LESSON 6 TANGENT LINES AND INSTANTANEOUS VELOCITY
Example Find the equation of the tangent line to the graph of the function
y  f ( x ) at the point ( a , f ( a ) ) .
(a  h, f ( a  h))
(a, f (a))
 y  f ( x)
NOTE: We call the point ( a , f ( a ) ) the tangent point. This point is on the graph
of y  f ( x ) and it is on the graph of the tangent line. Thus, if we can find the
slope of the tangent line, then we can use the point-slope form for the equation of a
line to write the equation of the tangent line.
Recall that the point-slope form for the equation of a line is given by the equation
y  y 1  m ( x  x 1 ) , where m is the slope of the line and ( x 1 , y 1 ) is a point on
the line.
Copyrighted by James D. Anderson, The University of Toledo
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If we knew a second point, that was on the tangent line, then we could use the slope
formula to calculate the slope of the tangent line.
Recall that if ( x 1 , y 1 ) and ( x 2 , y 2 ) are two points on a line, then the slope of the
y 2  y1
y1  y 2
m

m

line is given by
x 2  x 1 or
x1  x 2 .
Notice that the two points ( a , f ( a ) ) and ( a  h , f ( a  h ) ) are on the secant
line. Thus, we can calculate the slope of the secant line. Thus,
m sec 
f (a  h)  f (a)
f (a  h)  f (a)
=
aha
h
Does this last expression look familiar? We saw it back in Lesson 2.
NOTE: If a  h was not far from a, then the slope of the secant line would be a
good approximation to the slope of the tangent line. Let’s look at the following
animation.
In order to find the slope of the tangent line, we need h to get closer and closer to
zero. That is, we need h to approach zero ( h  0 ). We can not let h get to zero
because if h = 0, then the second point of ( a  h , f ( a  h ) ) becomes the point
( a , f ( a ) ) . Then we would only has one point of ( a , f ( a ) ) instead of two
points of ( a , f ( a ) ) and ( a  h , f ( a  h ) ) . Then our calculation for the slope
of the secant line would be invalid.
Thus, to find the slope of the tangent line, we want h to approach 0 but not get to 0.
Does this sound familiar? This is the concept of a limit. Thus, to find the slope of
the tangent line, we take the limit of the slope of the secant line as h approaches
zero ( h  0 ). That is,
m tan  lim m sec = lim
h0
h0
f (a  h)  f (a)
h
Since the tangent point of ( a , f ( a ) ) is on the tangent line, then the equation of
the tangent line is given by y  f ( a )  m tan ( x  a ) .
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Recall: In mathematics, a rate is a quantity whose unit is the quotient of two
different units.
NOTE: The slope of the tangent line to the graph of y  f ( x ) at the point
( a , f ( a ) ) gives the instantaneous rate of change of y with respect to x when
x  a and the slope of the secant line passing through the two points of ( a , f ( a ) )
and ( a  h , f ( a  h ) ) on the graph of y  f ( x ) gives the average rate of
change of y with respect to x on the interval [ a , a  h ] .
( unit y )
 m tan 
dy
dx

unit y
xa
unit x
 m sec 
 y unit y

 x unit x
(a  h, f ( a  h))
(a, f (a))
( unit x )
 y  f ( x)
dy
COMMENT: The expression dx
xa
is called the derivative of y with respect to x
dy
dy
x

