Kenneth Kuttler

16.2. THE DERIVATIVE AND INTEGRAL 333

Proof: Say f (t) = ( f1 (t) , · · · , fp (t)). Then it follows

∫ t+h

af (s) ds− 1

∫ t

af (s) ds =

(1h

∫ t+h

tf1 (s) ds, · · · , 1

∫ t+h

tfp (s) ds

)and limh→0

1h∫ t+h

t fi (s) ds = fi (t) for each i = 1, · · · , p from the fundamental theorem ofcalculus for scalar valued functions. Therefore,

limh→0

∫ t+h

af (s) ds− 1

∫ t

af (s) ds = ( f1 (t) , · · · , fp (t)) = f (t) .

Example 16.2.5 Let f (x) = c where c is a constant. Find f ′ (x).

The difference quotient,

f (x+h)−f (x)h

=c−c

h= 0

Therefore,

limh→0

f (x+h)−f (x)h

= limh→0

0= 0

Example 16.2.6 Let f (t) = (at,bt) where a,b are constants. Find f ′ (t).

From the above discussion this derivative is just the vector valued functions whosecomponents consist of the derivatives of the components of f . Thus f ′ (t) = (a,b).

16.2.1 Geometric and Physical Significance of the DerivativeSuppose r is a vector valued function of a parameter t not necessarily time and considerthe following picture of the points traced out by r.

r(t)r(t +h)

In this picture there are unit vectors in the direction of the vector from r (t) to r (t +h).You can see that it is reasonable to suppose these unit vectors, if they converge, convergeto a unit vector T which is tangent to the curve at the point r (t). Now each of these unitvectors is of the form

r (t +h)−r (t)|r (t +h)−r (t)|

≡ T h.

Thus T h → T, a unit tangent vector to the curve at the point r (t). Therefore,

r′ (t) ≡ limh→0

r (t +h)−r (t)h

= limh→0

|r (t +h)−r (t)|h

r (t +h)−r (t)|r (t +h)−r (t)|

= limh→0

|r (t +h)−r (t)|h

T h =∣∣r′ (t)∣∣T.

16.2. THE DERIVATIVE AND INTEGRAL 333Proof: Say f (t) = (fi (t).--: fp (t)). Then it followst+h t +h ihLye fo)as-7[reas=(F [hod 7 [Has]and lim;_,9 i pen fi(s) ds = fj (t) for each i= 1,--- ,p from the fundamental theorem ofcalculus for scalar valued functions. Therefore,1 pithtim = [ f6)as—F [ FO) = (AO fO=FO- Ohoo0h JaExample 16.2.5 Let f (x) = where c is a constant. Find f' (x).The difference quotient,firth)—fx) ene _= =0h hTherefore,lim f(x+h)— f(a) = lim0O=—0h>0 h h>0Example 16.2.6 Let f (t) = (at, bt) where a,b are constants. Find f' (t).From the above discussion this derivative is just the vector valued functions whosecomponents consist of the derivatives of the components of f. Thus f’ (t) = (a,b).16.2.1 Geometric and Physical Significance of the DerivativeSuppose r is a vector valued function of a parameter f not necessarily time and considerthe following picture of the points traced out by r.In this picture there are unit vectors in the direction of the vector from r (t) to r(t+h).You can see that it is reasonable to suppose these unit vectors, if they converge, convergeto a unit vector T which is tangent to the curve at the point r (t). Now each of these unitvectors is of the formr(t+h)—r(t) _In +h)—r()]Thus T, > T, a unit tangent vector to the curve at the point r (t). Therefore,kim r(t+h)—r(t) — lim Ir (t+h)—r(t)| r(t+h)—r(t)h0 h AO h Ir (t+h)—r(t)|In (¢ +h) —r (|hh-r'(t)= limanlim Tp = |r’ (t)|T.