Section 2.1 Derivatives and Rates of Change
2010 Kiryl Tsishchanka
Derivatives and Rates of Change
The Tangent Problem
EXAMPLE: Graph the parabola y = x2 and the tangent line at the point P (1, 1).
Solution: We have:
DEFINITION: The tangent line to the curve y = f (x) at the point P (a, f (a)) is the line
through P with slope
f (x) − f (a)
m = lim
(1)
x→a
x−a
provided that this limit exists.
There is another (equivalent) expression for the slope of the tangent line:
f (a + h) − f (a)
h→0
h
m = lim
1
(2)
Section 2.1 Derivatives and Rates of Change
2010 Kiryl Tsishchanka
EXAMPLE: Find an equation of the tangent line to the hyperbola y = 3/x at the point (3, 1).
Solution 1: Let f (x) = 3/x. Then the slope of the tangent line at (3, 1) is
3 3
3
3 3
3
xa ·
−
xa
·
−
−
xa
·
f (x) − f (a) f (x)=3/x
x a
x
a
= lim
= lim x a = lim
m = lim
x→a
x→a
x→a
x→a
x−a
x−a
xa · (x − a)
xa(x − a)
3(a − x)
−3(x − a)
−3
3a − 3x
−3 a=3 −3
1
= lim
= lim
= lim
= 2 = 2 =−
x→a xa(x − a)
x→a xa(x − a)
x→a xa
x→a xa(x − a)
a
3
3
= lim
Recall, that the point-slope equation of a line is
y − y0 = m(x − x0 )
Therefore, an equation of the tangent line at the point (3, 1) is
1
y − 1 = − (x − 3)
3
which simplifies to
x + 3y − 6 = 0
Solution 2: Equivalently, if we use formula (2), we get
3
3
3
3
−
(a + h)a ·
−
f (a + h) − f (a) f (x)=3/x
a+h a
a
+
h
a
m = lim
= lim
= lim
h→0
h→0
h→0
h
h
(a + h)a · h
(a + h)a ·
= lim
h→0
3
3
− (a + h)a ·
a+h
a = lim 3a − 3(a + h) = lim 3a − 3a − 3h
h→0
h→0 (a + h)ah
(a + h)ah
(a + h)ah
−3h
−3
−3 a=3 −3
1
−3
= lim
=
= 2 = 2 =−
h→0 (a + h)ah
h→0 (a + h)a
(a + 0)a
a
3
3
= lim
and the same result follows.
EXAMPLE: Find an equation of the tangent line to y = x3 − x at the point (1/3, f (1/3)).
2
Section 2.1 Derivatives and Rates of Change
2010 Kiryl Tsishchanka
EXAMPLE: Find an equation of the tangent line to the hyperbola y = x3 − x at the point
(1/3, f (1/3)).
Solution 1: Let f (x) = x3 − x. Then the slope of the tangent line at (1/3, f (1/3)) is
f (x) − f (a)
x→a
x−a
m = lim
f (x)=x3 −x
=
x 3 − x − a3 + a
(x3 − x) − (a3 − a)
= lim
x→a
x→a
x−a
x−a
lim
(x3 − a3 ) − (x − a)
(x − a)(x2 + xa + a2 ) − (x − a) · 1
x 3 − a3 − x + a
= lim
= lim
x→a
x→a
x→a
x−a
x−a
x−a
= lim
(x − a)(x2 + xa + a2 − 1)
= lim (x2 + xa + a2 − 1)
x→a
x→a
x−a
2
2
1
a=1/3
2
2
2
2
−1=−
= a + a + a − 1 = 3a − 1 = 3 ·
3
3
Recall, that the point-slope equation of a line is
= lim
y − y0 = m(x − x0 )
Therefore, an equation of the tangent line at the point (1/3, f (1/3)) is
2
1
8
1
2
y − y0 = m(x − x0 ) =⇒ y − f (1/3) = −
x−
=⇒ y +
x−
=−
3
3
27
3
3
which simplifies to
18x + 27y + 2 = 0
Solution 2: Equivalently, if we use formula (2), we get
f (a + h) − f (a)
h→0
h
m = lim
f (x)=x3 −x
=
[(a + h)3 − (a + h)] − [a3 − a]
h→0
h
lim
3a2 h + 3ah2 + h3 − h
[a3 + 3a2 h + 3ah2 + h3 − a − h] − [a3 − a]
= lim
h→0
h→0
h
h
2
h(3a2 + 3ah + h2 − 1)
2
1
a=1/3
2
2
2
= lim
−1=−
= lim (3a + 3ah + h − 1) = 3a − 1 = 3 ·
h→0
h→0
h
3
3
and the same result follows.
