1. Section 3.4
Exponential Growth and Decay
V63.0121.021, Calculus I
New York University
October 28, 2010
Announcements
Quiz 3 next week in recitation on 2.6, 2.8, 3.1, 3.2
. . . . . .
2. Announcements
Quiz 3 next week in
recitation on 2.6, 2.8, 3.1,
3.2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 2 / 40
3. Objectives
Solve the ordinary
differential equation
y′ (t) = ky(t), y(0) = y0
Solve problems involving
exponential growth and
decay
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 3 / 40
4. Outline
Recall
The differential equation y′ = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 4 / 40
5. Derivatives of exponential and logarithmic functions
y y′
ex ex
ax (ln a) · ax
1
ln x
x
1 1
loga x ·
ln a x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 5 / 40
6. Outline
Recall
The differential equation y′ = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 6 / 40
7. What is a differential equation?
Definition
A differential equation is an equation for an unknown function which
includes the function and its derivatives.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
8. What is a differential equation?
Definition
A differential equation is an equation for an unknown function which
includes the function and its derivatives.
Example
Newton’s Second Law F = ma is a differential equation, where
a(t) = x′′ (t).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
9. What is a differential equation?
Definition
A differential equation is an equation for an unknown function which
includes the function and its derivatives.
Example
Newton’s Second Law F = ma is a differential equation, where
a(t) = x′′ (t).
In a spring, F(x) = −kx, where x is displacement from equilibrium
and k is a constant. So
k
−kx(t) = mx′′ (t) =⇒ x′′ (t) + x(t) = 0.
m
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
10. What is a differential equation?
Definition
A differential equation is an equation for an unknown function which
includes the function and its derivatives.
Example
Newton’s Second Law F = ma is a differential equation, where
a(t) = x′′ (t).
In a spring, F(x) = −kx, where x is displacement from equilibrium
and k is a constant. So
k
−kx(t) = mx′′ (t) =⇒ x′′ (t) + x(t) = 0.
m
The √ general solution is x(t) = A sin ωt + B cos ωt, where
most
ω = k/m.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
11. Showing a function is a solution
Example (Continued)
Show that x(t) = A sin ωt + B cos ωt satisfies the differential equation
k √
x′′ + x = 0, where ω = k/m.
m
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 8 / 40
12. Showing a function is a solution
Example (Continued)
Show that x(t) = A sin ωt + B cos ωt satisfies the differential equation
k √
x′′ + x = 0, where ω = k/m.
m
Solution
We have
x(t) = A sin ωt + B cos ωt
x′ (t) = Aω cos ωt − Bω sin ωt
x′′ (t) = −Aω 2 sin ωt − Bω 2 cos ωt
Since ω 2 = k/m, the last line plus k/m times the first line result in zero.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 8 / 40
13. The Equation y′ = 2
Example
Find a solution to y′ (t) = 2.
Find the most general solution to y′ (t) = 2.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
14. The Equation y′ = 2
Example
Find a solution to y′ (t) = 2.
Find the most general solution to y′ (t) = 2.
Solution
A solution is y(t) = 2t.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
15. The Equation y′ = 2
Example
Find a solution to y′ (t) = 2.
Find the most general solution to y′ (t) = 2.
Solution
A solution is y(t) = 2t.
The general solution is y = 2t + C.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
16. The Equation y′ = 2
Example
Find a solution to y′ (t) = 2.
Find the most general solution to y′ (t) = 2.
Solution
A solution is y(t) = 2t.
The general solution is y = 2t + C.
Remark
If a function has a constant rate of growth, it’s linear.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
17. The Equation y′ = 2t
Example
Find a solution to y′ (t) = 2t.
Find the most general solution to y′ (t) = 2t.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 10 / 40
18. The Equation y′ = 2t
Example
Find a solution to y′ (t) = 2t.
Find the most general solution to y′ (t) = 2t.
Solution
A solution is y(t) = t2 .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 10 / 40
19. The Equation y′ = 2t
Example
Find a solution to y′ (t) = 2t.
Find the most general solution to y′ (t) = 2t.
Solution
A solution is y(t) = t2 .
The general solution is y = t2 + C.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 10 / 40
20. The Equation y′ = y
Example
Find a solution to y′ (t) = y(t).
Find the most general solution to y′ (t) = y(t).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 40
21. The Equation y′ = y
Example
Find a solution to y′ (t) = y(t).
Find the most general solution to y′ (t) = y(t).
Solution
A solution is y(t) = et .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 40
22. The Equation y′ = y
Example
Find a solution to y′ (t) = y(t).
