00, −0.25);

I when (x, y) = (−0.50, 0.87) we have u = (0.12, −0.22).

The vector ﬁeld u = (−x/4, −y/4)

I When (x, y) = (0.87, 0.50) we have u = (−0.22, −0.12); I when (x, y) = (0.00, 1.00) we have u = (0.00, −0.25);

I when (x, y) = (−0.50, 0.87) we have u = (0.12, −0.22).

The vector ﬁeld u = (−x/4, −y/4)

I When (x, y) = (0.87, 0.50) we have u = (−0.22, −0.12);

I when (x, y) = (0.50, 0.87) we have u = (−0.12, −0.22); I when (x, y) = (−0.50, 0.87) we have u = (0.12, −0.22).

The vector ﬁeld u = (−x/4, −y/4)

I When (x, y) = (0.87, 0.50) we have u = (−0.22, −0.12);

I when (x, y) = (0.50, 0.87) we have u = (−0.12, −0.22);

I when (x, y) = (0.00, 1.00) we have u = (0.00, −0.25); The vector ﬁeld u = (−x/4, −y/4)

I When (x, y) = (0.87, 0.50) we have u = (−0.22, −0.12);

I when (x, y) = (0.50, 0.87) we have u = (−0.12, −0.22);

I when (x, y) = (0.00, 1.00) we have u = (0.00, −0.25);

I when (x, y) = (−0.50, 0.87) we have u = (0.12, −0.22). The vector ﬁeld u = (−x/4, −y/4)

I When (x, y) = (0.87, 0.50) we have u = (−0.22, −0.12);

I when (x, y) = (0.50, 0.87) we have u = (−0.12, −0.22);

I when (x, y) = (0.00, 1.00) we have u = (0.00, −0.25);

I when (x, y) = (−0.50, 0.87) we have u = (0.12, −0.22). The vector ﬁeld u = (−x/4, −y/4)

I When (x, y) = (0.87, 0.50) we have u = (−0.22, −0.12);

I when (x, y) = (0.50, 0.87) we have u = (−0.12, −0.22);

I when (x, y) = (0.00, 1.00) we have u = (0.00, −0.25);

I when (x, y) = (−0.50, 0.87) we have u = (0.12, −0.22). The vector ﬁeld u = (−x/4, −y/4)

I When (x, y) = (0.87, 0.50) we have u = (−0.22, −0.12);

I when (x, y) = (0.50, 0.87) we have u = (−0.12, −0.22);

I when (x, y) = (0.00, 1.00) we have u = (0.00, −0.25);

I when (x, y) = (−0.50, 0.87) we have u = (0.12, −0.22). The vector ﬁeld u = (−x/4, −y/4)

I When (x, y) = (0.87, 0.50) we have u = (−0.22, −0.12);

I when (x, y) = (0.50, 0.87) we have u = (−0.12, −0.22);

I when (x, y) = (0.00, 1.00) we have u = (0.00, −0.25);

I when (x, y) = (−0.50, 0.87) we have u = (0.12, −0.22). The vector ﬁeld u = (−x/4, −y/4)

I When (x, y) = (0.87, 0.50) we have u = (−0.22, −0.12);

I when (x, y) = (0.50, 0.87) we have u = (−0.12, −0.22);

I when (x, y) = (0.00, 1.00) we have u = (0.00, −0.25);

I when (x, y) = (−0.50, 0.87) we have u = (0.12, −0.22). The vector ﬁeld u = (−x/4, −y/4)

I When (x, y) = (0.87, 0.50) we have u = (−0.22, −0.12);

I when (x, y) = (0.50, 0.87) we have u = (−0.12, −0.22);

I when (x, y) = (0.00, 1.00) we have u = (0.00, −0.25);

I when (x, y) = (−0.50, 0.87) we have u = (0.12, −0.22). The vector ﬁeld u = (−x/4, −y/4)

I When (x, y) = (0.87, 0.50) we have u = (−0.22, −0.12);

I when (x, y) = (0.50, 0.87) we have u = (−0.12, −0.22);

I when (x, y) = (0.00, 1.00) we have u = (0.00, −0.25);

I when (x, y) = (−0.50, 0.87) we have u = (0.12, −0.22). 1 x 2 − y 2 −2xy The formula for this vector ﬁeld is u = 1 − , . I 5 (x 2 + y 2)2 (x 2 + y 2)2 2 2 I Far from the origin where x + y is large we have u ' (1/5, 0). 2 2 I On the unit circle where x + y = 1 we have 2 2 2y u = (1 − x + y , −2xy)/5 = − 5 (−y, x) (tangent to the circle).

Fluid ﬂow around a cylinder 2 2 I Far from the origin where x + y is large we have u ' (1/5, 0). 2 2 I On the unit circle where x + y = 1 we have 2 2 2y u = (1 − x + y , −2xy)/5 = − 5 (−y, x) (tangent to the circle).

Fluid ﬂow around a cylinder

1 x 2 − y 2 −2xy The formula for this vector ﬁeld is u = 1 − , . I 5 (x 2 + y 2)2 (x 2 + y 2)2 2 2 I On the unit circle where x + y = 1 we have 2 2 2y u = (1 − x + y , −2xy)/5 = − 5 (−y, x) (tangent to the circle).

