What is GRADIENT ∇ in MATHEMATICS ?

GRADIENT, ∇

To provide a motivation for the vector nature of partial derivatives, we now introduce the

total variation of a function F(x, y),

dF =

∂F

∂x

dx +

∂F

∂y

dy.

It consists of independent variations in the x- and y-directions. We write dF as a sum of

two increments, one purely in the x- and the other in the y-direction,

dF(x,y) ≡ F(x +dx,y +dy)− F(x, y)

=

F(x +dx,y +dy)− F(x,y +dy)

+

F(x,y + dy)−F(x, y)

=

∂F

∂x

dx +

∂F

∂y

dy,

by adding and subtracting F(x,y +dy). The mean value theorem (that is, continuity of F)

tells us that here ∂F/∂x, ∂F/∂y are evaluated at some point ξ,η between x and x + dx, y

1.6 Gradient, ∇ 33

and y + dy, respectively. As dx→0 and dy→0, ξ →x and η→y. This result generalizes

to three and higher dimensions. For example, for a function ϕ of three variables,

dϕ(x, y, z) ≡

ϕ(x +dx,y + dy, z +dz) −ϕ(x, y +dy, z +dz)

+

ϕ(x, y +dy, z +dz)− ϕ(x, y, z + dz)

+

ϕ(x, y, z +dz)− ϕ(x, y, z)

(1.57)

=

∂ϕ

∂x

dx +

∂ϕ

∂y

dy +

∂ϕ

∂z

dz.

Algebraically, dϕ in the total variation is a scalar product of the change in position dr and

the directional change of ϕ. And now we are ready to recognize the three-dimensional

partial derivative as a vector, which leads us to the concept of gradient.

Suppose that ϕ(x, y, z) is a scalar point function, that is, a function whose value depends

on the values of the coordinates (x, y, z). As a scalar, it must have the same value at a given

fixed point in space, independent of the rotation of our coordinate system, or

ϕ′(x′1, x′2, x′3) = ϕ(x1, x2, x3). (1.58)

By differentiating with respect to x′i we obtain

∂ϕ′(x′1, x′2, x′3)

∂x′i =

∂ϕ(x1, x2, x3)

∂x′i =

j

∂ϕ

∂xj

∂xj

∂x′i =

j

aij

∂ϕ

∂xj

(1.59)

by the rules of partial differentiation and Eqs. (1.16a) and (1.16b). But comparison with

Eq. (1.17), the vector transformation law, now shows that we have constructed a vector

with components ∂ϕ/∂xj . This vector we label the gradient of ϕ.

A convenient symbolism is

∇ϕ = ˆx

∂ϕ

∂x + ˆy

∂ϕ

∂y + ˆz

∂ϕ

∂z

(1.60)

or

∇ = ˆx

∂

∂x + ˆy

∂

∂y + ˆz

∂

∂z

. (1.61)

∇ϕ (or del ϕ) is our gradient of the scalar ϕ, whereas ∇ (del) itself is a vector differential

operator (available to operate on or to differentiate a scalar ϕ). All the relationships for ∇

(del) can be derived from the hybrid nature of del in terms of both the partial derivatives

and its vector nature.

The gradient of a scalar is extremely important in physics and engineering in expressing

the relation between a force field and a potential field,

force F=−∇(potential V ), (1.62)

which holds for both gravitational and electrostatic fields, among others. Note that the

minus sign in Eq. (1.62) results in water flowing downhill rather than uphill! If a force can

be described, as in Eq. (1.62), by a single function V (r) everywhere, we call the scalar

function V its potential. Because the force is the directional derivative of the potential, we

can find the potential, if it exists, by integrating the force along a suitable path. Because the

34 Chapter 1 Vector Analysis

total variation dV = ∇V · dr=−F · dr is the work done against the force along the path

dr, we recognize the physical meaning of the potential (difference) as work and energy.

Moreover, in a sum of path increments the intermediate points cancel,

V (r +dr1 +dr2)− V (r+ dr1)

+

V (r +dr1)− V (r)

= V (r +dr2 +dr1)− V (r),

so the integrated work along some path from an initial point ri to a final point r is given by

the potential difference V (r) − V (ri ) at the endpoints of the path. Therefore, such forces

are especially simple and well behaved: They are called conservative.When there is loss of

energy due to friction along the path or some other dissipation, the work will depend on the

path, and such forces cannot be conservative: No potential exists. We discuss conservative

forces in more detail in Section 1.13.

