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In this section, we will examine: Some applications of the rate of change to physics, chemistry, biology, economics, and other sciences.
1.
3
DIFFERENTIATION RULES
2.
DIFFERENTIATION RULES
We know that, if y = f(x), then the
derivative dy/dx can be interpreted as
the rate of change of y with respect to x.
3.
DIFFERENTIATION RULES
3.7
Rates of Change in the
Natural and Social Sciences
In this section, we will examine:
Some applications of the rate of change to physics,
chemistry, biology, economics, and other sciences.
4.
RATES OF CHANGE
Let’s recall from Section 2.7 the basic
idea behind rates of change.
If x changes from x1 to x2, then the change in x
is ∆x = x2 – x1
The corresponding change in y is ∆y = f(x2) – f(x1)
5.
AVERAGE RATE
The difference quotient y f ( x2 ) f ( x1 )
x x2 x1
is the average rate of change of y with
respect to x over the interval [x1, x2].
It can be interpreted as
the slope of the secant
line PQ.
6.
INSTANTANEOUS RATE
Its limit as ∆x → 0 is the derivative
This can therefore be interpreted as the instantaneous
rate of change of y with respect to x or the slope of
the tangent line at P(x1,f(x1)).
7.
RATES OF CHANGE
Using Leibniz notation, we write the
process in the form dy y
lim
dx x 0 x
8.
RATES OF CHANGE
Whenever the function y = f(x) has a specific
interpretation in one of the sciences, its
derivative will have a specific interpretation
as a rate of change.
As we discussed in Section 2.7, the units for dy/dx
are the units for y divided by the units for x.
9.
NATURAL AND SOCIAL SCIENCES
We now look at some of these
interpretations in the natural and
social sciences.
10.
Let s = f(t) be the position function of
a particle moving in a straight line.
∆s/∆t represents the average velocity over
a time period ∆t
v = ds/dt represents the instantaneous velocity
(velocity is the rate of change of displacement
with respect to time)
The instantaneous rate of change of velocity with
respect to time is acceleration: a(t) = v’(t) = s’’(t)
11.
These were discussed in Sections
2.7 and 2.8
However, now that we know the differentiation
formulas, we are able to solve problems involving
the motion of objects more easily.
12.
PHYSICS Example 1
The position of a particle is given by
the equation s = f(t) = t3 – 6t2 + 9t where
t is measured in seconds and s in meters.
a) Find the velocity at time t.
b) What is the velocity after 2 s? After 4 s?
c) When is the particle at rest?
13.
PHYSICS Example 1
a) When is the particle moving forward
(that is, in the positive direction)?
b) Draw a diagram to represent the motion
of the particle.
c) Find the total distance traveled by the particle
during the first five seconds.
14.
PHYSICS Example 1
a) Find the acceleration at time t and after 4 s.
b) Graph the position, velocity, and acceleration
functions for 0 ≤ t ≤ 5.
c) When is the particle speeding up?
When is it slowing down?
15.
PHYSICS Example 1 a
The velocity function is the derivative
of the position function.
s = f(t) = t3 – 6t2 + 9t
v(t) = ds/dt = 3t2 – 12t + 9
16.
PHYSICS Example 1 b
The velocity after 2 s means the
instantaneous velocity when t = 2, that is,
ds 2
v(2) 3(2) 12(2) 9 3 m / s
dt t 2
The velocity after 4 s is:
2
v(4) 3(4) 12(4) 9 9 m / s
17.
PHYSICS Example 1 c
The particle is at rest when v(t) = 0,
that is,
3t2 - 12t + 9 = 3(t2 - 4t + 3) = 3(t - 1)(t - 3) = 0
This is true when t = 1 or t = 3.
Thus, the particle is at rest after 1 s and after 3 s.
18.
PHYSICS Example 1 d
The particle moves in the positive direction
when v(t) > 0, that is,
3t2 – 12t + 9 = 3(t – 1)(t – 3) > 0
This inequality is true when both factors are positive
(t > 3) or when both factors are negative (t < 1).
Thus the particle moves in the positive direction
in the time intervals t < 1 and t > 3.
It moves backward (in the negative direction)
when 1 < t < 3.
19.