Fluid ﬂow around a cylinder

1 x 2 − y 2 −2xy The formula for this vector ﬁeld is u = 1 − , . I 5 (x 2 + y 2)2 (x 2 + y 2)2 2 2 I Far from the origin where x + y is large we have u ' (1/5, 0). Fluid ﬂow around a cylinder

1 x 2 − y 2 −2xy The formula for this vector ﬁeld is u = 1 − , . I 5 (x 2 + y 2)2 (x 2 + y 2)2 2 2 I Far from the origin where x + y is large we have u ' (1/5, 0). 2 2 I On the unit circle where x + y = 1 we have 2 2 2y u = (1 − x + y , −2xy)/5 = − 5 (−y, x) (tangent to the circle). ,

and similarly ry = y/r and rz = z/r. This means that ∇(r) = (x/r, y/r, z/r). More generally, for any n we have n n−1 n−1 n−2 (r )x = nr rx = nr x/r = nr x. The other two derivatives work in the same way, so ∇(r n) = nr n−2 (x, y, z).

(A vector ﬁeld, the gradient of f , sometimes written grad(f ) rather than ∇(f ).) (a) For the function f = x 3 + y 4 + z 5, we have ∇(f ) = (3x 2, 4y 3, 5z 4). (b) For the function f = sin(x) sin(y) sin(z) we have ∇(f ) = (cos(x) sin(y) sin(z), sin(x) cos(y) sin(z), sin(x) sin(y) cos(z)). 1 (c) For the function r = (x 2 + y 2 + z 2) 2 we have 1 2 2 2 − x 1 2 rx = 2 (x + y + z ) .2x = 1 = x/r (x 2 + y 2 + z 2) 2

The gradient of a scalar ﬁeld

∂f ∂f ∂f If f is a scalar ﬁeld, then we deﬁne ∇(f ) = (f , f , f ) = , , . x y z ∂x ∂y ∂z ,

(a) For the function f = x 3 + y 4 + z 5, we have ∇(f ) = (3x 2, 4y 3, 5z 4). (b) For the function f = sin(x) sin(y) sin(z) we have ∇(f ) = (cos(x) sin(y) sin(z), sin(x) cos(y) sin(z), sin(x) sin(y) cos(z)). 1 (c) For the function r = (x 2 + y 2 + z 2) 2 we have 1 2 2 2 − x 1 2 rx = 2 (x + y + z ) .2x = 1 = x/r (x 2 + y 2 + z 2) 2

The gradient of a scalar ﬁeld

∂f ∂f ∂f If f is a scalar field, then we define ∇(f ) = (f , f , f ) = , , . x y z ∂x ∂y ∂z (A vector field, the gradient of f , sometimes written grad(f ) rather than ∇(f ).) ,

(b) For the function f = sin(x) sin(y) sin(z) we have ∇(f ) = (cos(x) sin(y) sin(z), sin(x) cos(y) sin(z), sin(x) sin(y) cos(z)). 1 (c) For the function r = (x 2 + y 2 + z 2) 2 we have 1 2 2 2 − x 1 2 rx = 2 (x + y + z ) .2x = 1 = x/r (x 2 + y 2 + z 2) 2

The gradient of a scalar ﬁeld

1 (c) For the function r = (x 2 + y 2 + z 2) 2 we have 1 2 2 2 − x 1 2 rx = 2 (x + y + z ) .2x = 1 = x/r (x 2 + y 2 + z 2) 2

The gradient of a scalar ﬁeld

∂f ∂f ∂f If f is a scalar field, then we define ∇(f ) = (f , f , f ) = , , . x y z ∂x ∂y ∂z (A vector field, the gradient of f , sometimes written grad(f ) rather than ∇(f ).) (a) For the function f = x 3 + y 4 + z 5, we have ∇(f ) = (3x 2, 4y 3, 5z 4). (b) For the function f = sin(x) sin(y) sin(z) we have ∇(f ) = (cos(x) sin(y) sin(z), sin(x) cos(y) sin(z), sin(x) sin(y) cos(z)). 1 (c) For the function r = (x 2 + y 2 + z 2) 2 we have 1 2 2 2 − x 1 2 rx = 2 (x + y + z ) .2x = 1 = x/r (x 2 + y 2 + z 2) 2 This means that ∇(r) = (x/r, y/r, z/r). More generally, for any n we have n n−1 n−1 n−2 (r )x = nr rx = nr x/r = nr x. The other two derivatives work in the same way, so ∇(r n) = nr n−2 (x, y, z).

The gradient of a scalar ﬁeld

and similarly ry = y/r and rz = z/r. More generally, for any n we have n n−1 n−1 n−2 (r )x = nr rx = nr x/r = nr x. The other two derivatives work in the same way, so ∇(r n) = nr n−2 (x, y, z).

The gradient of a scalar ﬁeld

and similarly ry = y/r and rz = z/r. This means that ∇(r) = (x/r, y/r, z/r). The other two derivatives work in the same way, so ∇(r n) = nr n−2 (x, y, z).

The gradient of a scalar ﬁeld

and similarly ry = y/r and rz = z/r. This means that ∇(r) = (x/r, y/r, z/r). More generally, for any n we have n n−1 n−1 n−2 (r )x = nr rx = nr x/r = nr x. The gradient of a scalar ﬁeld

The picture below illustrates the two-dimensional version of this fact in the case where f = px 2/9 + y 2/4.