Example 1.6.1 THE GRADIENT OF A POTENTIAL V (r)

Let us calculate the gradient of V (r) = V (

x2 +y2 +z2 ), so

∇V (r) = ˆx

∂V (r)

∂x + ˆy

∂V (r)

∂y + ˆz

∂V (r)

∂z

.

Now, V (r) depends on x through the dependence of r on x. Therefore14

∂V (r)

∂x =

dV (r)

dr ·

∂r

∂x

.

From r as a function of x, y, z,

∂r

∂x =

∂(x2 +y2 +z2)1/2

∂x =

x

(x2 +y2 +z2)1/2 =

x

r

.

Therefore

∂V (r)

∂x =

dV (r)

dr ·

x

r

.

Permuting coordinates (x→y,y→z, z→x) to obtain the y and z derivatives, we get

∇V (r) = (ˆxx + ˆyy + ˆzz)

1

r

dV

dr

=

r

r

dV

dr = ˆr

dV

dr

.

Here ˆr is a unit vector (r/r) in the positive radial direction. The gradient of a function of

r is a vector in the (positive or negative) radial direction. In Section 2.5, ˆr is seen as one

of the three orthonormal unit vectors of spherical polar coordinates and ˆr∂/∂r as the radial

component of ∇.

14This is a special case of the chain rule of partial differentiation:

∂V(r, θ,ϕ)

∂x =

∂V

∂r

∂r

∂x +

∂V

∂θ

∂θ

∂x +

∂V

∂ϕ

∂ϕ

∂x

,

where ∂V/∂θ = ∂V/∂ϕ = 0, ∂V/∂r →dV/dr.

1.6 Gradient, ∇ 35

A Geometrical Interpretation

One immediate application of ∇ϕ is to dot it into an increment of length

dr = ˆxdx + ˆy dy + ˆz dz.

Thus we obtain

∇ϕ · dr =

∂ϕ

∂x

dx +

∂ϕ

∂y

dy +

∂ϕ

∂z

dz = dϕ,

the change in the scalar function ϕ corresponding to a change in position dr. Now consider

P and Q to be two points on a surface ϕ(x, y, z) = C, a constant. These points are chosen

so thatQ is a distance dr from P . Then, moving from P to Q, the change in ϕ(x, y, z) = C

is given by

dϕ = (∇ϕ) · dr = 0 (1.63)

since we stay on the surface ϕ(x, y, z) = C. This shows that ∇ϕ is perpendicular to dr.

Since dr may have any direction from P as long as it stays in the surface of constant ϕ,

point Q being restricted to the surface but having arbitrary direction, ∇ϕ is seen as normal

to the surface ϕ = constant (Fig. 1.19).

If we now permit dr to take us from one surface ϕ = C1 to an adjacent surface ϕ = C2

(Fig. 1.20),

dϕ = C1 −C2 = C = (∇ϕ) · dr. (1.64)

For a given dϕ, |dr| is a minimum when it is chosen parallel to ∇ϕ (cos θ = 1); or, for

a given |dr|, the change in the scalar function ϕ is maximized by choosing dr parallel to

FIGURE 1.19 The length increment dr has to stay on the surface ϕ = C.

36 Chapter 1 Vector Analysis

FIGURE 1.20 Gradient.

∇ϕ. This identifies ∇ϕ as a vector having the direction of the maximum space rate

of change of ϕ, an identification that will be useful in Chapter 2 when we consider non-

Cartesian coordinate systems. This identification of ∇ϕ may also be developed by using

the calculus of variations subject to a constraint, Exercise 17.6.9.

Example 1.6.2 FORCE AS GRADIENT OF A POTENTIAL

As a specific example of the foregoing, and as an extension of Example 1.6.1, we consider

the surfaces consisting of concentric spherical shells, Fig. 1.21. We have

ϕ(x, y, z) =

x2 + y2 +z2 1/2

= r = C,

where r is the radius, equal to C, our constant. C = ϕ = r, the distance between two

shells. From Example 1.6.1

∇ϕ(r) = ˆr

dϕ(r)

dr = ˆr.

The gradient is in the radial direction and is normal to the spherical surface ϕ = C.

Example 1.6.3 INTEGRATION BY PARTS OF GRADIENT

Let us prove the formula

A(r) ·∇f (r)d3r =−

f (r)∇·A(r)d3r, where A or f or both

vanish at infinity so that the integrated parts vanish. This condition is satisfied if, for example,

A is the electromagnetic vector potential and f is a bound-state wave function ψ(r).