= lim
Solution 3: Similarly, if we use formula (2), we get
f (a + h) − f (a)
h→0
h
m = lim
f (x)=x3 −x
=
[(a + h)3 − (a + h)] − [a3 − a]
h→0
h
lim
[(a + h)3 − a3 ] − [(a + h) − a]
[((a + h) − a)((a + h)2 + (a + h)a + a2 )] − [(a + h) − a]
= lim
h→0
h→0
h
h
= lim
h((a + h)2 + (a + h)a + a2 ) − h
h[(a + h)2 + (a + h)a + a2 − 1]
= lim
h→0
h→0
h
h
2
2
1
a=1/3
−1=−
= lim ((a + h)2 + (a + h)a + a2 − 1) = a2 + a2 + a2 − 1 = 3a2 − 1 = 3 ·
h→0
3
3
and the same result follows.
= lim
3
Section 2.1 Derivatives and Rates of Change
2010 Kiryl Tsishchanka
The Velocity Problem
Suppose an object moves along a straight line according to an equation of motion s = f (t),
where s is the displacement (directed distance) of the object from the origin at time t. The
function f that describes the motion is called the position function of the object. In the time
interval from t = a to t = a + h the change in position is f (a + h) − f (a). The average velocity
over this time interval is
average velocity =
displacement
f (a + h) − f (a)
=
time
h
which is the same as the slope of the secant line P Q in the second figure.
Now suppose we compute the average velocities over shorter and shorter time intervals [a, a+h].
In other words, we let h approach 0. We define velocity (or instantaneous velocity) v(a) at
time t = a to be the limit of these average velocities:
f (a + h) − f (a)
h→0
h
v(a) = lim
REMARK: Equivalently,
f (x) − f (a)
x→a
x−a
v(a) = lim
EXAMPLE: Suppose that a ball is dropped from the upper observation deck of the CN Tower,
450 m above the ground.
(a) What is the velocity of the ball after 5 seconds?
(b) How fast is the ball traveling when it hits the ground?
4
Section 2.1 Derivatives and Rates of Change
2010 Kiryl Tsishchanka
EXAMPLE: Suppose that a ball is dropped from the upper observation deck of the CN Tower,
450 m above the ground.
(a) What is the velocity of the ball after 5 seconds?
(b) How fast is the ball traveling when it hits the ground?
Solution:
(a) We use the equation of motion
s = f (t) = 4.9t2
where t is time (in seconds) and s is the displacement (in meters) to find the velocity v(a) after
a seconds:
f (a + h) − f (a)
f (x) − f (a)
v(a) = lim
v(a) = lim
x→a
h→0
h
x−a
4.9(a + h)2 − 4.9a2
h→0
h
4.9x2 − 4.9a2
x→a
x−a
= lim
4.9(x2 − a2 )
x→a
x−a
= lim
4.9(x − a)(x + a)
x→a
x−a
= lim
= lim
4.9(a2 + 2ah + h2 ) − 4.9a2
h→0
h
= lim
4.9(a2 + 2ah + h2 − a2 )
h→0
h
= lim
= lim 4.9(x + a)
x→a
4.9(2ah + h2 )
= lim
h→0
h
or
= 4.9(a + a)
4.9h(2a + h)
h→0
h
= lim
= 4.9(2a)
= lim 4.9(2a + h)
= 9.8a
h→0
= 4.9(2a + 0)
= 4.9(2a)
= 9.8a
Therefore the velocity after 5 seconds is v(5) = 9.8 · 5 = 49 m/s.
(b) Since the observation deck is 450 m above the ground, the ball will hit the ground at the
time t1 when s(t1 ) = 450, that is,
r
450
450
2
2
=⇒ t1 =
≈ 9.6 s
4.9t1 = 450 =⇒ t1 =
4.9
4.9
The velocity of the ball as it hits the ground is therefore
r
450
≈ 94 m/s
v(t1 ) = 9.8t1 = 9.8
4.9
5
Section 2.1 Derivatives and Rates of Change
2010 Kiryl Tsishchanka
Derivatives
We have seen that limits of the form
f (x) − f (a)
x→a
x−a
lim
f (a + h) − f (a)
h→0
h
or
lim
arise in finding the slope of a tangent line or the velocity of an object. Moreover, the same
type of limit arises whenever we calculate a rate of change in any of the sciences or engineering,
such as a rate of reaction in chemistry or a marginal cost in economics. Since this type of limit
occurs so widely, it is given a special name and notation.
DEFINITION: The derivative of a function f at a number a, denoted by f ′ (a), is
f (a + h) − f (a)
h→0
h
f ′ (a) = lim
if this limit exists.
REMARK: Equivalently,
f (x) − f (a)
x→a
x−a
f ′ (a) = lim
EXAMPLE: Find the derivative of the function y = x2 − 8x + 9 at the number a. Then find an
equation of the tangent line to the parabola y = x2 − 8x + 9 at the point (3, −6).