The four red ovals are given by f = 1, f = 2, f = 3 and f = 4. The blue arrows show the vector ﬁeld ∇(f ), which is perpendicular to the red ovals as expected.

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . The picture below illustrates the two-dimensional version of this fact in the case where f = px 2/9 + y 2/4.

The four red ovals are given by f = 1, f = 2, f = 3 and f = 4. The blue arrows show the vector ﬁeld ∇(f ), which is perpendicular to the red ovals as expected.

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant. The four red ovals are given by f = 1, f = 2, f = 3 and f = 4. The blue arrows show the vector ﬁeld ∇(f ), which is perpendicular to the red ovals as expected.

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

The picture below illustrates the two-dimensional version of this fact in the case where f = px 2/9 + y 2/4. The blue arrows show the vector ﬁeld ∇(f ), which is perpendicular to the red ovals as expected.

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

The picture below illustrates the two-dimensional version of this fact in the case where f = px 2/9 + y 2/4.

The four red ovals are given by f = 1, f = 2, f = 3 and f = 4. , which is perpendicular to the red ovals as expected.

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

The picture below illustrates the two-dimensional version of this fact in the case where f = px 2/9 + y 2/4.

The four red ovals are given by f = 1, f = 2, f = 3 and f = 4. The blue arrows show the vector ﬁeld ∇(f ) Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

The picture below illustrates the two-dimensional version of this fact in the case where f = px 2/9 + y 2/4.

The four red ovals are given by f = 1, f = 2, f = 3 and f = 4. The blue arrows show the vector ﬁeld ∇(f ), which is perpendicular to the red ovals as expected. = |∇(f )||δr| cos(θ),

where θ is the angle between δr and ∇(f ).

To see why the above fact is true, remember that if we make small changes δx, δy and δz to x, y and z, then the resulting change in f is approximately given by δf = fx δx + fy δy + fz δz. If we write r for the vector (x, y, z), this becomes

δf = ∇(f ).δr

If we move along a surface where f is constant, then δf will be zero so we must have cos(θ) = 0, so θ = ±π/2, so δr is perpendicular to ∇(f ). This means that ∇(f ) is perpendicular to the surfaces of constant f , as we stated before. On the other hand, to make δf as large as possible (for a ﬁxed step size |δr|) we need to maximise cos(θ), which means taking θ = 0, so that δr is in the same direction as ∇(f ). In other words, ∇(f ) points in the direction of maximum increase of f .

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant. = |∇(f )||δr| cos(θ),

where θ is the angle between δr and ∇(f ).

If we write r for the vector (x, y, z), this becomes

δf = ∇(f ).δr

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

To see why the above fact is true, remember that if we make small changes δx, δy and δz to x, y and z, then the resulting change in f is approximately given by δf = fx δx + fy δy + fz δz. = |∇(f )||δr| cos(θ),

where θ is the angle between δr and ∇(f ). If we move along a surface where f is constant, then δf will be zero so we must have cos(θ) = 0, so θ = ±π/2, so δr is perpendicular to ∇(f ). This means that ∇(f ) is perpendicular to the surfaces of constant f , as we stated before. On the other hand, to make δf as large as possible (for a ﬁxed step size |δr|) we need to maximise cos(θ), which means taking θ = 0, so that δr is in the same direction as ∇(f ). In other words, ∇(f ) points in the direction of maximum increase of f .

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

δf = ∇(f ).δr If we move along a surface where f is constant, then δf will be zero so we must have cos(θ) = 0, so θ = ±π/2, so δr is perpendicular to ∇(f ). This means that ∇(f ) is perpendicular to the surfaces of constant f , as we stated before. On the other hand, to make δf as large as possible (for a ﬁxed step size |δr|) we need to maximise cos(θ), which means taking θ = 0, so that δr is in the same direction as ∇(f ). In other words, ∇(f ) points in the direction of maximum increase of f .

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

δf = ∇(f ).δr = |∇(f )||δr| cos(θ),

where θ is the angle between δr and ∇(f ). so we must have cos(θ) = 0, so θ = ±π/2, so δr is perpendicular to ∇(f ). This means that ∇(f ) is perpendicular to the surfaces of constant f , as we stated before. On the other hand, to make δf as large as possible (for a ﬁxed step size |δr|) we need to maximise cos(θ), which means taking θ = 0, so that δr is in the same direction as ∇(f ). In other words, ∇(f ) points in the direction of maximum increase of f .

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

δf = ∇(f ).δr = |∇(f )||δr| cos(θ),

where θ is the angle between δr and ∇(f ). If we move along a surface where f is constant, then δf will be zero , so θ = ±π/2, so δr is perpendicular to ∇(f ). This means that ∇(f ) is perpendicular to the surfaces of constant f , as we stated before. On the other hand, to make δf as large as possible (for a ﬁxed step size |δr|) we need to maximise cos(θ), which means taking θ = 0, so that δr is in the same direction as ∇(f ). In other words, ∇(f ) points in the direction of maximum increase of f .

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

δf = ∇(f ).δr = |∇(f )||δr| cos(θ),

where θ is the angle between δr and ∇(f ). If we move along a surface where f is constant, then δf will be zero so we must have cos(θ) = 0 , so δr is perpendicular to ∇(f ). This means that ∇(f ) is perpendicular to the surfaces of constant f , as we stated before. On the other hand, to make δf as large as possible (for a ﬁxed step size |δr|) we need to maximise cos(θ), which means taking θ = 0, so that δr is in the same direction as ∇(f ). In other words, ∇(f ) points in the direction of maximum increase of f .