1.6 Gradient, ∇ 37

FIGURE 1.21 Gradient for

ϕ(x, y, z) = (x2 + y2 + z2)1/2, spherical

shells: (x2

2 + y2

2 + z2

2)1/2 = r2 = C2,

(x2

1 + y2

1 + z2

1)1/2 = r1 = C1.

Writing the inner product in Cartesian coordinates, integrating each one-dimensional

integral by parts, and dropping the integrated terms, we obtain

A(r) · ∇f (r)d3r =

 

Axf |∞x =−∞ −

f

∂Ax

∂x

dx

dy dz+···

=−

f

∂Ax

∂x

dx dy dz −

f

∂Ay

∂y

dy dx dz −

f

∂Az

∂z

dz dx dy

=−

f (r)∇ · A(r)d3r.

If A = eikz ˆe describes an outgoing photon in the direction of the constant polarization unit

vector ˆe and f = ψ(r) is an exponentially decaying bound-state wave function, then

eikz ˆe · ∇ψ(r)d3r =−ez

ψ(r)

deikz

dz

d3r =−ikez

ψ(r)eikz d3r,

because only the z-component of the gradient contributes.

Exercises

1.6.1 If S(x, y, z) = (x2 + y2 + z2)−3/2, find

(a) ∇S at the point (1, 2, 3);

(b) the magnitude of the gradient of S, |∇S| at (1, 2, 3); and

(c) the direction cosines of ∇S at (1, 2, 3).

38 Chapter 1 Vector Analysis

1.6.2 (a) Find a unit vector perpendicular to the surface

x2 +y2 + z2 = 3

at the point (1, 1, 1). Lengths are in centimeters.

(b) Derive the equation of the plane tangent to the surface at (1, 1, 1).

ANS. (a) (ˆx + ˆy+ ˆz)/√3, (b) x +y +z = 3.

1.6.3 Given a vector r12 = ˆx(x1 − x2) + ˆy(y1 − y2) + ˆz(z1 − z2), show that ∇1r12 (gradient

with respect to x1, y1, and z1 of the magnitude r12) is a unit vector in the direction of

r12.

1.6.4 If a vector function F depends on both space coordinates (x, y, z) and time t , show that

dF = (dr · ∇)F+

∂F

∂t

dt.

1.6.5 Show that ∇(uv) = v∇u + u∇v, where u and v are differentiable scalar functions of

x, y, and z.

(a) Show that a necessary and sufficient condition that u(x, y, z) and v(x, y, z) are

related by some function f (u,v) = 0 is that (∇u)×(∇v) = 0.

(b) If u = u(x, y) and v = v(x, y), show that the condition (∇u) ×(∇v) = 0 leads to

the two-dimensional Jacobian

J

u, v

x,y

=

∂

u

∂x

∂u

∂y

∂v

∂x

∂v

∂y

= 0.

The functions u and v are assumed differentiable.

1.7 DIVERGENCE, ∇

Differentiating a vector function is a simple extension of differentiating scalar quantities.

Suppose r(t) describes the position of a satellite at some time t . Then, for differentiation

with respect to time,

dr(t)

dt = lim

→0

r(t + t)− r(t)

t = v, linear velocity.

Graphically, we again have the slope of a curve, orbit, or trajectory, as shown in Fig. 1.22.

If we resolve r(t) into its Cartesian components, dr/dt always reduces directly to a

vector sum of not more than three (for three-dimensional space) scalar derivatives. In other

coordinate systems (Chapter 2) the situation is more complicated, for the unit vectors are

no longer constant in direction. Differentiation with respect to the space coordinates is

handled in the same way as differentiation with respect to time, as seen in the following

paragraphs.

1.7 Divergence, ∇ 39

FIGURE 1.22 Differentiation of a vector.

In Section 1.6, ∇ was defined as a vector operator. Now, paying attention to both its

vector and its differential properties, we let it operate on a vector. First, as a vector we dot

it into a second vector to obtain

∇ · V =

∂Vx

∂x +

∂Vy

∂y +

∂Vz

∂z

, (1.65a)

known as the divergence of V. This is a scalar, as discussed in Section 1.3.