a
at
. We will discuss derivatives in Lesson 7. The expression
in dx
Copyrighted by James D. Anderson, The University of Toledo
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xa
is called the differential of y and the expression dx is called the differential of x.
We will discuss differentials in Lesson 20.
Examples Find the equation (in point-slope form) of the tangent line to the graph
of the following functions for the indicated point.
1.
f ( x )  x2  5 ; x   2
Tangent Point: (  2 , f (  2 ) )  (  2 ,  1)
f (  2)  4  5   1
m tan  lim
h0
f ( 2  h)  f ( 2)
h
f (  2  h )  f ( h  2 )  ( h  2 ) 2  5 = h 2  4h  4  5 = h 2  4h  1
f (  2 )   1 (from above)
f (  2  h )  f (  2 ) = h 2  4h = h ( h  4 )
f ( 2  h)  f ( 2)
= h4
h
m tan  lim ( h  4 )   4
h0
Answer: y  1   4 ( x  2 )
NOTE: If you wanted the equation in slope-intercept form, which is
y  m x  b , then you could solve y  1   4 ( x  2 ) for y:
y  1   4 ( x  2 )  y  1   4x  8  y   4x  9
2.
f ( x )  3x 2  7 x ; x  6
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Tangent Point: ( 6 , f ( 6 ) )  ( 6 , 150 )
f ( 6 )  108  42  150
m tan  lim
h0
f (6  h)  f (6)
h
f ( 6  h )  f ( h  6 )  3( h  6 )2  7 ( h  6 ) =
3 ( h 2  12h  36 )  7h  42 = 3h 2  36h  108  7h  42 =
3h 2  43h  150
f ( 6 )  150 (from above)
f ( 6  h )  f ( 6 ) = 3h 2  43h = h ( 3h  43 )
f (6  h)  f (6)
= 3h  43
h
m tan  lim ( 3h  43 )  43
h0
Answer: y  150  43 ( x  6 )
NOTE: If you wanted the equation in slope-intercept form, which is
y  m x  b , then you could solve y  150  43 ( x  6 ) for y:
y  150  43 ( x  6 )  y  150  43x  258  y  43x  108
3.
g( x )  x3 ; x  2
Tangent Point: ( 2 , g ( 2 ) )  ( 2 , 8 )
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m tan  lim
h0
g( 2  h )  g( 2 )
h
g ( 2  h )  g ( h  2 )  ( h  2 ) 3 = h 3  3h 2 2  3h ( 2 ) 2  8 =
h 3  6h 2  12h  8 using the Binomial Expansion Theorem
g ( 2 )  8 (from above)
g ( 2  h )  g ( 2 ) = h 3  6h 2  12h = h ( h 2  6h  12 )
g( 2  h )  g( 2 )
2
= h  6h  12
h
m tan  lim ( h 2  6h  12 )  12
h0
Answer: y  8  12 ( x  2 )
NOTE: Solving y  8  12 ( x  2 ) for y, we have that y  12 x  16 .
4.
f ( x )  x 3  4 x 2  12 x  28 ; x   3
Tangent Point: (  3 , f (  3 ) )  (  3 , 17 )
f (  3 )   27  36  36  28  17
m tan  lim
h0
f (  3  h )  f (  3)
h
f (  3  h )  f ( h  3 ) = ( h  3 ) 3  4 ( h  3 ) 2  12 ( h  3 )  28
3
Using the Binomial Expansion Theorem to find ( h  3 ) , we have that
( h  3 ) 3 = h 3  3h 2 (  3 )  3h (  3 ) 2  (  3) 3 = h 3  9h 2  27h  27
Copyrighted by James D. Anderson, The University of Toledo
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f (  3  h ) = h 3  9h 2  27h  27  4 ( h 2  6h  9 )  12h  36  28 =
h 3  9h 2  27h  27  4h 2  24h  36  12h  36  28
_______ ________ ______ _______ ________ ______ ________ ______ ______
________ ______
________ ______ ________ ______ ______
______
______
______ ______
=
h 3  5h 2  9h  17
f (  3 )  17 (from above)
f (  3  h )  f (  3 ) = h 3  5h 2  9h = h ( h 2  5h  9 )
f (  3  h )  f (  3)
2
= h  5h  9
h
m tan  lim ( h 2  5h  9 )   9
h0
Answer: y  17   9 ( x  3 )
NOTE: Solving y  17   9 ( x  3 ) for y, we have that y   9 x  10 .
5.
s( t ) 
6t ; t  3
Tangent Point: ( 3 , s( 3 ) )  ( 3 ,
m tan  lim
h0
18 )
s( 3  h )  s( 3 )
h
s( 3  h )  s( h  3 ) 
6 ( h  3) 
6h  18
s( 3 )  18 (from above)
Copyrighted by James D. Anderson, The University of Toledo
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s( 3  h )  s( 3 ) =
6h  18 
18
s( 3  h )  s( 3 )
=
h
6h  18 
h
18
s( 3  h )  s( 3 )
as h  0 , you will
h
find that it is an indeterminate form. Thus, you will need to rationalize the
numerator of the fraction. Thus, we have that
NOTE: If you try to take the limit of
6h  18 
h
s( 3  h )  s( 3 )
=
h
6h  18  18
h ( 6h  18 
3 2
18 )
6h  18 
18
6h  18 
18
h ( 6h  18 
18
=
=
6
18 )
6
6h  18 
h0

6h
=
6
m tan  lim
3
18
18 
=
6h  18 
6
18
=
2 18
18
3
=
18
=
1
=
2
Answer: y 
18 
1
2
( t  3)
NOTE: In order to solve y 
1
to write
2
as
18 
1
2
( t  3 ) for y, you probably want
2
by rationalizing the denominator. You probably want
2
Copyrighted by James D. Anderson, The University of Toledo
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18 as 3 2 so that you can combine like terms.
to write
y
18 
1
2
( t  3)  y  3 2 
Thus,
2
( t  3) 
2
2
3 2
6 2
2
3 2
t
 y
t