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

δf = ∇(f ).δr = |∇(f )||δr| cos(θ),

where θ is the angle between δr and ∇(f ). If we move along a surface where f is constant, then δf will be zero so we must have cos(θ) = 0, so θ = ±π/2 This means that ∇(f ) is perpendicular to the surfaces of constant f , as we stated before. On the other hand, to make δf as large as possible (for a ﬁxed step size |δr|) we need to maximise cos(θ), which means taking θ = 0, so that δr is in the same direction as ∇(f ). In other words, ∇(f ) points in the direction of maximum increase of f .

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

δf = ∇(f ).δr = |∇(f )||δr| cos(θ),

where θ is the angle between δr and ∇(f ). If we move along a surface where f is constant, then δf will be zero so we must have cos(θ) = 0, so θ = ±π/2, so δr is perpendicular to ∇(f ). On the other hand, to make δf as large as possible (for a ﬁxed step size |δr|) we need to maximise cos(θ), which means taking θ = 0, so that δr is in the same direction as ∇(f ). In other words, ∇(f ) points in the direction of maximum increase of f .

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

δf = ∇(f ).δr = |∇(f )||δr| cos(θ),

where θ is the angle between δr and ∇(f ). If we move along a surface where f is constant, then δf will be zero so we must have cos(θ) = 0, so θ = ±π/2, so δr is perpendicular to ∇(f ). This means that ∇(f ) is perpendicular to the surfaces of constant f , as we stated before. , which means taking θ = 0, so that δr is in the same direction as ∇(f ). In other words, ∇(f ) points in the direction of maximum increase of f .

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

δf = ∇(f ).δr = |∇(f )||δr| cos(θ),

where θ is the angle between δr and ∇(f ). If we move along a surface where f is constant, then δf will be zero so we must have cos(θ) = 0, so θ = ±π/2, so δr is perpendicular to ∇(f ). This means that ∇(f ) is perpendicular to the surfaces of constant f , as we stated before. On the other hand, to make δf as large as possible (for a ﬁxed step size |δr|) we need to maximise cos(θ) , so that δr is in the same direction as ∇(f ). In other words, ∇(f ) points in the direction of maximum increase of f .

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

δf = ∇(f ).δr = |∇(f )||δr| cos(θ),

where θ is the angle between δr and ∇(f ). If we move along a surface where f is constant, then δf will be zero so we must have cos(θ) = 0, so θ = ±π/2, so δr is perpendicular to ∇(f ). This means that ∇(f ) is perpendicular to the surfaces of constant f , as we stated before. On the other hand, to make δf as large as possible (for a ﬁxed step size |δr|) we need to maximise cos(θ), which means taking θ = 0 In other words, ∇(f ) points in the direction of maximum increase of f .

Geometry of the gradient

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

δf = ∇(f ).δr = |∇(f )||δr| cos(θ),

Fact: The vector ∇(f ) points in the direction of maximum increase of f . It is perpendicular to the surfaces where f is constant.

δf = ∇(f ).δr = |∇(f )||δr| cos(θ),

= (y 2, 2xy, 3z 2)

= (22, 2 × 1 × 2, 3 × 32) at (1, 2, 3) n.∇(f ) = (0.6, 0, 0.8).(4, 4, 27) = 0.6 × 4 + 0.8 × 27 = 24.

In other words, it is the component of ∇(f ) in the direction n.

Then ∂f ∂f ∂f ∇(f ) = , , ∂x ∂y ∂z

I The directional derivative of f at a in direction n is the rate of change of f when we move away from a in direction n at speed one.

I It can be calculated as n.∇(f ) (where ∇(f ) must be evaluated at a).

2 3 I Example: take f (x, y, z) = xy + z and a = (1, 2, 3) and n = (0.6, 0, 0.8).

Directional derivatives

I Suppose we have a function f (x, y, z), a point a and a unit vector n. = (4, 4, 27)

= (y 2, 2xy, 3z 2)

= (22, 2 × 1 × 2, 3 × 32) at (1, 2, 3) n.∇(f ) = (0.6, 0, 0.8).(4, 4, 27) = 0.6 × 4 + 0.8 × 27 = 24.

In other words, it is the component of ∇(f ) in the direction n.

Then ∂f ∂f ∂f ∇(f ) = , , ∂x ∂y ∂z

I It can be calculated as n.∇(f ) (where ∇(f ) must be evaluated at a).

2 3 I Example: take f (x, y, z) = xy + z and a = (1, 2, 3) and n = (0.6, 0, 0.8).

Directional derivatives

I Suppose we have a function f (x, y, z), a point a and a unit vector n.

I The directional derivative of f at a in direction n is the rate of change of f when we move away from a in direction n at speed one. = (4, 4, 27)

= (y 2, 2xy, 3z 2)

= (22, 2 × 1 × 2, 3 × 32) at (1, 2, 3) n.∇(f ) = (0.6, 0, 0.8).(4, 4, 27) = 0.6 × 4 + 0.8 × 27 = 24.

