Mathematical Physics Vol 1

3.2 Vector analysis

71

Then

d F = d F 1 i + d F 2 j + d F 3 k =

∂ F 1 ∂ t ∂ F 2 ∂ t

d z i + d z j + d z k =

∂ F 1 ∂ x ∂ F 2 ∂ x

∂ F 1 ∂ y ∂ F 2 ∂ y

∂ F 1 ∂ z ∂ F 2 ∂ z

d t +

d x +

d y +

d t +

d x +

d y +

∂ F 3 ∂ t ∂ F 2 ∂ t

∂ F 3 ∂ x

∂ F 3 ∂ y

∂ F 3 ∂ z

d t +

d x +

d y +

∂ F 1 ∂ t ∂ F 1 ∂ y

k d t + k d y +

k d x + k d z =

∂ F 3 ∂ t ∂ F 3 ∂ y

∂ F 1 ∂ x ∂ F 1 ∂ z

∂ F 2 ∂ x ∂ F 2 ∂ z

∂ F 3 ∂ x ∂ F 3 ∂ z

i +

j +

i +

j +

∂ F 2 ∂ y

i +

j +

i +

j +

∂ F ∂ t

∂ F ∂ x

∂ F ∂ y

∂ F ∂ z

d t +

d x +

d y +

d z ,

and we obtain

∂ F ∂ t

∂ F ∂ x

∂ F ∂ y

∂ F ∂ z

d F d t

d x d t

d y d t

d z d t

=

+

+

+

.

Problem29 Find the derivative of a vector with constant magnitude.

Solution Observe a vector function ρ of the scalar variable t , where | ρ | = ρ = const. The hodograph of this vector is a curve that lies on a sphere with radius ρ (Fig. 3.10a).

(a)

(b)

Figure 3.10: Figure in Problem 29.

Given that ρ = const . , it follows that ρ ( t + ∆ t )= ρ ( t ) , an the triangle △ OAB is isosceles (Fig. 3.10b). The derivative of a vector function is a vector, which can be represented, the same as any other vector, by its magnitude and its unit direction vector d ρ d t = d ρ d t · p .