Rate of change and tangent lines

2.4 Rates of Change and Tangent Lines
Devil’s Tower, Wyoming

2 1
2 1
Average rate of change
change in y y y y
change in x x x x
 
  
 
If f(t) represents the position of an object as a function of time,
then the rate of change is the velocity of the object.
Average rate of change (from bc)
Average rate of change of f(x) over the interval [a,b]
a
b
a
f
b
f



)
(
)
(

Find the average rate of change of f (t) = 2 + cost on [0, π]
F(b)= f(π) = 2 + cos (π) = 2 – 1 = 1
F(a)= f(0) = 2 + cos (0) = 2 + 1 = 3
1. Calculate the function value (position) at each
endpoint of the interval
The average velocity on [0, π] is 0.63366
2. Use the slope formula
63366
.
0
2
0
3
1
)
(
)
(











a
b
a
f
b
f

Consider a graph of displacement (distance traveled) vs. time.
time (hours)
distance
(miles)
Average velocity can be
found by taking:
change in position
change in time
s
t



t

s

A
B
   
ave
f t t f t
s
V
t t
  

 
 
The speedometer in your car does not measure average
velocity, but instantaneous velocity.
(The velocity at one
moment in time.)
The velocity problem

The slope of a line is given by:
y
m
x


 x

y

The slope of a curve at (1,1) can be
approximated by the slope of the secant line
through (1,1) and (4,16).
5

We could get a better approximation if we
move the point closer to (1,1). ie: (3,9)
y
x


9 1
3 1



8
2
 4

Even better would be the point (2,4).
y
x


4 1
2 1



3
1
 3


y
x


16 1
4 1



15
3


The slope of a line is given by:
y
m
x


 x

y

If we got really close to (1,1), say (1.1,1.21),
the approximation would get better still
y
x


1.21 1
1.1 1



.21
.1
 2.1

How far can we go?
  2
f x x

0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
1 2 3 4


If we try to apply the same formula to find The instantaneous
velocity
and evaluate the velocity at an instant (a,f(a)) not an interval
, we will find it
0
0
)
(
)
(
t






a
a
a
f
a
f
s
Which is undefined, so the best is to make Δx as small as
experimentally possible Δx 0
a
x
a
f
x
f
a
x 


)
(
)
(
lim
The instantaneous velocity at the point (a,f(a)) =
 
   
0
lim
t
f t t f t
ds
V t
dt t
 
  
 


The slope of the secant line= the average rate of change
= The average velocity
The slope of the curve at a point
= the slope of the tangent line of the curve at this point
= instantaneous velocity
=
2 1
2 1
Average rate of change
change in y y y y
change in x x x x
 
  
 
h
x
f
h
x
f
h
)
(
)
(
lim
0






 a
x
a
f
x
f
a
x
)
(
)
(
lim

Other form for Slope of secant
line of tangent line
sec
( ) ( )
y f a h f a
m
x h
  
 

tan 0
( ) ( )
limh
f a h f a
m
h

 

Let h = x - a Then x = a + h

Rates of Change:
Average rate of change =
   
f x h f x
h
 
Instantaneous rate of change =  
   
0
lim
h
f x h f x
f x
h

 
 
These definitions are true for any function.
( x does not have to represent time. )

Analytic Techniques
Rewrite algebraically if direct substitution
produces an indeterminate form such as
0/0
• Factor and reduce
• Rationalize a numerator or denominator
• Simplify a complex fraction
When you rewrite you are often producing another function that
agrees with the original in all but one point. When this happens
the limits at that point are equal.

Find the indicated limit
2
3
6
lim
3
x
x x
x

 

3
lim ( 2)
x
x


3
( 3)( 2)
lim
3
x
x x
x

 

= - 5
direct substitution fails
Rewrite and cancel
now use direct sub.
0
0

 
1
f
1 1 h

 
1
f h

h
slope
y
x



   
1 1
f h f
h
 

slope at  
1,1
 
2
0
1 1
lim
h
h
h

 

2
0
1 2 1
lim
h
h h
h

  

 
0
2
lim
h
h h
h


 2

The slope of the curve at the point is:
 
y f x
  
 
,
P a f a
   
0
lim
h
f a h f a
m
h

 



The slope of the curve at the point is:
 
y f x
  
 
,
P a f a
   
0
lim
h
f a h f a
m
h

 

   
f a h f a
h
 
is called the difference quotient of f at a.
If you are asked to find the slope using the definition or using
the difference quotient, this is the technique you will use.


In the previous example, the tangent line could be found
using .
 
1 1
y y m x x
  
The slope of a curve at a point is the same as the slope of
the tangent line at that point.
If you want the normal line, use the negative reciprocal of
the slope. (in this case, )
1
2

(The normal line is perpendicular.)


Example 4:
a Find the slope at .
x a

   
0
lim
h
f a h f a
m
h

 

0
1 1
lim
h
a h a
h




 
 
0
1
lim
h
h
a a h
a a h

 
 

 
0
lim
h
a a h
h a a h

 

 
2
1
a
 
Let  
1
f x
x

On the TI-89:
limit ((1/(a + h) – 1/ a) / h, h, 0)
F3 Calc
Note:
If it says “Find the limit”
on a test, you must
show your work!

 
a a h

 
a a h

 
a a h

0

Example 4:
b Where is the slope ?
1
4

Let  
1
f x
x

2
1 1
4 a
  
2
4
a 
2
a  
On the TI-89:
Y= y = 1 / x
WINDOW
6 6
3 3
scl 1
scl 1
x
y
x
y
  
  


GRAPH


Example 4:
b Where is the slope ?
1
4

Let  
1
f x
x

On the TI-89:
Y= y = 1 / x
WINDOW
6 6
3 3
scl 1
scl 1
x
y
x
y
  
  


GRAPH
We can let the calculator
plot the tangent:
F5 Math
A: Tangent ENTER
2 ENTER
Repeat for x = -2
tangent equation


Review:
average slope:
y
m
x



slope at a point:
   
0
lim
h
f a h f a
m
h

 

average velocity: ave
total distance
total time
V 
instantaneous velocity:
   
0
lim
h
f t h f t
V
h

 

If is the position function:
 
f t
These are
often
mixed up
by
Calculus
students!
So are these!
velocity = slope


Rate of change and tangent lines

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