Then ∂f ∂f ∂f ∇(f ) = , , ∂x ∂y ∂z

In other words, it is the component of ∇(f ) in the direction n. 2 3 I Example: take f (x, y, z) = xy + z and a = (1, 2, 3) and n = (0.6, 0, 0.8).

Directional derivatives

I Suppose we have a function f (x, y, z), a point a and a unit vector n.

I The directional derivative of f at a in direction n is the rate of change of f when we move away from a in direction n at speed one.

I It can be calculated as n.∇(f ) (where ∇(f ) must be evaluated at a). = (4, 4, 27)

= (y 2, 2xy, 3z 2)

= (22, 2 × 1 × 2, 3 × 32) at (1, 2, 3) n.∇(f ) = (0.6, 0, 0.8).(4, 4, 27) = 0.6 × 4 + 0.8 × 27 = 24.

Then ∂f ∂f ∂f ∇(f ) = , , ∂x ∂y ∂z

2 3 I Example: take f (x, y, z) = xy + z and a = (1, 2, 3) and n = (0.6, 0, 0.8).

Directional derivatives

I Suppose we have a function f (x, y, z), a point a and a unit vector n.

I The directional derivative of f at a in direction n is the rate of change of f when we move away from a in direction n at speed one.

I It can be calculated as n.∇(f ) (where ∇(f ) must be evaluated at a). In other words, it is the component of ∇(f ) in the direction n. = (4, 4, 27)

= (y 2, 2xy, 3z 2)

= (22, 2 × 1 × 2, 3 × 32) at (1, 2, 3) n.∇(f ) = (0.6, 0, 0.8).(4, 4, 27) = 0.6 × 4 + 0.8 × 27 = 24.

Then ∂f ∂f ∂f ∇(f ) = , , ∂x ∂y ∂z

Directional derivatives

I Suppose we have a function f (x, y, z), a point a and a unit vector n.

I The directional derivative of f at a in direction n is the rate of change of f when we move away from a in direction n at speed one.

= (y 2, 2xy, 3z 2)

= (22, 2 × 1 × 2, 3 × 32) at (1, 2, 3) n.∇(f ) = (0.6, 0, 0.8).(4, 4, 27) = 0.6 × 4 + 0.8 × 27 = 24.

Directional derivatives

I Suppose we have a function f (x, y, z), a point a and a unit vector n.

I The directional derivative of f at a in direction n is the rate of change of f when we move away from a in direction n at speed one.

= (22, 2 × 1 × 2, 3 × 32) at (1, 2, 3) n.∇(f ) = (0.6, 0, 0.8).(4, 4, 27) = 0.6 × 4 + 0.8 × 27 = 24.

Directional derivatives

I Suppose we have a function f (x, y, z), a point a and a unit vector n.

I The directional derivative of f at a in direction n is the rate of change of f when we move away from a in direction n at speed one.

I It can be calculated as n.∇(f ) (where ∇(f ) must be evaluated at a). In other words, it is the component of ∇(f ) in the direction n. 2 3 I Example: take f (x, y, z) = xy + z and a = (1, 2, 3) and n = (0.6, 0, 0.8). Then ∂f ∂f ∂f ∇(f ) = , , = (y 2, 2xy, 3z 2) ∂x ∂y ∂z = (4, 4, 27) n.∇(f ) = (0.6, 0, 0.8).(4, 4, 27) = 0.6 × 4 + 0.8 × 27 = 24.

Directional derivatives

I Suppose we have a function f (x, y, z), a point a and a unit vector n.

I The directional derivative of f at a in direction n is the rate of change of f when we move away from a in direction n at speed one.

I It can be calculated as n.∇(f ) (where ∇(f ) must be evaluated at a). In other words, it is the component of ∇(f ) in the direction n. 2 3 I Example: take f (x, y, z) = xy + z and a = (1, 2, 3) and n = (0.6, 0, 0.8). Then ∂f ∂f ∂f ∇(f ) = , , = (y 2, 2xy, 3z 2) ∂x ∂y ∂z = (22, 2 × 1 × 2, 3 × 32) at (1, 2, 3) n.∇(f ) = (0.6, 0, 0.8).(4, 4, 27) = 0.6 × 4 + 0.8 × 27 = 24.

Directional derivatives

I Suppose we have a function f (x, y, z), a point a and a unit vector n.

I The directional derivative of f at a in direction n is the rate of change of f when we move away from a in direction n at speed one.

I It can be calculated as n.∇(f ) (where ∇(f ) must be evaluated at a). In other words, it is the component of ∇(f ) in the direction n. 2 3 I Example: take f (x, y, z) = xy + z and a = (1, 2, 3) and n = (0.6, 0, 0.8). Then ∂f ∂f ∂f ∇(f ) = , , = (y 2, 2xy, 3z 2) ∂x ∂y ∂z = (22, 2 × 1 × 2, 3 × 32) = (4, 4, 27) at (1, 2, 3) = 0.6 × 4 + 0.8 × 27 = 24.

Directional derivatives

I Suppose we have a function f (x, y, z), a point a and a unit vector n.

I The directional derivative of f at a in direction n is the rate of change of f when we move away from a in direction n at speed one.