PHYSICS Example 1 e
Using the information from (d), we make a
schematic sketch of the motion of the particle
back and forth along a line (the s -axis).
20.
PHYSICS Example 1 f
Due to what we learned in (d) and (e),
we need to calculate the distances traveled
during the time intervals [0, 1], [1, 3], and
[3, 5] separately.
21.
PHYSICS Example 1 f
The distance traveled in the first second is:
|f(1) – f(0)| = |4 – 0| = 4 m
From t = 1 to t = 3, it is:
|f(3) – f(1)| = |0 – 4| = 4 m
From t = 3 to t = 5, it is:
|f(5) – f(3)| = |20 – 0| = 20 m
The total distance is 4 + 4 + 20 = 28 m
22.
PHYSICS Example 1 g
The acceleration is the derivative of
the velocity function:
2
d s dv
a(t ) 2 6t 12
dt dt
2
a(4) 6(4) 12 12 m / s
23.
PHYSICS Example 1 h
The figure shows the graphs
of s, v, and a.
24.
PHYSICS Example 1 i
The particle speeds up when the velocity is
positive and increasing (v and a are both
positive) and when the velocity is negative
and decreasing (v and a are both negative).
In other words, the particle speeds up when
the velocity and acceleration have the same sign.
The particle is pushed in the same direction it is moving.
25.
PHYSICS Example 1 i
From the figure, we see that this
happens when 1 < t < 2 and when t > 3.
26.
PHYSICS Example 1 i
The particle slows down when v and a have
opposite signs—that is, when 0 ≤ t < 1 and
when 2 < t < 3.
27.
PHYSICS Example 1 i
This figure summarizes the motion
of the particle.
28.
PHYSICS Example 2
If a rod or piece of wire is homogeneous,
then its linear density is uniform and is
defined as the mass per unit length (ρ = m/l)
and measured in kilograms per meter.
29.
PHYSICS Example 2
However, suppose that the rod is not
homogeneous but that its mass
measured from its left end to a point x
is m = f(x).
30.
PHYSICS Example 2
The mass of the part of the rod that lies
between x = x1 and x = x2 is given by
∆m = f(x2) – f(x1)
31.
PHYSICS Example 2
So, the average density of that part
m
is: average density
x
f ( x2 ) f ( x1 )
x2 x1
32.
PHYSICS Example 2
If we now let ∆x → 0 (that is, x2 → x1),
we are computing the average density
over smaller and smaller intervals.
33.
LINEAR DENSITY
The linear density ρ at x1 is the limit of
these average densities as ∆x → 0.
That is, the linear density is the rate of change
of mass with respect to length.
34.
LINEAR DENSITY Example 2
m dm
lim
x 0 x dx
Thus, the linear density of the rod is
the derivative of mass with respect to length.
35.
PHYSICS Example 2
For instance, if m = f(x) = x , where x is
measured in meters and m in kilograms,
then the average density of the part of the rod
given by 1≤ x ≤ 1.2 is: m f (1.2) f (1)
x 1.2 1
1.2 1
0.2
0.48 kg / m
36.
PHYSICS Example 2
The density right at x = 1 is:
dm 1
0.50 kg / m
dx x 1 2 x x 1
37.
PHYSICS Example 3
A current exists whenever electric
charges move.
The figure shows part of a wire and electrons
moving through a shaded plane surface.
38.
PHYSICS Example 3
If ∆Q is the net charge that passes through
this surface during a time period ∆t, then
the average current during this time interval
is defined as:
Q Q2 Q1
average current
t t2 t1
39.
PHYSICS Example 3
If we take the limit of this average current over
smaller and smaller time intervals, we get
what is called the current I at a given time t1:
Q dQ
I lim
t 0 t dt
Thus, the current is the rate at which charge
flows through a surface.
It is measured in units of charge per unit time
(often coulombs per second—called amperes).
40.
Velocity, density, and current are not the only
rates of change important in physics.
Others include:
Power (the rate at which work is done)
Rate of heat flow
Temperature gradient (the rate of change of
temperature with respect to position)
Rate of decay of a radioactive substance in
nuclear physics
41.
CHEMISTRY Example 4
A chemical reaction results in the formation
of one or more substances (products) from
one or more starting materials (reactants).
