# Solving real life problems using differentiation

Contributed by:
In this section, we will: Use differentiation to solve real-life problems involving exponentially growing quantities.
1. 3
DIFFERENTIATION RULES
2. DIFFERENTIATION RULES
3.8
Exponential Growth
and Decay
In this section, we will:
Use differentiation to solve real-life problems
involving exponentially growing quantities.
3. EXPONENTIAL GROWTH & DECAY
In many natural phenomena,
quantities grow or decay at a rate
proportional to their size.
4. For instance, suppose y = f(t) is
the number of individuals in a population
of animals or bacteria at time t.
 Then, it seems reasonable to expect that the rate
of growth f’(t) is proportional to the population f(t).
 That is, f’(t) = kf(t) for some constant k.
5. EXPONENTIAL GROWTH & DECAY
Indeed, under ideal conditions—unlimited
environment, adequate nutrition, and
immunity to disease—the mathematical
model given by the equation f’(t) = kf(t)
predicts what actually happens fairly
6. Another example occurs in nuclear
physics where the mass of a radioactive
substance decays at a rate proportional
to the mass.
7. In chemistry, the rate of a unimolecular
first-order reaction is proportional to
the concentration of the substance.
8. In finance, the value of a savings
account with continuously compounded
interest increases at a rate proportional
to that value.
9. EXPONENTIAL GROWTH & DECAY Equation 1
In general, if y(t) is the value of a quantity y
at time t and if the rate of change of y with
respect to t is proportional to its size y(t)
at any time, then
dy
ky
dt
where k is a constant.
10. EXPONENTIAL GROWTH & DECAY
Equation 1 is sometimes called the law of
natural growth (if k > 0) or the law of natural
decay (if k < 0).
It is called a differential equation because
it involves an unknown function and its
derivative dy/dt.
11. EXPONENTIAL GROWTH & DECAY
It’s not hard to think of a solution of
Equation 1.
 The equation asks us to find a function whose
derivative is a constant multiple of itself.
 We have met such functions in this chapter.
 Any exponential function of the form y(t) = Cekt,
where C is a constant, satisfies
y '(t ) C (ke kt ) k (Ce kt ) ky (t )
12. EXPONENTIAL GROWTH & DECAY
We will see in Section 9.4 that any
function that satisfies dy/dt = ky must be
of the form y = Cekt.
 To see the significance of the constant C,
we observe that
k 0
y (0) Ce C
 Therefore, C is the initial value of the function.
13. EXPONENTIAL GROWTH & DECAY Theorem 2
The only solutions of the differential
equation dy/dt = ky are the exponential
y(t) = y(0)ekt
14. POPULATION GROWTH
What is the significance of
the proportionality constant k?
15. POPULATION GROWTH Equation 3
In the context of population growth,
where P(t) is the size of a population
at time t, we can write:
dP 1 dP
kP or k
dt P dt
16. RELATIVE GROWTH RATE
The quantity 1 dP
P dt
is the growth rate divided by
the population size.
 It is called the relative growth rate.
17. RELATIVE GROWTH RATE
According to Equation 3, instead of saying
“the growth rate is proportional to population
size,” we could say “the relative growth rate
is constant.”
 Then, Theorem 2 states that a population
with constant relative growth rate must grow
exponentially.
18. RELATIVE GROWTH RATE
Notice that the relative growth rate k
appears as the coefficient of t in the
exponential function Cekt.
19. RELATIVE GROWTH RATE
For instance, if dP
0.02 P
dt
and t is measured in years, then the relative
growth rate is k = 0.02 and the population
grows at a relative rate of 2% per year.
 If the population at time 0 is P0, then the
expression for the population is:
P(t) = P0e0.02t
20. POPULATION GROWTH Example 1
Use the fact that the world population was
2,560 million in 1950 and 3,040 million in
1960 to model the population in the second
half of the 20th century. (Assume the growth
rate is proportional to the population size.)
 What is the relative growth rate?
 Use the model to estimate the population in 1993
and to predict the population in 2020.
21. POPULATION GROWTH Example 1
We measure the time t in years and let
t = 0 in 1950.
We measure the population P(t) in millions
of people.
