Engineering Math

9.2 The Derivative And Integral

The following definition is on the derivative and integral of a vector valued function of one variable.

Definition 9.2.1 The derivative of a function f^′

, is defined as the following limit whenever the limit exists. If the limit does not exist, then neither does f^′

lim f (t+-h)--f-(t)\equiv f'(t) h\to0 h

As before,

f (s)- f (t) f'(t) = lis\tomt ----------. s- t

The function of h on the left is called the difference quotient just as it was for a scalar valued function. If f

and ∫ _a^bf_i

dt exists for each i = 1,

,p, then ∫ _a^bf

dt is defined as the vector

( ) \int b \int b f1(t)dt,\cdot\cdot\cdot, fp(t)dt . a a

This is what is meant by saying f ∈ R

Here is a simple proposition which is useful to have.

Proposition 9.2.2 Let a ≤ b, f =

is vector valued and each f_i is continuous, then

| | ||\int b || \sqrt--\int b ||a f (t)dt|| \leq n a |f (t)|dt.

Proof: This follows from the Cauchy Schwarz inequality.

|\int | |( \int \int ) | || b || || b b || || a f (t) dt|| = || a f1 (t)dt,\cdot\cdot\cdot, a fn(t)dt ||

(| | | |) ( ) ||\int b || ||\int b || \int b \int b = || f1(t)dt||,\cdot\cdot\cdot,|| fn(t)dt|| \leq |f1(t)|dt,\cdot\cdot\cdot, |fn (t)|dt ( a a ) a a \int b \int b \sqrt -\int b \leq a |f (t)|dt,\cdot\cdot\cdot, a |f (t)|dt = n a |f (t)|dt. ■

As in the case of a scalar valued function differentiability implies continuity but not the other way around.

Theorem 9.2.3 If f^′

exists, then f is continuous at t.

Proof: Suppose ε > 0 is given and choose δ₁ > 0 such that if

< δ₁,

|| || ||f (t+-h)--f (t)-- f'(t)|| < 1. h

then for such h, the triangle inequality implies

. Now letting δ < min

it follows if

< δ, then

< ε. Letting y = h + t, this shows that if

< δ,

< ε which proves f is continuous at t. ■

As in the scalar case, there is a fundamental theorem of calculus.

Theorem 9.2.4 If f ∈ R

and if f is continuous at t ∈

, then

( \int t ) d- f (s)ds = f (t). dt a

Proof: Say f

. Then it follows

( ) 1-\int t+h 1-\int t 1\int t+h 1-\int t+h h a f (s)ds - h a f (s)ds = h t f1(s)ds,\cdot\cdot\cdot,h t fp(s)ds

and lim_h→0

∫ _t^t+hf_i

ds = f_i

for each i = 1,

,p from the fundamental theorem of calculus for scalar valued functions. Therefore,

1\int t+h 1\int t lhim\to0 h- f (s)ds- h- f (s)ds = (f1(t),\cdot\cdot\cdot,fp(t)) = f (t). ■ a a

Example 9.2.5 Let f

= c where c is a constant. Find f^′

The difference quotient,

f (x+ h) - f (x ) c - c --------------= -----= 0 h h

Therefore,

f (x+-h)--f-(x) lhim\to0 h = lhim\to00 = 0

Example 9.2.6 Let f

where a,b are constants. Find f^′

From the above discussion this derivative is just the vector valued functions whose components consist of the derivatives of the components of f. Thus f^′