6
Section 2.1 Derivatives and Rates of Change
2010 Kiryl Tsishchanka
EXAMPLE: Find the derivative of the function y = x2 − 8x + 9 at the number a. Then find an
equation of the tangent line to the parabola y = x2 − 8x + 9 at the point (3, −6).
Solution: We have
f (x) − f (a)
f ′ (a) = lim
x→a
x−a
f (a + h) − f (a)
h→0
h
f ′ (a) = lim
[(a + h)2 − 8(a + h) + 9] − [a2 − 8a + 9]
= lim
h→0
h
(x2 − 8x + 9) − (a2 − 8a + 9)
= lim
x→a
x−a
a2 + 2ah + h2 − 8a − 8h + 9 − a2 + 8a − 9
h→0
h
x2 − 8x + 9 − a2 + 8a − 9
x→a
x−a
= lim
x2 − 8x − a2 + 8a
x→a
x−a
= lim
= lim
2ah + h2 − 8h
h→0
h
= lim
or
x2 − a2 − 8x + 8a
x→a
x−a
= lim
h→0
(x − a)(x + a) − 8(x − a)
x→a
x−a
= lim
(x − a)[(x + a) − 8]
= lim
x→a
x−a
= lim
h(2a + h − 8)
h
= lim (2a + h − 8)
h→0
= 2a + 0 − 8
= 2a − 8
= lim [(x + a) − 8]
x→a
= 2a − 8
so
f ′ (a) = 2a − 8
To find an equation of the tangent line to the parabola y = x2 − 8x + 9 at the point (3, −6) we
will use the point-slope equation of a line:
y − y0 = m(x − x0 )
Since (x0 , y0 ) = (3, −6) and m = f ′ (3) = 2 · 3 − 8 = −2, we obtain
y − (−6) = −2(x − 3) or y = −2x
7
Section 2.1 Derivatives and Rates of Change
2010 Kiryl Tsishchanka
Rates of Change
Suppose y is a quantity that depends on another quantity x. Thus y is a function of x and we
write y = f (x). If x changes from x1 to x2 , then the change in x (also called the increment of
x) is
∆x = x2 − x1
and the corresponding change in y is
∆y = f (x2 ) − f (x1 )
The difference quotient
f (x2 ) − f (x1 )
∆y
=
∆x
x2 − x1
is called the average rate of change of y with respect to x over the interval [x1 , x2 ] and
can be interpreted as the slope of the secant line P Q in the figure above.
By analogy with velocity, we consider the average rate of change over smaller and smaller
intervals by letting x2 approach x1 and therefore letting ∆x approach 0. The limit of these
average rates of change is called the (instantaneous) rate of change of y with respect to
x at x = x1 , which is interpreted as the slope of the tangent curve y = f (x) at (P (x1 , f (x1 )):
f (x2 ) − f (x1 )
∆y
= lim
x2 →x1
∆x→0 ∆x
x2 − x1
instantaneous rate of change = lim
Since f ′ (x1 ) = lim
x2 →x1
f (x2 ) − f (x1 )
, we have a second interpretation of the rate of change:
x2 − x1
The derivative f ′ (a) is the instantaneous rate of change of y = f (x)
with respect to x when x = a
EXAMPLE: Let D(t) be the US national debt at time t. The table below gives approximate
values of this function by providing end of year estimates, in billions of dollars, from 1980 to
2000. Interpret and estimate the value of D′ (1990).
t
1980
1985
1990
1995
2000
D(t)
930.2
1945.9
3233.3
4974.0
5674.2
8
Section 2.1 Derivatives and Rates of Change
2010 Kiryl Tsishchanka
EXAMPLE: Let D(t) be the US national debt at time t. The table below gives approximate
values of this function by providing end of year estimates, in billions of dollars, from 1980 to
2000. Interpret and estimate the value of D′ (1990).
t
1980
1985
1990
1995
2000
D(t)
930.2
1945.9
3233.3
4974.0
5674.2
Solution: The derivative D′ (1990) means the rate of change of D with respect to t when
t = 1990, that is, the rate of increase of the national debt in 1990. We know that
D(t) − D(1990)
t→1990
t − 1990
D′ (1990) = lim
So we compute values of the difference quotient (the average rates of change) as follows:
t
1980
1985
1995
2000
D(t) − D(1990)
t − 1990
230.31
257.48
348.14
244.09
From this table we see that D′ (1990) lies somewhere between 257.48 and 348.14 billion dollars
per year. [Here we are making the reasonable assumption that the debt didn’t fluctuate wildly
between 1980 and 2000.] We estimate that the rate of increase of the national debt of the
United States in 1990 was the average of these two numbers, namely
D′ (1990) ≈
257.48 + 348.14
= 302.81 ≈ 303 billion dollars per year
2
Another method would be to plot the debt function and estimate the slope of the tangent line
when t = 1990.
9