Example 1.7.1 DIVERGENCE OF COORDINATE VECTOR

Calculate ∇ · r:

∇ · r =

ˆx

∂

∂x + ˆy

∂

∂y + ˆz

∂

∂z

· (ˆxx + ˆyy + ˆzz)

=

∂x

∂x +

∂y

∂y +

∂z

∂z

,

or ∇ · r = 3.

Example 1.7.2 DIVERGENCE OF CENTRAL FORCE FIELD

Generalizing Example 1.7.1,

∇ ·

rf (r)

=

∂

∂x

x f(r)

+

∂

∂y

y f(r)

+

∂

∂z

zf (r)

= 3f (r) +

x2

r

df

dr +

y2

r

df

dr +

z2

r

df

dr

= 3f (r) + r

df

dr

.

To provide a motivation for the vector nature of partial derivatives, we now introduce the

total variation of a function F(x, y),

dF =

∂F

∂x

dx +

∂F

∂y

dy.

It consists of independent variations in the x- and y-directions. We write dF as a sum of

two increments, one purely in the x- and the other in the y-direction,

dF(x,y) ≡ F(x +dx,y +dy)− F(x, y)

=

F(x +dx,y +dy)− F(x,y +dy)

+

F(x,y + dy)−F(x, y)

=

∂F

∂x

dx +

∂F

∂y

dy,

by adding and subtracting F(x,y +dy). The mean value theorem (that is, continuity of F)

tells us that here ∂F/∂x, ∂F/∂y are evaluated at some point ξ,η between x and x + dx, y

1.6 Gradient, ∇ 33

and y + dy, respectively. As dx→0 and dy→0, ξ →x and η→y. This result generalizes

to three and higher dimensions. For example, for a function ϕ of three variables,

dϕ(x, y, z) ≡

ϕ(x +dx,y + dy, z +dz) −ϕ(x, y +dy, z +dz)

+

ϕ(x, y +dy, z +dz)− ϕ(x, y, z + dz)

+

ϕ(x, y, z +dz)− ϕ(x, y, z)

(1.57)

=

∂ϕ

∂x

dx +

∂ϕ

∂y

dy +

∂ϕ

∂z

dz.

Algebraically, dϕ in the total variation is a scalar product of the change in position dr and

the directional change of ϕ. And now we are ready to recognize the three-dimensional

partial derivative as a vector, which leads us to the concept of gradient.

Suppose that ϕ(x, y, z) is a scalar point function, that is, a function whose value depends

on the values of the coordinates (x, y, z). As a scalar, it must have the same value at a given

fixed point in space, independent of the rotation of our coordinate system, or

ϕ′(x′1, x′2, x′3) = ϕ(x1, x2, x3). (1.58)

By differentiating with respect to x′i we obtain

∂ϕ′(x′1, x′2, x′3)

∂x′i =

∂ϕ(x1, x2, x3)

∂x′i =

j

∂ϕ

∂xj

∂xj

∂x′i =

j

aij

∂ϕ

∂xj

(1.59)

by the rules of partial differentiation and Eqs. (1.16a) and (1.16b). But comparison with

Eq. (1.17), the vector transformation law, now shows that we have constructed a vector

with components ∂ϕ/∂xj . This vector we label the gradient of ϕ.

A convenient symbolism is

∇ϕ = ˆx

∂ϕ

∂x + ˆy

∂ϕ

∂y + ˆz

∂ϕ

∂z

(1.60)

or

∇ = ˆx

∂

∂x + ˆy

∂

∂y + ˆz

∂

∂z

. (1.61)

∇ϕ (or del ϕ) is our gradient of the scalar ϕ, whereas ∇ (del) itself is a vector differential

operator (available to operate on or to differentiate a scalar ϕ). All the relationships for ∇

(del) can be derived from the hybrid nature of del in terms of both the partial derivatives

and its vector nature.

The gradient of a scalar is extremely important in physics and engineering in expressing

the relation between a force field and a potential field,

force F=−∇(potential V ), (1.62)

which holds for both gravitational and electrostatic fields, among others. Note that the

minus sign in Eq. (1.62) results in water flowing downhill rather than uphill! If a force can

be described, as in Eq. (1.62), by a single function V (r) everywhere, we call the scalar

function V its potential. Because the force is the directional derivative of the potential, we

can find the potential, if it exists, by integrating the force along a suitable path. Because the

34 Chapter 1 Vector Analysis

total variation dV = ∇V · dr=−F · dr is the work done against the force along the path

dr, we recognize the physical meaning of the potential (difference) as work and energy.