For instance, the ‘equation’
2H2 + O2 → 2H2O
indicates that two molecules of hydrogen and
one molecule of oxygen form two molecules of water.
42.
CONCENTRATION Example 4
Let’s consider the reaction A + B → C
where A and B are the reactants and C is
the product.
The concentration of a reactant A is the number
of moles (6.022 X 1023 molecules) per liter and
is denoted by [A].
The concentration varies during a reaction.
So, [A], [B], and [C] are all functions of time (t).
43.
AVERAGE RATE Example 4
The average rate of reaction of
the product C over a time interval
t1 ≤ t ≤ t2 is:
[C ] [C ](t2 ) [C ](t1 )
t t2 t1
44.
INSTANTANEOUS RATE Example 4
However, chemists are more interested
in the instantaneous rate of reaction.
This is obtained by taking the limit of
the average rate of reaction as the time
interval ∆t approaches 0:
[C ] d [C ]
lim
rate of reaction =
t 0 t dt
45.
PRODUCT CONCENTRATION Example 4
Since the concentration of the product
increases as the reaction proceeds,
the derivative d[C]/dt will be positive.
So, the rate of reaction of C is positive.
46.
REACTANT CONCENTRATION Example 4
However, the concentrations of the
reactants decrease during the reaction.
So, to make the rates of reaction of A and B
positive numbers, we put minus signs in front
of the derivatives d[A]/dt and d[B]/dt.
47.
CHEMISTRY Example 4
Since [A] and [B] each decrease
at the same rate that [C] increases,
we have:
d [C ] d [ A] d [ B]
rate of reaction
dt dt dt
48.
CHEMISTRY Example 4
More generally, it turns out that for a reaction
of the form
aA + bB → cC + dD
we have
1 d [ A] 1 d [ B] 1 d [C ] 1 d [ D]
a dt b dt c dt d dt
49.
CHEMISTRY Example 4
The rate of reaction can be determined
from data and graphical methods.
In some cases, there are explicit formulas for
the concentrations as functions of time—which
enable us to compute the rate of reaction.
50.
COMPRESSIBILITY Example 5
One of the quantities of interest in
thermodynamics is compressibility.
If a given substance is kept at a constant temperature,
then its volume V depends on its pressure P.
We can consider the rate of change of volume with
respect to pressure—namely, the derivative dV/dP.
As P increases, V decreases, so dV/dP < 0.
51.
COMPRESSIBILITY Example 5
The compressibility is defined by introducing
a minus sign and dividing this derivative by
the volume V:
1 dV
isothermal compressibility
V dP
Thus, β measures how fast, per unit volume,
the volume of a substance decreases as the pressure
on it increases at constant temperature.
52.
CHEMISTRY Example 5
For instance, the volume V (in cubic meters)
of a sample of air at 25ºC was found to be
related to the pressure P (in kilopascals) by
the equation 5.3
V
P
53.
CHEMISTRY Example 5
The rate of change of V with respect to P
when P = 50 kPa is:
dV 5.3
2
dP P 50 P P 50
5.3
2500
3
0.00212 m / kPa
54.
CHEMISTRY Example 5
The compressibility at that pressure is:
1 dV 0.00212
V dP P 50
5.3
50
3 3
0.02(m / kPa) / m
55.
BIOLOGY Example 6
Let n = f(t) be the number of
individuals in an animal or plant
population at time t.
The change in the population size between
the times t = t1 and t = t2 is ∆n = f(t2) – f(t1)
56.
AVERAGE RATE
So, the average rate of growth during
the time period t1 ≤ t ≤ t2 is:
n f (t2 ) f (t1 )
average rate of growth
t t2 t1
57.
INSTANTANEOUS RATE Example 6
The instantaneous rate of growth is obtained
from this average rate of growth by letting
the time period ∆t approach 0:
n dn
growth rate lim
t 0 t dt
58.
BIOLOGY Example 6
Strictly speaking, this is not quite
This is because the actual graph of a population
function n = f(t) would be a step function that is
discontinuous whenever a birth or death occurs
and, therefore, not differentiable.
59.
BIOLOGY Example 6
However, for a large animal or plant
population, we can replace the graph by
a smooth approximating curve.
60.
