lessons

1. Derivative as a Rate of Change

2. 2
Derivative as a Rate of Change
If y = f(x) and if x changes from the value x1 to x2, then y changes from
f(x1) to f(x2). So, the change in y, which we denote by ∆y, is f(x2) - f(x1)
when the change in x is ∆x = x2 – x1. The average rate of change of y with
respect to x, over the interval [x1, x2] , is then
This can also be interpreted as the slope of the secant line.
2 1
2 1
( ) ( )f x f x y
x x x
Δ
Δ
−
=
−
To illustrate, average velocity=
where the interval is from t = a to t = a + h.
h
afhaf
time
ntdisplaceme )()( −+
=

3. x
y Tangent line
Secant line
y = f(x)
P
Q
f(x2)-f(x1)
x2 – x1
Average rate of change is the slope of the secant line.
Instantaneous rate of change is the slope of the tangent line at P.

4. Definition of Instantaneous Rate of Change
Instantaneous Rate of Change=
f’(x1) is the instantaneous rate of change at x1.
Note that a positive rate means a quantity increases with respect to the
other quantity, that is y increases with x. If it is negative, then the
quantity decreases with respect to other quantity.
Note that heat, velocity, density, current, temperature, pressure, molar
concentration, fluid flow, bacterial growth, reaction rate, blood flow
and cost are just some of the few quantities that maybe analyzed
through derivatives.
We consider the average rate of change over a smaller and smaller
intervals by letting x2 approaches x1 and therefore letting ∆x approach 0.
The limit of this average rate 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 line to the curve y = f(x) at x = x1.
2 1
2 1
02 1
( ) ( )
x x x
f x f x y
Lim Lim
x x xΔ
Δ
Δ→ →
−
=
−

5. Given a set of data, we may approximate instantaneous rate of change in
values using average values or the graph representing the set of data.
For example:
Page 147 number 24
The population P (in thousands) of Canada from 1994 to 2002 is shown in
the table. (Midyear estimates are given).
a) Find the average rate of growth
i. From 1996 to 2000
ii. From 1998 to 2000
iii. From 2000 to 2002
b) Estimate the instantaneous rate of growth in 2000 by taking the
average of two average rates of change.
Year 1994 1996 1998 2000 2002
P 29036 29672 30248 30791 31414

6. p.147 #28
If a cylindrical tank holds 100,000 liters of water, which can be drained
from the bottom of the tank in an hour, then Torricelli’s law gives the
volume V of water remaining in the tank after t minutes as
a)Find the rate of change at which the water is flowing out of the tank (the
instantaneous rate of change of V with respect to t) as a function of t.
b)For times t = 0, 20, 40, and 60,find the flow rate and the amount of
water remaining in the tank.
2
60
1000,100)( 





−=
t
tV

7. 1. Using Ohm’s Law where V volts is the electromotive force, R ohms is
the resistance and I amperes is the current in an electric circuit, find the
rate of change in I with respect to R and find the instantaneous rate of
change of I with respect to R in an electric circuit having 120 volts when
the resistance is 20 ohms. (Take that V is constant)
Other Examples
Solution:
Ohm’s Law states that V = IR. Thus I = V/R = VR-1
. So, I= 120R-1
.
We have 2
2
R
120
R120
dR
dI −
=−= −
And so when R = 20 ohms, then = -0.30 ampere/ohm
The negative signs implies that current is decreasing at this conditions.
2
)20(
120
dR
dI −
=

8. 2. A solid consists of a right circular cylinder and a hemisphere on each
end, and the length of the cylinder is twice its radius. Let r units be the
radius of the cylinder and the two hemispheres, and V(r) cubic units be
the volume of the solid. Find the instantaneous rate of change in V(r)
with respect to r.
Solution:
If the height of the cylinder is twice its radius r, then h = 2r. Thus, the
volume of the cylinder is V1 = πr2
h = πr2
(2r) = 2πr3
.
Since two hemispheres are equal to a sphere, the volume is
V2 = 4/3 πr3
.
So the volume of the solid is V(r) = 2πr3
+ 4/3 πr3
= 10/3 πr3
. The
instantaneous rate of change is
V ’(r) = 10πr2
.
Note that the rate of change of the volume with respect to r can be
obtained when r is given.