 Then, P(0) = 2560 and P(10) = 3040
22. POPULATION GROWTH Example 1
Since we are assuming dP/dt = kP,
Theorem 2 gives:
kt kt
P (t ) P(0)e 2560e
10 k
P (10) 2560e 3040
1 3040
k  ln 0.017185
10 2560
23. POPULATION GROWTH Example 1
The relative growth rate is about 1.7%
per year and the model is:
0.017185 t
P (t ) 2560e
 We estimate that the world population in 1993 was:
P (43) 2560e0.017185(43) 5360 million
 The model predicts that the population in 2020 will be:
0.017185(70)
P(70) 2560e 8524 million
24. POPULATION GROWTH Example 1
The graph shows that the model is fairly
accurate to the end of the 20th century.
 The dots represent
the actual population.
25. POPULATION GROWTH Example 1
So, the estimate for 1993 is quite reliable.
However, the prediction for 2020 is riskier.
26. RADIOACTIVE DECAY
Radioactive substances decay by
spontaneously emitting radiation.
 If m(t) is the mass remaining from an initial
mass m0 of a substance after time t, then
the relative decay rate  1 dm
m dt
has been found experimentally to be constant.
 Since dm/dt is negative, the relative decay rate
is positive.
27. RADIOACTIVE DECAY
It follows that dm
km
dt
where k is a negative constant.
 In other words, radioactive substances decay at
a rate proportional to the remaining mass.
 This means we can use Theorem 2 to show that
the mass decays exponentially: kt
m(t ) m0 e
28. Physicists express the rate of decay
in terms of half-life.
 This is the time required for half of any given
quantity to decay.
29. RADIOACTIVE DECAY Example 2
The half-life of radium-226 is 1590 years.
a. A sample of radium-226 has a mass of 100 mg.
Find a formula for the mass of the sample that
remains after t years.
b. Find the mass after 1,000 years correct to
the nearest milligram.
c. When will the mass be reduced to 30 mg?
30. RADIOACTIVE DECAY Example 2 a
Let m(t) be the mass of radium-226
(in milligrams) that remains after t years.
 Then, dm/dt = km and y(0) = 100.
 So, Theorem 2 gives:
m(t) = m(0)ekt = 100ekt
31. RADIOACTIVE DECAY Example 2 a
To determine the value of k , we use the fact
that y(1590) = ½(100).
 Thus, 100e1590k = 50. So, e1590k = ½.
 Also, 1590k = ln ½ = -ln 2
ln 2
k 
1590
 So, m(t) = 100e-(ln 2)t/1590
32. RADIOACTIVE DECAY Example 2 a
We could use the fact that eln 2 = 2
to write the expression for m(t) in
the alternative form
m(t) = 100 x 2-t/1590
33. RADIOACTIVE DECAY Example 2 b
The mass after 1,000 years is:
m(1000) = 100e-(ln 2)1000/1590
≈ 65 mg
34. RADIOACTIVE DECAY Example 2 c
We want to find the value of t such that
m(t) = 30, that is,
100e-(ln 2)t/1590 = 30 or e-(ln 2)t/1590 = 0.3
 We solve this equation for t by taking the natural
logarithm of both sides: ln 2
 t ln 0.3
1590
ln 0.3
 Thus, t  1590
ln 2
2762 years
35. RADIOACTIVE DECAY
As a check on our work in the example, we
use a graphing device to draw the graph of
m(t) together with the horizontal line m = 30.
 These curves
intersect when
t ≈ 2800.
 This agrees with
the answer to (c).
36. NEWTON’S LAW OF COOLING
Newton’s Law of Cooling states:
The rate of cooling of an object is proportional
to the temperature difference between the
object and its surroundings—provided the
difference is not too large.
 The law also applies to warming.
37. NEWTON’S LAW OF COOLING
If we let T(t) be the temperature of the object
at time t and Ts be the temperature of the
surroundings, then we can formulate the law
as a differential equation: dT
k  T  Ts 
dt
where k is a constant.
38. NEWTON’S LAW OF COOLING
This equation is not quite the same as
Equation 1.
So, we make the change of variable
y(t) = T(t) - Ts.
 As Ts is constant, we have y’(t) = T’(t).
 So, the equation becomes dy
ky
dt
 We can then use Theorem 2 to find an expression
for y, from which we can find T.
39. NEWTON’S LAW OF COOLING Example 3
A bottle of soda pop at room temperature
(72°F) is placed in a refrigerator, where
the temperature is 44°F. After half an hour,
the soda pop has cooled to 61°F.
a) What is the temperature of the soda pop
after another half hour?
b) How long does it take for the soda pop
to cool to 50°F?