Moreover, in a sum of path increments the intermediate points cancel,

V (r +dr1 +dr2)− V (r+ dr1)

+

V (r +dr1)− V (r)

= V (r +dr2 +dr1)− V (r),

so the integrated work along some path from an initial point ri to a final point r is given by

the potential difference V (r) − V (ri ) at the endpoints of the path. Therefore, such forces

are especially simple and well behaved: They are called conservative.When there is loss of

energy due to friction along the path or some other dissipation, the work will depend on the

path, and such forces cannot be conservative: No potential exists. We discuss conservative

forces in more detail in Section 1.13.

Example 1.6.1 THE GRADIENT OF A POTENTIAL V (r)

Let us calculate the gradient of V (r) = V (

x2 +y2 +z2 ), so

∇V (r) = ˆx

∂V (r)

∂x + ˆy

∂V (r)

∂y + ˆz

∂V (r)

∂z

.

Now, V (r) depends on x through the dependence of r on x. Therefore14

∂V (r)

∂x =

dV (r)

dr ·

∂r

∂x

.

From r as a function of x, y, z,

∂r

∂x =

∂(x2 +y2 +z2)1/2

∂x =

x

(x2 +y2 +z2)1/2 =

x

r

.

Therefore

∂V (r)

∂x =

dV (r)

dr ·

x

r

.

Permuting coordinates (x→y,y→z, z→x) to obtain the y and z derivatives, we get

∇V (r) = (ˆxx + ˆyy + ˆzz)

1

r

dV

dr

=

r

r

dV

dr = ˆr

dV

dr

.

Here ˆr is a unit vector (r/r) in the positive radial direction. The gradient of a function of

r is a vector in the (positive or negative) radial direction. In Section 2.5, ˆr is seen as one

of the three orthonormal unit vectors of spherical polar coordinates and ˆr∂/∂r as the radial

component of ∇.

14This is a special case of the chain rule of partial differentiation:

∂V(r, θ,ϕ)

∂x =

∂V

∂r

∂r

∂x +

∂V

∂θ

∂θ

∂x +

∂V

∂ϕ

∂ϕ

∂x

,

where ∂V/∂θ = ∂V/∂ϕ = 0, ∂V/∂r →dV/dr.

1.6 Gradient, ∇ 35

A Geometrical Interpretation

One immediate application of ∇ϕ is to dot it into an increment of length

dr = ˆxdx + ˆy dy + ˆz dz.

Thus we obtain

∇ϕ · dr =

∂ϕ

∂x

dx +

∂ϕ

∂y

dy +

∂ϕ

∂z

dz = dϕ,

the change in the scalar function ϕ corresponding to a change in position dr. Now consider

P and Q to be two points on a surface ϕ(x, y, z) = C, a constant. These points are chosen

so thatQ is a distance dr from P . Then, moving from P to Q, the change in ϕ(x, y, z) = C

is given by

dϕ = (∇ϕ) · dr = 0 (1.63)

since we stay on the surface ϕ(x, y, z) = C. This shows that ∇ϕ is perpendicular to dr.

Since dr may have any direction from P as long as it stays in the surface of constant ϕ,

point Q being restricted to the surface but having arbitrary direction, ∇ϕ is seen as normal

to the surface ϕ = constant (Fig. 1.19).

If we now permit dr to take us from one surface ϕ = C1 to an adjacent surface ϕ = C2

(Fig. 1.20),

dϕ = C1 −C2 = C = (∇ϕ) · dr. (1.64)

For a given dϕ, |dr| is a minimum when it is chosen parallel to ∇ϕ (cos θ = 1); or, for

a given |dr|, the change in the scalar function ϕ is maximized by choosing dr parallel to

FIGURE 1.19 The length increment dr has to stay on the surface ϕ = C.

36 Chapter 1 Vector Analysis

FIGURE 1.20 Gradient.

∇ϕ. This identifies ∇ϕ as a vector having the direction of the maximum space rate

of change of ϕ, an identification that will be useful in Chapter 2 when we consider non-

Cartesian coordinate systems. This identification of ∇ϕ may also be developed by using

the calculus of variations subject to a constraint, Exercise 17.6.9.