9. 3.Sand is being dropped onto a conical pile in such a way that the height of
the pile is always twice the base radius. Find the rate of change of the
volume of the pile with respect to the radius when the height of the pile is
(a) 4 m and (b) 8 m.
Solution:
If the height of the cone is twice its radius, then h = 2r. So, the
volume of the cone is
V = 1/3πr2
h
V = 1/3 πr2
(2r)
So,
V(r) = 2/3 πr3
.
The rate of change of the volume with respect to r is
V ’(r) = 2 πr2
.
When a) h = 4m, r = 2 then v’(2) = 2 π (2)2
= 8π m2
.
b) h = 8m, r = 4, then v’(4) = 2 π (4)2
= 32π m2
.

10. 4. If water is being drained from a swimming pool and V(t) liters is the
volume of water in the pool t minutes after draining starts, where V(t)
= 250(1600 – 80t + t2
). Find
a. The average rate at which the water leaves the pool during the first 5
minutes.
b. How fast the water is flowing out of the pool 5 min after the draining
starts?
Solution:
b) The rate of change of the volume with respect to t is V ’(t) = 250(-80 + 2t).
So, at t = 5mins, the rate of change of the volume is
V ’(5) = 250[-80 + 2(5)] = -17,500 liters/min.
a) We find V(0) = 250(1600) = 400,000 and V(5) = 250(1600 – 80(5) + 52
)
= 306,250. The average rate at which the water leaves the pool during
the first 5 minutes is
min/liters18750
5
000,400250,306
05
)0(V)5(V
t
V
−=
−
=
−
−
=
∆
∆

11. 11
In manufacturing companies, the costs of producing their products is a
major concern. The following terms are useful in dealing with problems
involving costs.
Total Cost Function, C(x) – expression giving the total amount needed to
produce x units of a certain product.
Marginal Cost Function, C’(x) – Rate of change in cost when x units of
product is produced.
In a similar sense, we may consider the revenue of the company. Thus, we
have
Total Revenue Function, R(x) – expression giving the total amount earned
in the sales of x units of a certain product.
Marginal Revenue Function, R’(x) – Rate of change in revenue when x
units of product is sold.

12. Solution:
a. Since C(x) = 1500 + 3x + x2
then C’(x) = 3 + 2x
b. When x = 40, C’(x) = 3 + 2(40) = 83 dollars/watch.
5. The number of dollars in the total cost of manufacturing x watches in a
certain plant is given by C(x) = 1500 + 3x + x2
. Find (a) the marginal
cost function and (b) the marginal cost when x = 40.

13. Other Problems:
1. A wave produced by a simple sound has the equation P(t) = 0.003 sin
1800πt where P(t) dynes/cm2
is the difference between the
atmospheric pressure and the air pressure at the eardrum at t
seconds. Find the instantaneous rate of change of P(t) with respect to t
at (a) 1/9 sec and (b) 1/8 sec.
2. A Cepheid variable star is a star whose brightness alternately increases
and decreases. The most easily visible such star is Delta Cephei, for
which the interval between times of maximum brightness is 5.4 days.
The average brightness of this star is 4.0 and changes by ± 0.35. Its
brightness B is modeled by
where t is measured in days. Find the rate of change of the brightness
after 1day. (page 229, #61)
2
( ) 4.0 0.35sin
5.4
t
B t
π 
= +  ÷
 