I It can be calculated as n.∇(f ) (where ∇(f ) must be evaluated at a). In other words, it is the component of ∇(f ) in the direction n. 2 3 I Example: take f (x, y, z) = xy + z and a = (1, 2, 3) and n = (0.6, 0, 0.8). Then ∂f ∂f ∂f ∇(f ) = , , = (y 2, 2xy, 3z 2) ∂x ∂y ∂z = (22, 2 × 1 × 2, 3 × 32) = (4, 4, 27) at (1, 2, 3) n.∇(f ) = (0.6, 0, 0.8).(4, 4, 27) = 24.

Directional derivatives

I Suppose we have a function f (x, y, z), a point a and a unit vector n.

I The directional derivative of f at a in direction n is the rate of change of f when we move away from a in direction n at speed one.

I It can be calculated as n.∇(f ) (where ∇(f ) must be evaluated at a). In other words, it is the component of ∇(f ) in the direction n. 2 3 I Example: take f (x, y, z) = xy + z and a = (1, 2, 3) and n = (0.6, 0, 0.8). Then ∂f ∂f ∂f ∇(f ) = , , = (y 2, 2xy, 3z 2) ∂x ∂y ∂z = (22, 2 × 1 × 2, 3 × 32) = (4, 4, 27) at (1, 2, 3) n.∇(f ) = (0.6, 0, 0.8).(4, 4, 27) = 0.6 × 4 + 0.8 × 27 Directional derivatives

I Suppose we have a function f (x, y, z), a point a and a unit vector n.

I The directional derivative of f at a in direction n is the rate of change of f when we move away from a in direction n at speed one.

I It can be calculated as n.∇(f ) (where ∇(f ) must be evaluated at a). In other words, it is the component of ∇(f ) in the direction n. 2 3 I Example: take f (x, y, z) = xy + z and a = (1, 2, 3) and n = (0.6, 0, 0.8). Then ∂f ∂f ∂f ∇(f ) = , , = (y 2, 2xy, 3z 2) ∂x ∂y ∂z = (22, 2 × 1 × 2, 3 × 32) = (4, 4, 27) at (1, 2, 3) n.∇(f ) = (0.6, 0, 0.8).(4, 4, 27) = 0.6 × 4 + 0.8 × 27 = 24. , and the gravitational force ﬁeld is proportional to ∇(ψ).

These are related by the equation E = ∇(φ). (All this is valid only when there are no signiﬁcant time-varying magnetic ﬁelds.) (b) Similarly, there is a gravitational potential function ψ

Applications of the gradient

(a) We write E for the electric field (which is a vector field) and φ for the electric potential (which is a scalar field). , and the gravitational force field is proportional to ∇(ψ).

(All this is valid only when there are no signiﬁcant time-varying magnetic ﬁelds.) (b) Similarly, there is a gravitational potential function ψ

Applications of the gradient

(a) We write E for the electric field (which is a vector field) and φ for the electric potential (which is a scalar field). These are related by the equation E = ∇(φ). , and the gravitational force field is proportional to ∇(ψ).

(b) Similarly, there is a gravitational potential function ψ

Applications of the gradient

(a) We write E for the electric field (which is a vector field) and φ for the electric potential (which is a scalar field). These are related by the equation E = ∇(φ). (All this is valid only when there are no significant time-varying magnetic fields.) , and the gravitational force field is proportional to ∇(ψ). (c) The net force on a particle of air involves ∇(p), where p is the pressure.

Applications of the gradient

(a) We write E for the electric field (which is a vector field) and φ for the electric potential (which is a scalar field). These are related by the equation E = ∇(φ). (All this is valid only when there are no significant time-varying magnetic fields.) (b) Similarly, there is a gravitational potential function ψ (c) The net force on a particle of air involves ∇(p), where p is the pressure.

Applications of the gradient

(a) We write E for the electric field (which is a vector field) and φ for the electric potential (which is a scalar field). These are related by the equation E = ∇(φ). (All this is valid only when there are no significant time-varying magnetic fields.) (b) Similarly, there is a gravitational potential function ψ, and the gravitational force field is proportional to ∇(ψ). Applications of the gradient

(a) We write E for the electric field (which is a vector field) and φ for the electric potential (which is a scalar field). These are related by the equation E = ∇(φ). (All this is valid only when there are no significant time-varying magnetic fields.) (b) Similarly, there is a gravitational potential function ψ, and the gravitational force field is proportional to ∇(ψ). (c) The net force on a particle of air involves ∇(p), where p is the pressure. = −x/r 3

= −Ar −3(x, y, z) = −Ar/r 3.

Note that 3 −1 1 2 2 2 − 2 (r )x = − 2 (x + y + z ) .2x −1 3 −1 3 and similarly (r )y = −y/r and (r )z = −z/r . This gives the electric ﬁeld:

E = ∇(φ)

Electric ﬁeld of a point charge

If we have a single charge at the origin, then the resulting electric potential 1 function is φ = Ar −1 for some constant A, where r = (x 2 + y 2 + z 2) 2 as usual. = −Ar −3(x, y, z) = −Ar/r 3.

= −x/r 3 −1 3 −1 3 and similarly (r )y = −y/r and (r )z = −z/r . This gives the electric ﬁeld:

E = ∇(φ)

Electric ﬁeld of a point charge

If we have a single charge at the origin, then the resulting electric potential 1 function is φ = Ar −1 for some constant A, where r = (x 2 + y 2 + z 2) 2 as usual. Note that 3 −1 1 2 2 2 − 2 (r )x = − 2 (x + y + z ) .2x = −Ar −3(x, y, z) = −Ar/r 3.