2
2
2
2
2
2
3 2
2
y
t
. Now factor out the common
, obtaining that
2
2
2
2
y
( t  3) .
2
y3 2 
6.
g( w) 
9w  85 ; w   4
Tangent Point: (  4 , g (  4 ) )  (  4 , 7 )
g(  4 ) 
m tan  lim
h0
 36  85 
49  7
g(  4  h )  g(  4 )
h
g(  4  h )  g( h  4 ) 
9 ( h  4 )  85 
9h  36  85 
g (  4 )  7 (from above)
g(  4  h )  g(  4 ) =
9h  49  7
g(  4  h )  g(  4 )
=
h
9h  49  7
h
Copyrighted by James D. Anderson, The University of Toledo
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9h  49
g(  4  h )  g(  4 )
as h  0 , you
h
will find that it is an indeterminate form. Thus, you will need to rationalize
the numerator of the fraction. Thus, we have that
NOTE: If you try to take the limit of
9h  49  7
g(  4  h )  g(  4 )
=
h
h
9h  49  49
h0
=
f ( x) 
=
9h  49  7
9
9h  49  7
=
9
=
49  7 14
9
( w  4)
14
NOTE: Solving y  7 
7.
9
9
67
( w  4 ) for y, we have that y  w 
.
14
14
7
3
x7
x  15 ;

Tangent Point: (  7 , f (  7 ) )    7 , 

m tan  lim
h0
=
9
h ( 9h  49  7 )
9
Answer: y  7 
9h  49  7
9h
h ( 9h  49  7 )
m tan  lim
9h  49  7

3

8
f ( 7  h)  f ( 7)
h
f ( 7  h)  f (h  7) 
3
3

h  7  15 h  8
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f ( 7)  
3
(from above)
8
f ( 7  h)  f ( 7) =
3

 
h8

3
3
3

=
8 =
8 h  8
 24
3( h  8)
 24  3 ( h  8 )
 24  3h  24
3h

=
=
=
8( h  8)
8( h  8)
8( h  8)
8( h  8)
8( h  8)
3
f ( 7  h)  f ( 7)
=
8( h  8)
h
m tan  lim
h0
3
3
3


8 ( h  8 ) 8 ( 8 ) 64
Answer: y 
3
3

( x  7)
8 64
3
3
3
3

( x  7 ) for y, we have that y 
x
.
8 64
64
64
3
y

( x  1) .
This can be written as
64
NOTE: Solving y 
8.
g( x ) 
21
x5
9 x  17 ;
 3
(
5
,
g
(
5
)
)

 5, 
Tangent Point:
 4
21
21 3
g( 5) 


45  17 28 4
m tan  lim
h0
g(5  h )  g(5)
h
Copyrighted by James D. Anderson, The University of Toledo
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g(5  h )  g( h  5) 
g(5) 
21
21
21


9 ( h  5 )  17 9h  45  17 9h  28
3
(from above)
4
3 (9h  28 )
84
21
3


g( 5  h )  g( 5) =
=
4 ( 9h  28 )
4 ( 9h  28 ) =
9h  28
4
84  3 (9h  28 ) 84  27h  84
 27 h
=
=
4 ( 9h  28 )
4 ( 9h  28 )
4 ( 9h  28 )
 27
g( 5  h )  g( 5)
=
4 ( 9h  28 )
h
m tan  lim
h0
 27
 27
27


4 ( 9h  28 ) 4 ( 28 )
112
Answer: y 
3
27

( x  5)
4
112
3
27

( x  5 ) for y, we have that
4
112
27
135
84
27
219
y
x

 y
x
. We can write this last
112
112 112
112
112
3
( x  73 )
equation as y  
112
NOTE: Solving y 
9.
y  3x  20 ; x  4
Tangent Point: ( 4 , y ( 4 ) )  ( 4 ,  8 )
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m tan  lim
h0
y( 4  h )  y( 4 )
h
y ( 4  h )  y ( h  4 )  3 ( h  4 )  20  3h  12  20  3h  8
y ( 4 )   8 (from above)
y ( 4  h )  y ( 4 ) = 3h
y( 4  h )  y( 4 )
= 3
h
m tan  lim 3  3
h0
Answer: y  8  3 ( x  4 )
Solving y  8  3 ( x  4 ) for y, we have that y  3 x  20 .
NOTE: The equation of the tangent line to the graph of the linear function
y  3x  20 at the point ( 4 ,  8 ) is itself.
Now, we will consider the instantaneous velocity of a particle moving along a
straight line.
Let s be a function of time t which gives the position of a particle at time t, which
moves along a straight line. Thus, s( t1 ) is the position of the particle at time t1 ,
s( t 2 ) is the position of the particle at time t 2 , s( t 3 ) is the position of the particle at
time t 3 , and s( t 4 ) is the position of the particle at time t 4 and might look like the
following:

s( t 3 )

s( t 4 )