14. 3. A company estimates that in t years, the number of its employees will be
N(t) = 1000 (0.8)t/2
(a) How many employees do the company expect to
have in 4 years. (b) At what rate is the number of employees expected to
be changing in 4 years?
4. Consider a blood vessel with radius 0.01 cm, length 3 cm, pressure
difference 3000 dyne/cm2
and viscosity η = 0.027. Using the law of
laminar flow:
(a) Find the velocity of the blood along the center line, at r = 0.005 cm and
at the wall. (b) Find the velocity gradient (instantaneous rate of change of
v with respect to r) at r = 0, r = 0.005 and r = 0.01 cm. (page 212, #25)
( )2 2
4
P
v R r
lη
= −
5. If R denotes the reaction of the body to some stimulus of strength x, the
sensitivity S is defined to be the rate of change of reaction with respect to
x. A particular example is that when the brightness x of a light source is
increased, the eye reacts by decreasing the area R of the pupil. The
experimental formula
has been used to model the dependence of R on x, when R is measured in
mm2
and x measured in brightness, find the sensitivity. (page 212, #30)
0.4
0.4
40 24
1 4
x
R
x
+
=
+

15. Another application of derivatives is rectilinear motion.
Rectilinear Motion is classified as horizontal motion or vertical motion
(free fall)
For Horizontal Motion:
We let x = f(t) be the distance function.
Now, velocity is v = dx/dt and acceleration is a = dv/dt = d2
x/dt2
.
The rate of change in acceleration is called jerk and this is
j = da/dt = d2
v/dt2
= d3
x/dt3
.
In some cases, s is used instead of x.
Example 1:
If a ball is given a push so that it has an initial velocity of 5m/s down a
certain inclined plane, then the distance it has rolled after t seconds is
s = 5t + 3t2
.
(a) Find the velocity after 2 sec.
(b) How long does it take for the velocity to reach 35m/s? (page 210,
number 6)

16. Example 3.
The motion of a spring that is subject to a frictional force or a damping
force is often modeled by the following function:
where s is measured in cm and t in seconds. Find the velocity after t
seconds. (page 229, #63)
1.5
( ) 2 sin2t
s t e tπ−
=
Example 2:
A particle moves according to a law of motion s = t3
– 12t2
+ 36t, t ≥ 0
where t is measured in seconds and s in meters.
(a) Find the velocity at time t.
(b) What is the velocity after 3 sec?
(c) When is the particle at rest?
(d) When is the particle moving forward?
(e) Find the total distance travelled during the first 8 sec.
(f) Find the acceleration at time t and after 3 sec.
(page 210, #1)

17. 17
Vertical Motion
Here, the velocity is still the derivative of position function y = f(t), given by v =
If a particle is projected straight upward from an initial height y0 (ft) above the ground at
time t = 0 (sec) and with initial velocity v0 (ft/sec) and if air resistance is negligible, then
its height y = f(t) (in feet above the ground) at time t is given by a formula known from
Physics,
y = f(t) = ½ gt2
+ v0t + y0.
where g denotes the acceleration due to the force of gravity. At the surface of the earth,
g ≈ -32 ft/s2
(or -9.8 m/s2
). Thus, y= f(t) = -16t2
+ v0t + y0. (or y= f(t) = -4.9t2
+ v0t + y0)
y is increasing y is decreasing
v > 0, decreasing v < 0, increasing
y=f(t)
Ground level
y = 0

18. 18
Exercises:
1. A stone is dropped from a building 256 ft high.
a. Write the equation of motion of the stone.
b. Find the instantaneous velocity of the stone at 1 sec and 2 sec.
c. Find how long it takes the stone to reach the ground.
d. What is the speed of the stone when it reaches the ground?
2. A ball is thrown vertically upward from the ground with an initial velocity of 32 ft/sec.
a. Estimate how high the ball will go and how long it takes the ball to reach the
highest point?
b. Find the instantaneous velocity of the ball at t = 0.75 sec and t = 1.25 sec.
c. Find speed of the ball at t = 0.75 sec and t = 1.25 sec.
d. Find the speed of the ball when it reaches the ground.
3. In an opera house, the base of a chandelier is 160 ft above the lobby floor. Suppose the
phantom of the opera dislocates the chandelier and is able to give the chandelier an initial
velocity of 48 ft/sec and causes it to fall and crash on the floor below.
a. Write the equation of motion of the chandelier.
b. Find the instantaneous velocity of the chandelier at 1 sec.
c. Find how long it takes the chandelier to hit the floor.
d. What is the speed of the chandelier when it hits the floor?