BIOLOGY Example 6
To be more specific, consider
a population of bacteria in
a homogeneous nutrient medium.
Suppose that, by sampling the population
at certain intervals, it is determined that
the population doubles every hour.
61.
BIOLOGY Example 6
If the initial population is n0 and the time t
is measured in hours, then f (1) 2 f (0) 2n
0
2
f (2) 2 f (1) 2 n0
f (3) 2 f (2) 23 n0
and, in general, f (t ) 2t n
0
The population function is n = n02t
62.
BIOLOGY Example 6
In Section 3.4, we showed that:
d x x
(a ) a ln a
dx
So, the rate of growth of the bacteria
population at time t is:
dn d t t
(n0 2 ) n0 2 ln 2
dt dt
63.
BIOLOGY Example 6
For example, suppose that we start with
an initial population of n0 = 100 bacteria.
Then, the rate of growth after 4 hours is:
dn
t 4 100 24 ln 2 1600 ln 2 1109
dt
This means that, after 4 hours, the bacteria population
is growing at a rate of about 1109 bacteria per hour.
64.
BIOLOGY Example 7
When we consider the flow of blood through
a blood vessel, such as a vein or artery, we
can model the shape of the blood vessel by
a cylindrical tube with radius R and length l.
65.
BIOLOGY Example 7
Due to friction at the walls of the tube,
the velocity v of the blood is greatest along
the central axis of the tube and decreases
as the distance r from the axis increases
until v becomes 0 at the wall.
66.
BIOLOGY Example 7
The relationship between v and r is given by
the law of laminar flow discovered by the
French physician Jean-Louis-Marie Poiseuille
in 1840.
67.
LAW OF LAMINAR FLOW E. g. 7—Eqn. 1
The law states that P 2 2
v (R r )
4 l
where η is the viscosity of the blood and P
is the pressure difference between the ends
of the tube.
If P and l are constant, then v is a function of r
with domain [0, R].
68.
AVERAGE RATE Example 7
The average rate of change of the velocity
as we move from r = r1 outward to r = r2
is given by:
v v(r2 ) v(r1 )
r r2 r1
69.
VELOCITY GRADIENT Example 7
If we let ∆r → 0, we obtain the velocity
gradient—that is, the instantaneous rate
of change of velocity with respect to r:
v dv
velocity gradient lim
r 0 r dr
70.
BIOLOGY Example 7
Using Equation 1, we obtain:
dv P Pr
(0 2r )
dr 4 l 2 l
71.
BIOLOGY Example 7
For one of the smaller human arteries,
we can take η = 0.027, R = 0.008 cm,
l = 2 cm, and P = 4000 dynes/cm2.
4000 2
This gives: v (0.000064 r )
4(0.027)2
4 5 2
1.85 10 (6.4 10 r )
72.
BIOLOGY Example 7
At r = 0.002 cm, the blood is flowing at:
4 6 6
v(0.002) 1.85 10 (64 10 4 10 )
1.11cm / s
The velocity gradient at that point is:
dv 4000(0.002)
74 (cm / s) / cm
dr r 0.002 2(0.027)2
73.
BIOLOGY Example 7
To get a feeling of what this statement means,
let’s change our units from centimeters to
micrometers (1 cm = 10,000 μm).
Then, the radius of the artery is 80 μm.
The velocity at the central axis is 11,850 μm/s,
which decreases to 11,110 μm/s at a distance
of r = 20 μm.
74.
BIOLOGY Example 7
The fact that dv/dr = -74 (μm/s)/μm means
that, when r = 20 μm, the velocity is
decreasing at a rate of about 74 μm/s
for each micrometer that we proceed away
from the center.
75.
ECONOMICS Example 8
Suppose C(x) is the total cost that
a company incurs in producing x units
of a certain commodity.
The function C is called a cost function.
76.
AVERAGE RATE Example 8
If the number of items produced is increased
from x1 to x2, then the additional cost is
∆C = C(x2) - C(x1) and the average rate
of change of the cost is:
C C ( x2 ) C ( x1 ) C ( x1 x) C ( x1 )
x x2 x1 x
77.
MARGINAL COST Example 8
The limit of this quantity as ∆x → 0, that is,
the instantaneous rate of change of cost with
respect to the number of items produced,
is called the marginal cost by economists:
C dC
marginal cost = lim
x 0 x dx
78.
