DIFFERENTIATION OF VECTORS

4. DIFFERENTIATION OF VECTORS

If A = iA x + jA y + kA z , where i, j, k are fixed unit vectors and A x ,A y ,A z are functions of t, then we define the derivative dA/dt by the equation

Thus the derivative of a vector A means a vector whose components are the deriva- tives of the components of A.

286 Vector Analysis Chapter 6

Example 1.

Let (x, y, z) be the coordinates of a moving particle at time t; then x, y, z are functions of t. The vector displacement of the particle from the origin at time t is

r = ix + jy + kz,

where r is a vector from the origin to the particle at time t. We say that r is the position vector or vector coordinate of the particle. The components of the velocity of the particle at time t are dx/dt, dy/dt, dz/dt so the velocity vector is

The acceleration vector is

The product of a scalar and a vector and the dot and cross products of vectors are differentiated by the ordinary calculus rules for differentiating a product, with one word of caution: The order of the factors must be kept in a cross product. You can easily prove equations (4.5) below by writing out components (Problem 1) and using (4.1).

The second term in (d/dt)(A · B) can be written B · dA/dt if you like since A · B =

B · A. But the corresponding term in (d/dt)(A × B) must not be turned around unless you put a minus sign in front of it since A × B = −B × A.

Example 2. Consider the motion of a particle in a circle at constant speed. We can then write

r 2 = r · r = const.,

v 2 = v · v = const.

If we differentiate these two equations using (4.5), we get

dr

2r ·

=0 or

r · v = 0,

dt dv

2v ·

=0 or

v · a = 0.

dt

Section 4 Differentiation of Vectors 287

Also differentiating r · v = 0, we get (4.8)

r·a+v·v=0

or

r · a = −v 2 .

The first of equations (4.7) says that r is perpendicular to v; the second says that

a is perpendicular to v. Therefore a and r are either parallel or antiparallel (since the motion is in a plane) and the angle θ between a and r is either 0 ◦ or 180 ◦ . From (4.8) and the definition of scalar product, we have

r · a = |r| |a| cos θ = −v 2 .

Thus we see that cos θ is negative, so θ = 180 ◦ . Then from (4.9) we get

|r| |a|(−1) = −v

or a=

We have just given a vector proof that for motion in a circle at constant speed the acceleration is toward the center of the circle and of magnitude v 2 /r.

So far we have written vectors only in terms of their rectangular components using the unit basis vectors i, j, k. It is often convenient to use other coordinate systems, for example polar coordinates in two dimensions and spherical or cylin- drical coordinates in three dimensions (see Chapter 5, Section 4, and Chapter 10, Sections 8 and 9). We shall consider using vectors in various coordinate systems in detail in Chapter 10, but it will be useful to discuss briefly here the use of plane polar coordinates. In Figure 4.1, think of starting at the point (x, y) or (r, θ) and moving along the line θ = const. in the direction of increasing r. We call this the “r direction”; we draw a unit vector (that is, a vector of length 1) in this direction

Figure 4.1 Figure 4.2 and label it e r . Similarly, think of moving along the circle r = const. in the direction

of increasing θ. We call this the “θ direction”; we draw a unit vector tangent to the circle and label it e θ . These two vectors e r and e θ are the polar coordinate unit basis vectors just as i and j are the rectangular unit basis vectors. We can now write any given vector in terms of its components in the directions e r and

e θ (by finding its projections in these directions). There is a complication here, however. In rectangular coordinates, the vectors i and j are constant in magnitude and direction. The polar coordinate unit basis vectors are constant in magnitude but their directions change from point to point (Figure 4.2). Thus in calculating the derivative of a vector written in polar coordinates, we must differentiate the basis vectors as well as the components [compare (4.1) where we differentiate the

288 Vector Analysis Chapter 6

components only.] One straightforward way to do this is to express the vectors e r and e θ in terms of i and j. From Figure 4.3, we see that the x and y components of e r are cos θ and sin θ. Thus we have

Similarly (Problem 7) we find

e θ = −i sin θ + j cos θ.

Differentiating e r and e θ with respect to t, we get

We can now use (4.13) in calculating the derivative of any vector which is written in terms of its polar components.

Example 3. Given A = A r e r +A θ e θ , where A r and A θ are functions of t, find dA/dt. We get

dt Using (4.13), we find

dt We can find higher-order derivatives if we like by differentiating again using (4.13)

each time to evaluate the derivatives of e r and e θ .

Section 5 Fields 289

PROBLEMS, SECTION 4

1. Verify equations (4.5) by writing out the components. 2. Let the position vector (with its tail at the origin) of a moving particle be r = r(t) =

t 2 i − 2tj + (t 2 + 2t)k, where t represents time.

(a) Show that the particle goes through the point (4, −4, 8). At what time does it do this?

(b) Find the velocity vector and the speed of the particle at time t; at the time when it passes though the point (4, −4, 8).

(c) Find the equations of the line tangent to the curve described by the particle and the plane normal to this curve, at the point (4, −4, 8).

3. As in Problem 2, if the position vector of a particle is r = (4 + 3t)i + t 3 j − 5tk, at what time does it pass through the point (1, −1, 5)? Find its velocity at this time. Find the equations of the line tangent to its path and the plane normal to the path, at (1, −1, 5).

4. Let r = r(t) be a vector whose length is always 1 (it may vary in direction). Prove that either r is a constant vector or dr/dt is perpendicular to r. Hint: Differentiate r · r.

5. The position of a particle at time t is given by r = i cos t + j sin t + kt. Show that both the speed and the magnitude of the acceleration are constant. Describe the motion.

6. The force acting on a moving charged particle in a magnetic field B is F = q(v × B) where q is the electric charge of the particle, and v is its velocity. Suppose that a particle moves in the (x, y) plane with a uniform B in the z direction. Assuming Newton’s second law, m dv/dt = F, show that the force and velocity are perpendic- ular and that both have constant magnitude. Hint: Find (d/dt)(v · v).

7. Sketch a figure and verify equation (4.12). 8. In polar coordinates, the position vector of a particle is r = re r . Using (4.13), find

the velocity and acceleration of the particle. 9. The angular momentum of a particle m is defined by L = mr × (dr/dt) (see end of

Section 3). Show that

10. If V(t) is a vector function of t, find the indefinite integral

V×

2 dt dt.