0

s( t1 )
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
s( t 2 )
We would like to be able to find the velocity of the particle at these times. We
have a function which gives us the position of the particle at any time t. We would
like to have a function which gives us the velocity of the particle at any time t. We
have enough information to find such a function. This function will be called the
instantaneous velocity function of the particle and will be name v. Thus, v( t ) will
be the velocity of the particle at time t. Thus, v( t1 ) will be the velocity of the
particle at time t1 , v( t 2 ) will be the velocity of the particle at time t 2 , v( t 3 ) will
be the velocity of the particle at time t 3 , and v( t 4 ) will be the velocity of the
particle at time t 4 . Now, let’s find this instantaneous velocity function.
Consider the position of the particle at the two times of t and t  h , where h  0 .
The position of particle on the line at time t is s ( t ) and the position of particle on
the line at time t  h is s ( t  h ) and might look like the following:

s( t )

s( t  h )
The average velocity of the particle between the times of t and t  h is given by
the following:
Average velocity =
change in position  s s ( t  h )  s ( t ) s ( t  h )  s( t )



change in time
t
t ht
h
The instantaneous velocity of the particle at time t is obtained by taking the limit as
s( t  h )  s( t )
h  0 . That is, v( t )  hlim
.
0
h
COMMENT: Instantaneous velocity is a special case of instantaneous rate with the
horizontal axis being the t-axis with unit of time and the vertical axis being the saxis with unit of distance. Thus, the slope of the tangent line to the graph of the
function s  s(t ) at the point ( t , s( t ) ) is the velocity of the particle at time t,
v( t ) .
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2
Example If s( t )  4t  18t is the position function which gives the position (in
meters) of a particle at time t (in hours), then find
a.
b.
c.
the instantaneous velocity function,
the position and velocity of the particle when t = 0, 2, 3, 6, 7, and 100 hours,
the position(s) when the particle is stopped.
For discussion purposes, we will assume the particle is traveling on a horizontal
line. Thus, the motion is either to the left or to the right.
a.
Find the instantaneous velocity function.
v( t )  lim
h0
s( t  h )  s( t )
h
s( t  h )  4 ( t  h ) 2  18 ( t  h ) = 4 ( t 2  2th  h 2 )  18t  18h =
4t 2  8th  4h 2  18t  18h
s( t )  4t 2  18t (from above)
s( t  h )  s( t ) = 8th  4h 2  18h = h ( 8t  4h  18 )
s( t  h )  s( t )
= 8t  4h  18
h
v( t )  lim ( 8t  4h  18 )  8t  18
h0
Answer: v( t )  8t  18
b.
Find the position and velocity of the particle when t = 0, 2, 3, 6, 7, and 100
hours.
s( 0 )  0 m
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v( 0 )   18
m
hr
NOTE: At time t  0 hr, the position of the particle is 0 m. Since the
m
velocity at t  0 hr is negative 18
, then the particle is traveling to the left
hr
m
18
on the horizontal line at a speed of
. This might be the initial position
hr
and the initial velocity of the particle.
s( 2 )  16  36   20 m
m
v( 2 )  16  18   2
hr
NOTE: At time t  2 hr, the position of the particle is negative 20 m .
Thus, the particle is 20 m to the left of 0 on the horizontal line. Since the
m
2
t

2
velocity at
hr is negative
, then the particle is traveling to the left
hr
m
2
at a speed of
.
hr
s( 3 )  36  54   18 m
m
v( 3 )  24  18  6
hr
NOTE: At time t  3 hr, the position of the particle is negative 18 m .
Thus, the particle is 18 m to the left of 0 on the horizontal line. Since the
m
velocity at t  3 hr is positive 6
, then the particle is traveling to the right
hr
m
at a speed of 6
.
hr
s( 6 )  144  108  36 m
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
v( 6 )  48  18  30
m
hr
NOTE: At time t  6 hr, the position of the particle is positive 36 m . Thus,
the particle is 36 m to the right of 0 on the horizontal line. Since the
m
velocity at t  6 hr is positive 30
, then the particle is traveling to the
hr
m
right at a speed of 30
.
hr
2
Since s( t )  4t  18t  2t ( 2t  9 ) , then s( 7 )  14 (5 )  70 m
m
v( 7 )  56  18  38
hr
NOTE: At time t  7 hr, the position of the particle is positive 70 m . Thus,
the particle is 70 m to right of 0 on the horizontal line. Since the velocity at
m
t  7 hr is positive 38
, then the particle is traveling to the right at a
hr
m
speed of 38
.
hr
s(100 )  200 (191)  38200 m
m
v(100 )  800  18  782
hr
NOTE: At time t  100 hr, the position of the particle is positive 38200 m .
Thus, the particle is 38200 m to right of 0 on the horizontal line. Since the
m
782
t