ECONOMICS Example 8
As x often takes on only integer values,
it may not make literal sense to let ∆x
approach 0.
However, we can always replace C(x) by a
smooth approximating function—as in Example 6.
79.
ECONOMICS Example 8
Taking ∆x = 1 and n large (so that ∆x is
small compared to n), we have:
C’(n) ≈ C(n + 1) – C(n)
Thus, the marginal cost of producing n units
is approximately equal to the cost of producing
one more unit [the (n + 1)st unit].
80.
ECONOMICS Example 8
It is often appropriate to represent a total cost
function by a polynomial
C(x) = a + bx + cx2 + dx3
where a represents the overhead cost (rent,
heat, and maintenance) and the other terms
represent the cost of raw materials, labor,
and so on.
81.
ECONOMICS Example 8
The cost of raw materials may be
proportional to x.
However, labor costs might depend partly on
higher powers of x because of overtime costs
and inefficiencies involved in large-scale
82.
ECONOMICS Example 8
For instance, suppose a company
has estimated that the cost (in dollars)
of producing x items is:
C(x) = 10,000 + 5x + 0.01x2
Then, the marginal cost function is:
C’(x) = 5 +
0.02x
83.
ECONOMICS Example 8
The marginal cost at the production level
of 500 items is:
C’(500) = 5 + 0.02(500) = $15/item
This gives the rate at which costs are increasing
with respect to the production level when x = 500
and predicts the cost of the 501st item.
84.
ECONOMICS Example 8
The actual cost of producing the 501st item is:
C(501) – C(500) =
[10,000 + 5(501) + 0.01(501)2]
– [10,000 + 5(500) + 0.01(500)2]
=$15.01
Notice that C’(500) ≈ C(501) – C(500)
85.
ECONOMICS Example 8
Economists also study marginal demand,
marginal revenue, and marginal profit—which
are the derivatives of the demand, revenue,
and profit functions.
These will be considered in Chapter 4—after we have
developed techniques for finding the maximum and
minimum values of functions.
86.
GEOLOGY AND ENGINEERING
Rates of change occur in all
the sciences.
A geologist is interested in knowing the rate
at which an intruded body of molten rock cools
by conduction of heat into surrounding rocks.
An engineer wants to know the rate at which
water flows into or out of a reservoir.
87.
GEOGRAPHY AND METEOROLOGY
An urban geographer is interested in the rate
of change of the population density in a city as
the distance from the city center increases.
A meteorologist is concerned with the rate
of change of atmospheric pressure with respect
to height.
88.
In psychology, those interested in learning
theory study the so-called learning curve.
This graphs the performance P(t) of someone learning
a skill as a function of the training time t.
Of particular interest is the rate at which performance
improves as time passes—that is, dP/dt.
89.
In sociology, differential calculus is
used in analyzing the spread of rumors
(or innovations or fads or fashions).
If p(t) denotes the proportion of a population that
knows a rumor by time t, then the derivative dp/dt
represents the rate of spread of the rumor.
90.
A SINGLE IDEA, MANY INTERPRETATIONS
You have learned about many special
cases of a single mathematical concept,
the derivative.
Velocity, density, current, power, and temperature
gradient in physics
Rate of reaction and compressibility in chemistry
Rate of growth and blood velocity gradient in biology
Marginal cost and marginal profit in economics
Rate of heat flow in geology
Rate of improvement of performance in psychology
Rate of spread of a rumor in sociology
91.
A SINGLE IDEA, MANY INTERPRETATIONS
This is an illustration of the fact that
part of the power of mathematics lies
in its abstractness.
A single abstract mathematical concept (such as
the derivative) can have different interpretations
in each of the sciences.
92.
A SINGLE IDEA, MANY INTERPRETATIONS
When we develop the properties of
the mathematical concept once and for all,
we can then turn around and apply these
results to all the sciences.
This is much more efficient than developing properties
of special concepts in each separate science.
93.
A SINGLE IDEA, MANY INTERPRETATIONS
The French mathematician Joseph Fourier
(1768–1830) put it succinctly:
“Mathematics compares the most
diverse phenomena and discovers
the secret analogies that unite them.”
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