−1 3 −1 3 and similarly (r )y = −y/r and (r )z = −z/r . This gives the electric ﬁeld:

E = ∇(φ)

Electric ﬁeld of a point charge

If we have a single charge at the origin, then the resulting electric potential 1 function is φ = Ar −1 for some constant A, where r = (x 2 + y 2 + z 2) 2 as usual. Note that 3 −1 1 2 2 2 − 2 3 (r )x = − 2 (x + y + z ) .2x = −x/r = −Ar −3(x, y, z) = −Ar/r 3.

This gives the electric ﬁeld:

E = ∇(φ)

Electric ﬁeld of a point charge

E = ∇(φ) = −Ar/r 3.

Electric ﬁeld of a point charge

E = ∇(φ) = −Ar −3(x, y, z) Electric ﬁeld of a point charge

E = ∇(φ) = −Ar −3(x, y, z) = −Ar/r 3. Ax = − . x 2 + y 2

It works out that the corresponding electric potential function is 1 2 2 φ = − 2 A ln(x + y ) for some constant A. This is independent of z, so φz = 0. On the other hand, we have − 1 A φ = 2 .2x x x 2 + y 2

2 2 By a similar calculation we have φy = −Ay/(x + y ), so

Ax Ay E = ∇(φ) = − , − , 0 . x 2 + y 2 x 2 + y 2

Electric ﬁeld of a line charge

Suppose instead that we have a whole line of charges distributed along the z-axis. Ax = − . x 2 + y 2

This is independent of z, so φz = 0. On the other hand, we have − 1 A φ = 2 .2x x x 2 + y 2

2 2 By a similar calculation we have φy = −Ay/(x + y ), so

Ax Ay E = ∇(φ) = − , − , 0 . x 2 + y 2 x 2 + y 2

Electric ﬁeld of a line charge

On the other hand, we have − 1 A φ = 2 .2x x x 2 + y 2

2 2 By a similar calculation we have φy = −Ay/(x + y ), so

Ax Ay E = ∇(φ) = − , − , 0 . x 2 + y 2 x 2 + y 2

Electric ﬁeld of a line charge

Suppose instead that we have a whole line of charges distributed along the z-axis. It works out that the corresponding electric potential function is 1 2 2 φ = − 2 A ln(x + y ) for some constant A. This is independent of z, so φz = 0. Ax = − . x 2 + y 2

2 2 By a similar calculation we have φy = −Ay/(x + y ), so

Ax Ay E = ∇(φ) = − , − , 0 . x 2 + y 2 x 2 + y 2

Electric ﬁeld of a line charge

Ax Ay E = ∇(φ) = − , − , 0 . x 2 + y 2 x 2 + y 2

Electric ﬁeld of a line charge

− 1 A Ax φ = 2 .2x = − . x x 2 + y 2 x 2 + y 2 , so

Ax Ay E = ∇(φ) = − , − , 0 . x 2 + y 2 x 2 + y 2

Electric ﬁeld of a line charge

− 1 A Ax φ = 2 .2x = − . x x 2 + y 2 x 2 + y 2

2 2 By a similar calculation we have φy = −Ay/(x + y ) Electric ﬁeld of a line charge

− 1 A Ax φ = 2 .2x = − . x x 2 + y 2 x 2 + y 2

2 2 By a similar calculation we have φy = −Ay/(x + y ), so

Ax Ay E = ∇(φ) = − , − , 0 . x 2 + y 2 x 2 + y 2 The corresponding electric ﬁeld is

E = ∇(φ) = (a, b, c).

In other words, we have a uniform electric ﬁeld everywhere. If we put u = (a, b, c) we can write the above in vector notation as φ = u.r and ∇(φ) = ∇(u.r) = u.

Uniform electric ﬁeld

Suppose we have an electric potential of the form φ = ax + by + cz, where a, b and c are constant. In other words, we have a uniform electric ﬁeld everywhere. If we put u = (a, b, c) we can write the above in vector notation as φ = u.r and ∇(φ) = ∇(u.r) = u.

Uniform electric ﬁeld

Suppose we have an electric potential of the form φ = ax + by + cz, where a, b and c are constant. The corresponding electric ﬁeld is

E = ∇(φ) = (a, b, c). If we put u = (a, b, c) we can write the above in vector notation as φ = u.r and ∇(φ) = ∇(u.r) = u.

Uniform electric ﬁeld

Suppose we have an electric potential of the form φ = ax + by + cz, where a, b and c are constant. The corresponding electric ﬁeld is

E = ∇(φ) = (a, b, c).

In other words, we have a uniform electric ﬁeld everywhere. and ∇(φ) = ∇(u.r) = u.

Uniform electric ﬁeld

Suppose we have an electric potential of the form φ = ax + by + cz, where a, b and c are constant. The corresponding electric ﬁeld is

E = ∇(φ) = (a, b, c).

In other words, we have a uniform electric ﬁeld everywhere. If we put u = (a, b, c) we can write the above in vector notation as φ = u.r Uniform electric ﬁeld

Suppose we have an electric potential of the form φ = ax + by + cz, where a, b and c are constant. The corresponding electric ﬁeld is

E = ∇(φ) = (a, b, c).