100
velocity at
hr is positive
, then the particle is traveling to the
hr
m
782
right at a speed of
.
hr
c.
Find the position(s) when the particle is stopped.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
The particle is stopped when its velocity is zero. Since v is the instantaneous
velocity function for the particle, we want to find when v( t )  0 . Thus, we
want to solve the equation v ( t )  0 for t. Since v( t )  8t  18 , then
18 9

v(t )  0  8t  18  0  8t  18  t 
8
4
Thus, the particle is stopped when t 
9
hr.
4
2
Now, use the position function s( t )  4t  18t to find the position of the
81
9
 9  81 162
s




t



particle when
hr. Thus,
m.
4
4
4
4 4
Answer: 
81
m
4
Example If a ball is thrown upward with a velocity of 36 feet/second from the top
of a 90-foot building, then the height of the ball above ground at time t (in seconds)
2
is given by the position function s( t )  90  36t  16t , where t  0 . Find
a.
b.
c.
the instantaneous velocity function,
the maximum height reached by the ball,
the velocity of the ball when it strikes the ground.
a.
Find the instantaneous velocity function.
v( t )  lim
h0
s( t  h )  s( t )
h
s( t  h )  90  36 ( t  h )  16 ( t  h ) 2 =
90  36t  36h  16 ( t 2  2th  h 2 ) =
90  36t  36h  16t 2  32th  16h 2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
s( t )  90  36t  16t 2 (from above)
s( t  h )  s( t ) = 36h  32th  16h 2 = h ( 36  32t  16h )
s( t  h )  s( t )
= 36  32t  16h
h
v( t )  lim ( 36  32t  16h )  36  32t  4 ( 9  8t )
h0
Answer: v( t )  36  32t or v( t )  4 ( 9  8t )
b.
Find the maximum height reached by the ball.
At the moment the ball reaches its maximum height, the velocity of the ball is
zero. Since v is the instantaneous velocity function for the ball, we want to
find when v( t )  0 . Thus, we want to solve the equation v ( t )  0 for t.
This will give us the time when the ball reaches it maximum height. Then we
can use the position function s to find the position of the ball at this time.
Since v( t )  4 ( 9  8t ) , then v(t )  0  4 ( 9  8t )  0  t 
9
8
9
Thus, the ball is at its maximum height
seconds after the ball is thrown
8
upward. Thus, the maximum height, that the ball will reach, is given by
9
s .
8
81 81 360 162 81 441
9
s    90 




=
2
4
4
4
4
4
8
Thus, the ball reaches a maximum height of
441
feet.
4
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
441
Answer:
ft
4
c.
Find the velocity of the ball when it strikes the ground.
Surprisingly, I have some students who say that the velocity of the ball when
it strikes the ground is zero. Then I ask them if they would be willing to lay
down on the ground and let the ball strike them. I tell them that if the
velocity of the ball is zero, then it won’t hurt when the ball strikes you. So
far, I haven’t gotten any volunteers for this.
At the moment the ball strikes the ground, the position of the ball is 0 feet.
Since s is the position function for the ball, we want to find when s( t )  0 .
Thus, we want to solve the equation s( t )  0 for t. This will give us the time
when the ball strikes the ground. Then we can use the instantaneous velocity
function v to find the velocity of the ball at this time.
s (t )  0  90  36t  16t 2  0  2 ( 45  18t  8t 2 )  0 
2 ( 3  2t ) (15  4t )  0  t  
3
15
or t 
2
4
Since t  0 , then the only valid time is t 
15
seconds. Thus, the ball
4
15
strikes the ground
seconds after it is thrown upward. Thus, the velocity
4
 15 
.
of the ball when it strikes the ground is given by v 
4


 15
v
 4
ft

  4 ( 9  30 )  4 (  21)   84
sec

ft
. (This is
sec
probably why I haven’t gotten any volunteers to let the ball strike them.)
Thus, the velocity of the ball when it strikes the ground is  84
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Answer:  84
ft
sec
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850