Example 1.6.2 FORCE AS GRADIENT OF A POTENTIAL

As a specific example of the foregoing, and as an extension of Example 1.6.1, we consider

the surfaces consisting of concentric spherical shells, Fig. 1.21. We have

ϕ(x, y, z) =

x2 + y2 +z2 1/2

= r = C,

where r is the radius, equal to C, our constant. C = ϕ = r, the distance between two

shells. From Example 1.6.1

∇ϕ(r) = ˆr

dϕ(r)

dr = ˆr.

The gradient is in the radial direction and is normal to the spherical surface ϕ = C.

Example 1.6.3 INTEGRATION BY PARTS OF GRADIENT

Let us prove the formula

A(r) ·∇f (r)d3r =−

f (r)∇·A(r)d3r, where A or f or both

vanish at infinity so that the integrated parts vanish. This condition is satisfied if, for example,

A is the electromagnetic vector potential and f is a bound-state wave function ψ(r).

1.6 Gradient, ∇ 37

FIGURE 1.21 Gradient for

ϕ(x, y, z) = (x2 + y2 + z2)1/2, spherical

shells: (x2

2 + y2

2 + z2

2)1/2 = r2 = C2,

(x2

1 + y2

1 + z2

1)1/2 = r1 = C1.

Writing the inner product in Cartesian coordinates, integrating each one-dimensional

integral by parts, and dropping the integrated terms, we obtain

A(r) · ∇f (r)d3r =

 

Axf |∞x =−∞ −

f

∂Ax

∂x

dx

dy dz+···

=−

f

∂Ax

∂x

dx dy dz −

f

∂Ay

∂y

dy dx dz −

f

∂Az

∂z

dz dx dy

=−

f (r)∇ · A(r)d3r.

If A = eikz ˆe describes an outgoing photon in the direction of the constant polarization unit

vector ˆe and f = ψ(r) is an exponentially decaying bound-state wave function, then

eikz ˆe · ∇ψ(r)d3r =−ez

ψ(r)

deikz

dz

d3r =−ikez

ψ(r)eikz d3r,

because only the z-component of the gradient contributes.

Exercises

1.6.1 If S(x, y, z) = (x2 + y2 + z2)−3/2, find

(a) ∇S at the point (1, 2, 3);

(b) the magnitude of the gradient of S, |∇S| at (1, 2, 3); and

(c) the direction cosines of ∇S at (1, 2, 3).

38 Chapter 1 Vector Analysis

1.6.2 (a) Find a unit vector perpendicular to the surface

x2 +y2 + z2 = 3

at the point (1, 1, 1). Lengths are in centimeters.

(b) Derive the equation of the plane tangent to the surface at (1, 1, 1).

ANS. (a) (ˆx + ˆy+ ˆz)/√3, (b) x +y +z = 3.

1.6.3 Given a vector r12 = ˆx(x1 − x2) + ˆy(y1 − y2) + ˆz(z1 − z2), show that ∇1r12 (gradient

with respect to x1, y1, and z1 of the magnitude r12) is a unit vector in the direction of

r12.

1.6.4 If a vector function F depends on both space coordinates (x, y, z) and time t , show that

dF = (dr · ∇)F+

∂F

∂t

dt.

1.6.5 Show that ∇(uv) = v∇u + u∇v, where u and v are differentiable scalar functions of

x, y, and z.

(a) Show that a necessary and sufficient condition that u(x, y, z) and v(x, y, z) are

related by some function f (u,v) = 0 is that (∇u)×(∇v) = 0.

(b) If u = u(x, y) and v = v(x, y), show that the condition (∇u) ×(∇v) = 0 leads to

the two-dimensional Jacobian

J

u, v

x,y

=

∂

u

∂x

∂u

∂y

∂v

∂x

∂v

∂y

= 0.

The functions u and v are assumed differentiable.

1.7 DIVERGENCE, ∇

Differentiating a vector function is a simple extension of differentiating scalar quantities.

Suppose r(t) describes the position of a satellite at some time t . Then, for differentiation

with respect to time,

dr(t)

dt = lim

→0

r(t + t)− r(t)

t = v, linear velocity.

Graphically, we again have the slope of a curve, orbit, or trajectory, as shown in Fig. 1.22.

If we resolve r(t) into its Cartesian components, dr/dt always reduces directly to a

vector sum of not more than three (for three-dimensional space) scalar derivatives. In other

coordinate systems (Chapter 2) the situation is more complicated, for the unit vectors are

no longer constant in direction. Differentiation with respect to the space coordinates is

handled in the same way as differentiation with respect to time, as seen in the following

paragraphs.