2.1 Tangent Lines and Velocity
1. Slope of a Line:
Let L be a line passing through points Ÿx 1 , y 1 and Ÿx 2 , y 2 . Then
y "y
m x 22 " x 11
y " y 1 mŸx " x 1 the slope of L
the equation of L
Such a line is also called a secant line.
Example Find the equation of the line passing two points which are on the curve : y x 2 1 when
x "2 and x 0. Sketch the curve and the line.
First find two points: let fŸx x 2 1, x 1 "2 and x 2 0. Then y 1 fŸx 1 Ÿ"2 2 1 5, and
y 2 fŸx 2 Ÿ0 2 1 1.
Compute the slope of the line:
m 1 " 5 "2
0 " Ÿ"2 The equation of the line:
y " 5 Ÿ"2 Ÿx " Ÿ"2 "2x " 4, y "2x 1
10
8
6
4
-3
-2
-1
0
-2
1
x
2
3
-4
-6
-8
- y x 2 1, -.- y "2x 1
2. Difference Quotient and The General Case of a Secant Line:
Consider a curve y fŸx . Let h 0. The slope m sec of the secant line through the points a, fŸa and
a h, fŸa h is
fŸa h " fŸa fŸa h " fŸa m
ah"a
h
It is also called a difference quotient.
Example Let fŸx x 2 1.
a. Find the slope of the secant line through points 1, fŸ1 and 1 h, fŸ1 h .
b. What are the slopes of the secant lines when h 0. 1, h 0. 01 and h 0. 001?
c. What is the limit of the slope of the secant line as h v 0?
a.
1
m sec
Ÿ1 h 2 1 " Ÿ1 2 1
fŸ1 h " fŸ1 h
h
2
hŸ2 h 2h h 2h
h
h
2
1 2h h 1 " 2
h
b.
h
0. 1
0. 01
0. 001
m sec
2 0. 1 2. 1
2 0. 01 2. 01
2 0. 001 2. 001
c.
lim m sec lim Ÿ2 h 2
hv0
hv0
6
4
2
0
0.5
1
x
1.5
2
2.5
- y x 2 1, - - y 3Ÿx " 1 2, -.- y 2Ÿx " 1 2
3. Slope of a Tangent Line:
The tangent line to the curve y fŸx at x a is the line that touches the curve at only one point
a, fŸa when x is near a. The slope m tan of the tangent line to the curve y fŸx at x a is defined as
fŸa h " fŸa m tan lim m sec lim
hv0
hv0
h
provided the limit exists. The equation of the tangent line at the point a, fŸa is
y " fŸa m tan Ÿx " a Steps for computing m tan :
(i) Compute fŸa .
or
y m tan Ÿx " a fŸa fŸa h " fŸa .
(ii) Compute and simplify the difference quotient:
h
fŸa h " fŸa if the limit exists.
(iii) Compute m tan lim hv0
h
2
Example Problem 12 on Page 161.
m tan at D m tan at C m tan at B m tan at A
Example Find the equation of the tangent line to the curve y 2x 3 " x at x "1.
a. fŸ"1 2Ÿ"1 3 " Ÿ"1 " 1
b. Compute and simplify the difference quotient:
2Ÿ"1 h 3 " Ÿ"1 h " Ÿ"1 2Ÿh " 1 3 " Ÿh " 1 1
fŸ"1 h " fŸ"1 h
h
h
3
2
3
2
2Ÿh " 3h 3h " 1 " h 1 1
2h " 6h 6h " 2 " h 2
h
h
2
3
2
hŸ2h " 6h 5 2h " 6h 5h 2h 2 " 6h 5
h
h
c. Compute m tan
fŸ"1 h " fŸ"1 m tan lim
limŸ2h 2 " 6h 5 5
hv0
hv0
h
d. The equation of the tangent line:
y " Ÿ"1 5Ÿx " Ÿ"1 ® y 5x 5 " 1 ® y 5x 4
Example Find the equation of the tangent line to the curve y 3x at x 1.
x1
3Ÿ1 3.
2
Ÿ1 1
b. Compute and simplify the difference quotient:
3Ÿ2 h 3Ÿ1 h 3Ÿ1 h Ÿ2 3Ÿ1 h 3
" 3
"
"
2
2Ÿ2 h fŸ1 h " fŸ1 Ÿ1 h 1
2 Ÿ2 h Ÿ2 2h
h
h
h
h
6 6h " 6 " 3h
3h
2Ÿ2 h 2Ÿ2 h 3
h
h
2Ÿ2 h a. fŸ1 c. Compute m tan
m tan lim
hv0
3
fŸ1 h " fŸ1 lim
3
hv0 2Ÿ2 h 4
h
d. The equation of the tangent line:
y " 3 3 Ÿx " 1 ® y 3 x " 3 3 3 x 3
4
4
4
4
2
4
2
2x " 3 at x 2.
Example Find the equation of the tangent line to the curve y a. fŸ2 2Ÿ2 " 3 1 1.
b. Compute and simplify the difference quotient:
fŸ2 h " fŸ2 h
3
2Ÿ2 h " 3 " 1
h
2h 1
h
2
" 12
2h 1 1
2h 1 " 1
h
h
2h 1 " 1
2h 1 1
2h 1 " 1
h
h
2h 1 1
2h 1 1
2h
2h 1 1
2
2h 1 1
c. Compute m tan
m tan lim
hv0
fŸ2 h " fŸ2 lim
hv0
h
2
2 1
2
2h 1 1
d. The equation of the tangent line:
y " 1 Ÿ1 Ÿx " 2 ® y x " 2 1 x " 1
4. Velocity and Instantaneous Rate of Change:
If fŸt represents the position of an object at time t, then the velocity of the object at time t a is given by
fŸa h " fŸa vŸa lim
hv0
h
provided the limit exists. vŸa is also called the instantaneous rate of change of fŸt at t a, which is the
same as the slope of the tangent line at the point a, fŸa . The average rate of change of fŸt between
t a and t b is given by
fŸb " fŸa b"a
which is the same as the slope of the secant line passing through points a, fŸa and b, fŸb .
Example Suppose that the population of a city is estimated to be fŸt from now.
100 8t million people t years
a. Find the average rate of change of the population for the next 2 years .
b. Find the instantaneous rate of change of the population 2 years from now.
a. the average rate of change:
fŸ2 " fŸ0 100 16 " 100
116 " 10
X 0. 385 16 million/year
2
2
2"0
That means that in average the population of this city grows 0. 38516 million people every year for the
next 2 years.
b. the instantaneous rate of change:
i. fŸ2 100 16 116
ii. Compute and simplify the difference quotient:
100 8Ÿ2 h " 116
116 8h " 116
fŸ2 h " fŸ2 h
h
h
116 8h " 116
116 8h 116
h 116 8h 116
h
iii. the instantaneous rate of change:
fŸ2 h " fŸ2 lim
lim
hv0
hv0
h
116 8h " 116
116 8h 116
h
8h
116 8h 116
8
116 8h 116
8
8
0. 371 4 million/year
2 116
116 8h 116
That means that the population of this city grows 0. 3714 million people per year two years from
now.
4