In other words, we have a uniform electric ﬁeld everywhere. If we put u = (a, b, c) we can write the above in vector notation as φ = u.r and ∇(φ) = ∇(u.r) = u. 1 −y −y = = 1 + (y/x)2 x 2 x 2 + y 2 y ∂ y 1 1 x θ = arctan0 = = y x ∂y x 1 + (y/x)2 x x 2 + y 2

θz = 0,

It is a standard fact that arctan0(t) = 1/(1 + t2). Using this, we get y ∂ y θ = arctan0 x x ∂x x

so −y x ∇(θ) = , , 0 . x 2 + y 2 x 2 + y 2

grad(θ)

Consider the function

θ(x, y, z) = angle between the x-axis and (x, y, 0) = arctan(y/x)

(as used in polar coordinates). 1 −y −y = = 1 + (y/x)2 x 2 x 2 + y 2 y ∂ y 1 1 x θ = arctan0 = = y x ∂y x 1 + (y/x)2 x x 2 + y 2

θz = 0,

Using this, we get y ∂ y θ = arctan0 x x ∂x x

so −y x ∇(θ) = , , 0 . x 2 + y 2 x 2 + y 2

grad(θ)

Consider the function

θ(x, y, z) = angle between the x-axis and (x, y, 0) = arctan(y/x)

(as used in polar coordinates). It is a standard fact that arctan0(t) = 1/(1 + t2). 1 −y −y = = 1 + (y/x)2 x 2 x 2 + y 2 y ∂ y 1 1 x θ = arctan0 = = y x ∂y x 1 + (y/x)2 x x 2 + y 2

θz = 0,

so −y x ∇(θ) = , , 0 . x 2 + y 2 x 2 + y 2

grad(θ)

Consider the function

θ(x, y, z) = angle between the x-axis and (x, y, 0) = arctan(y/x)

(as used in polar coordinates). It is a standard fact that arctan0(t) = 1/(1 + t2). Using this, we get y ∂ y θ = arctan0 x x ∂x x −y = x 2 + y 2 y ∂ y 1 1 x θ = arctan0 = = y x ∂y x 1 + (y/x)2 x x 2 + y 2

θz = 0,

so −y x ∇(θ) = , , 0 . x 2 + y 2 x 2 + y 2

grad(θ)

Consider the function

θ(x, y, z) = angle between the x-axis and (x, y, 0) = arctan(y/x)

(as used in polar coordinates). It is a standard fact that arctan0(t) = 1/(1 + t2). Using this, we get y ∂ y 1 −y θ = arctan0 = x x ∂x x 1 + (y/x)2 x 2 y ∂ y 1 1 x θ = arctan0 = = y x ∂y x 1 + (y/x)2 x x 2 + y 2

θz = 0,

so −y x ∇(θ) = , , 0 . x 2 + y 2 x 2 + y 2

grad(θ)

Consider the function

θ(x, y, z) = angle between the x-axis and (x, y, 0) = arctan(y/x)

(as used in polar coordinates). It is a standard fact that arctan0(t) = 1/(1 + t2). Using this, we get y ∂ y 1 −y −y θ = arctan0 = = x x ∂x x 1 + (y/x)2 x 2 x 2 + y 2 1 1 x = = 1 + (y/x)2 x x 2 + y 2

θz = 0,

so −y x ∇(θ) = , , 0 . x 2 + y 2 x 2 + y 2

grad(θ)

Consider the function

θ(x, y, z) = angle between the x-axis and (x, y, 0) = arctan(y/x)

(as used in polar coordinates). It is a standard fact that arctan0(t) = 1/(1 + t2). Using this, we get y ∂ y 1 −y −y θ = arctan0 = = x x ∂x x 1 + (y/x)2 x 2 x 2 + y 2 y ∂ y θ = arctan0 y x ∂y x x = x 2 + y 2

θz = 0,

so −y x ∇(θ) = , , 0 . x 2 + y 2 x 2 + y 2

grad(θ)

Consider the function

θ(x, y, z) = angle between the x-axis and (x, y, 0) = arctan(y/x)

(as used in polar coordinates). It is a standard fact that arctan0(t) = 1/(1 + t2). Using this, we get y ∂ y 1 −y −y θ = arctan0 = = x x ∂x x 1 + (y/x)2 x 2 x 2 + y 2 y ∂ y 1 1 θ = arctan0 = y x ∂y x 1 + (y/x)2 x θz = 0,

so −y x ∇(θ) = , , 0 . x 2 + y 2 x 2 + y 2

grad(θ)

Consider the function

θ(x, y, z) = angle between the x-axis and (x, y, 0) = arctan(y/x)

(as used in polar coordinates). It is a standard fact that arctan0(t) = 1/(1 + t2). Using this, we get y ∂ y 1 −y −y θ = arctan0 = = x x ∂x x 1 + (y/x)2 x 2 x 2 + y 2 y ∂ y 1 1 x θ = arctan0 = = y x ∂y x 1 + (y/x)2 x x 2 + y 2 ,

so −y x ∇(θ) = , , 0 . x 2 + y 2 x 2 + y 2

grad(θ)

Consider the function

θ(x, y, z) = angle between the x-axis and (x, y, 0) = arctan(y/x)

θz = 0 grad(θ)

Consider the function

θ(x, y, z) = angle between the x-axis and (x, y, 0) = arctan(y/x)

θz = 0,

so −y x ∇(θ) = , , 0 . x 2 + y 2 x 2 + y 2