40. NEWTON’S LAW OF COOLING Example 3 a
Let T(t) be the temperature of the soda
after t minutes.
 The surrounding temperature is Ts = 44°F.
 So, Newton’s Law of Cooling states:
dT
k  T  44 
dt
41. NEWTON’S LAW OF COOLING
If we let y = T – 44, then y(0) = T(0) – 44
= 72 – 44
= 28
dy
 So, y satisfies ky y (0) 28
dt
 Also, by Theorem 2, we have:
y(t) = y(0)ekt = 28ekt
42. NEWTON’S LAW OF COOLING Example 3 a
We are given that T(30) = 61.
So, y(30) = 61 - 44 = 17
30 k 30 k
and 28e 17 e  2817
 Taking logarithms, we have: ln( 17
28 )
k
30
 0.01663
43. NEWTON’S LAW OF COOLING Example 3 a
 0.01663t
Thus, y (t ) 28e
 0.01663t
T (t ) 44  28e
 0.01663(60)
T (60) 44  28e
54.3
 So, after another half hour, the pop has cooled
to about 54°F.
44. NEWTON’S LAW OF COOLING Example 3 b
We have T(t) = 50 when
 0.01663t
44  28e 50
e  0.01663t  286
ln  286 
t
0.01663
92.6
 The pop cools to 50°F after
about 1 hour 33 minutes.
45. NEWTON’S LAW OF COOLING
In the example, notice that we have
lim T (t ) lim  44  28e  0.01663t  44  28 0 44
t  t 
which is to be expected.
 The graph of the
temperature function
is shown.
46. EXPONENTIAL GROWTH & DECAY
Finally, we will look at
an example of continuously
compounded interest.
47. CONTINUOUSLY COMPD. INT. Example 4
If \$1000 is invested at 6% interest,
compounded annually, then:
 After 1 year, the investment is worth \$1000(1.06)
= \$1060
 After 2 years, it’s worth \$[1000(1.06)] 1.06
= \$1123.60
 After t years, it’s worth \$1000(1.06)t
48. CONTINUOUSLY COMPD. INT. Example 4
In general, if an amount A0 is invested
at an interest rate r (r = 0.06 in this
example), then after t years it’s worth
A0(1 + r)t.
49. CONTINUOUSLY COMPD. INT. Example 4
Usually, however, interest is compounded
more frequently—say, n times a year.
 Then, in each compounding period, the interest
rate is r/n and there are nt compounding periods
in t years.
nt
 r
 So, the value of the investment is: A0  1  
 n
50. CONTINUOUSLY COMPD. INT. Example 4
For instance, after 3 years at 6% interest,
a \$1000 investment will be worth:
\$1000(1.06)3 \$1191.02 (annualcompounding)
6
\$1000(1.03) \$1194.05 (semiannualcompounding)
\$1000(1.015)12 \$1195.62 (quarterly compounding)
\$1000(1.005)36 \$1196.68 (monthly compounding)
3653
 0.06 
\$1000  1   \$1197.20 (daily compounding)
 365 
51. CONTINUOUSLY COMPD. INT. Example 4
You can see that the interest paid
increases as the number of
compounding periods (n) increases.
52. CONTINUOUSLY COMPD. INT. Example 4
If we let n → ∞, then we will be compounding
the interest continuously and the value of
the investment will be: rt
nt n/r
 r   r 
A(t ) lim A0  1   lim A0   1   
n 
 n n 
  n  
n/r rt
  r 
 A0  lim  1   
 n    n  
m rt
  1 
 A0  lim  1    (where m n / r )
 m   m  
53. CONTINUOUSLY COMPD. INT. Example 4
However, the limit in this expression is
equal to the number e. (See Equation 6
in Section 3.6)
 So, with continuous compounding of interest
at interest rate r, the amount after t years is:
A(t) =
A0ert
54. CONTINUOUSLY COMPD. INT. Example 4
If we differentiate this function,
we get:
dA rt
rA0 e rA(t )
dt
 This states that, with continuous compounding
of interest, the rate of increase of an investment
is proportional to its size.
55. CONTINUOUSLY COMPD. INT. Example 4
Returning to the example of \$1000 invested
for 3 years at 6% interest, we see that,
with continuous compounding of interest,
the value of the investment will be:
(0.06)3
A(3) \$1000e \$1197.22
 Notice how close this is to the amount we
calculated for daily compounding, \$1197.20
 However, it is easier to compute if we use
continuous compounding.