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VECTOR DIFFERENTIATION

Differentiation of vector functions. In calculus we compute derivatives of real functions of a real variable. In the case of functions of a single variable

y = f(x)

we compute the derivative of y with respect to x. In the case of a function of several variables

y = f(x_{1}, x_{2}, ... , x_{n})

we can compute the derivative of y with respect to any one of the independent variables, viewing the others as fixed.

In vector analysis we compute derivatives of vector functions of a real variable; that is we compute derivatives of functions of the type

F(t) = f_{1}(t) i + f_{2}(t) j + f_{3}(t) k

or, in different notation,

where f_{1}(t), f_{2}(t), and f_{3}(t) are real functions of the real variable t.

This function can be viewed as describing a space curve. Intuitively it can be regarded as a position vector, expressed as a function of t, that traces out a space curve with increasing values of t. Expressed in different notation it is the function

R(t) = x(t) i + y(t) j + z(t) k

or, equivalently,

that represents a position vector to a point P(x, y, z) in space which moves as t increases. It traces out a curve in space. This corresponds to the parametric representation of a space curve which is

x = x(t)

y = y(t)

z = z(t) .

In the case of a vector function of several variables

F = f_{1}(x_{1}, x_{2}, ... , x_{n}) i + f_{2}(x_{1}, x_{2}, ... , x_{n}) j + f_{3}(x_{1}, x_{2}, ... , x_{n}) k

or equivalently,

we can compute the derivative of F with respect to any one of the independent variables, viewing the others as fixed i.e. as parameters. Thus, in the case of several independent variables, the problem still reduces to that of computing the derivative of a function representing a space curve traced out by a position vector.

Derivative of a vector function of a single real variable. Let R(t) be a position vector, extending from the origin to some point P, depending on the single scalar variable t. Then R(t) traces out some curve in space with increasing values of t. Consider

where denotes an increment in t. See Fig. 1. The derivative of R(t) with respect to t is given by

if the limit exists. If

R(t) = x(t) i + y(t) j + z(t) k

then

See Fig. 2 .

So

Consequently,

Thus

if the derivative exists. If the curve makes sharp, abrupt turns the derivative won’t exist.

From what we have said we state the following theorem:

Theorem 1. The vector

F(t) = f_{1}(t) i + f_{2}(t) j + f_{3}(t) k

can be viewed as a position vector that traces out a curve in xyz-space. Its derivative with respect to t, if it exists, is given by

F'(t) = f_{1}'(t) i + f_{2}'(t) j + f_{3}'(t) k

or, equivalently,

where, with f_{1}(t), f_{2}(t) and f_{3}(t) viewed as distances in xyz-space,

f_{1}'(t) = dx/dt; f_{2}'(t) = dy/dt; f_{3}'(t) = dz/dt .

(the prime denotes differentiation with respect to t).

The derivative dF(t)/dt is a vector tangent to the space curve at the point in question. If the variable t represents time, then dF(t)/dt represents the velocity with which the terminal point of the radius vector describes the curve. Similarly, dv/dt represents its acceleration a along the curve.

A vector function

F(t) = f_{1}(t) i + f_{2}(t) j + f_{3}(t) k

is continuous at a point t if the three scalar functions f_{1}(t), f_{2}(t), and f_{3}(t) are continuous at t. A
scalar or vector function is called differentiable of order n if its n-th derivative exists. A
function which is differentiable is necessarily continuous but the converse is not true.

Example.

Problem. A particle moves along a curve

x = 2t^{2}

y = t^{2} - 4t

z = 3t - 5

where t is time. Find its velocity at time t.

Solution.

r = 2t^{2 }i + (t^{2} - 4t)j + (3t - 5)k

dr/dt = 4ti + (2t - 4)j + 3k

Singular points. A point on a space curve is called a singular point if | f '(t) | = 0 at the point . Otherwise it is called a non-singular point.

Def. Smooth vector function. A vector function that has a continuous derivative and no singular points.

Differentiation formulas. Let A = a_{1}(t) i + a_{2}(t) j + a_{3}(t) k , B = b_{1}(t) i + b_{2}(t) j +
b_{3}(t) k, and C = c_{1}(t) i + c_{2}(t) j + c_{3}(t) k be differentiable vector functions of a scalar t and let
be a differentiable scalar function of t. Then

7] Chain rule: If t = g(s) is differentiable, then

Partial derivatives. In the case of a vector function of several variables

6) F = f_{1}(x_{1}, x_{2}, ... , x_{n}) i + f_{2}(x_{1}, x_{2}, ... , x_{n}) j + f_{3}(x_{1}, x_{2}, ... , x_{n}) k

or equivalently,

we can compute the partial derivatives of the function with respect to a particular independent variable in the same way we do in ordinary calculus by considering all other independent variables to be fixed parameters. Thus

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