Find the most general solution to y′ (t) = y(t).
Solution
A solution is y(t) = et .
The general solution is y = Cet , not y = et + C.
(check this)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 40
23. Kick it up a notch: y′ = 2y
Example
Find a solution to y′ = 2y.
Find the general solution to y′ = 2y.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 12 / 40
24. Kick it up a notch: y′ = 2y
Example
Find a solution to y′ = 2y.
Find the general solution to y′ = 2y.
Solution
y = e2t
y = Ce2t
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 12 / 40
25. In general: y′ = ky
Example
Find a solution to y′ = ky.
Find the general solution to y′ = ky.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 13 / 40
26. In general: y′ = ky
Example
Find a solution to y′ = ky.
Find the general solution to y′ = ky.
Solution
y = ekt
y = Cekt
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 13 / 40
27. In general: y′ = ky
Example
Find a solution to y′ = ky.
Find the general solution to y′ = ky.
Solution
y = ekt
y = Cekt
Remark
What is C? Plug in t = 0:
y(0) = Cek·0 = C · 1 = C,
so y(0) = y0 , the initial value of y. . . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 13 / 40
28. Constant Relative Growth =⇒ Exponential Growth
Theorem
A function with constant relative growth rate k is an exponential
function with parameter k. Explicitly, the solution to the equation
y′ (t) = ky(t) y(0) = y0
is
y(t) = y0 ekt
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 14 / 40
29. Exponential Growth is everywhere
Lots of situations have growth rates proportional to the current
value
This is the same as saying the relative growth rate is constant.
Examples: Natural population growth, compounded interest,
social networks
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 15 / 40
30. Outline
Recall
The differential equation y′ = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 16 / 40
31. Bacteria
Since you need bacteria to
make bacteria, the amount
of new bacteria at any
moment is proportional to
the total amount of
bacteria.
This means bacteria
populations grow
exponentially.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 17 / 40
32. Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At
the end of 3 hours there are 10,000 bacteria. At the end of 5 hours
there are 40,000. How many bacteria were present initially?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
33. Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At
the end of 3 hours there are 10,000 bacteria. At the end of 5 hours
there are 40,000. How many bacteria were present initially?
Solution
Since y′ = ky for bacteria, we have y = y0 ekt . We have
10, 000 = y0 ek·3 40, 000 = y0 ek·5
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
34. Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At
the end of 3 hours there are 10,000 bacteria. At the end of 5 hours
there are 40,000. How many bacteria were present initially?
Solution
Since y′ = ky for bacteria, we have y = y0 ekt . We have
10, 000 = y0 ek·3 40, 000 = y0 ek·5
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
35. Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At
the end of 3 hours there are 10,000 bacteria. At the end of 5 hours
there are 40,000. How many bacteria were present initially?
Solution
Since y′ = ky for bacteria, we have y = y0 ekt . We have
10, 000 = y0 ek·3 40, 000 = y0 ek·5
Dividing the first into the second gives
4 = e2k =⇒ 2k = ln 4 =⇒ k = ln 2. Now we have
10, 000 = y0 eln 2·3 = y0 · 8
10, 000
So y0 = = 1250.
8 . . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
36. Could you do that again please?
We have
10, 000 = y0 ek·3
40, 000 = y0 ek·5
Dividing the first into the second gives
40, 000 y e5k
= 0 3k
10, 000 y0 e
=⇒ 4 = e2k
=⇒ ln 4 = ln(e2k ) = 2k
ln 4 ln 22 2 ln 2
=⇒ k = = = = ln 2
2 2 2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 19 / 40
37. Outline
Recall
The differential equation y′ = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 20 / 40
38. Modeling radioactive decay
Radioactive decay occurs because many large atoms spontaneously
give off particles.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 21 / 40
39. Modeling radioactive decay
Radioactive decay occurs because many large atoms spontaneously
give off particles.
This means that in a sample of
a bunch of atoms, we can
assume a certain percentage of
them will “go off” at any point.
(For instance, if all atom of a
certain radioactive element
have a 20% chance of decaying
at any point, then we can
expect in a sample of 100 that
20 of them will be decaying.)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 21 / 40
40. Radioactive decay as a differential equation
The relative rate of decay is constant:
y′
=k
y
where k is negative.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
41. Radioactive decay as a differential equation
The relative rate of decay is constant:
y′
=k
y
where k is negative. So
y′ = ky =⇒ y = y0 ekt
again!
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
42. Radioactive decay as a differential equation
The relative rate of decay is constant:
y′
=k
y
where k is negative. So
y′ = ky =⇒ y = y0 ekt
again!
It’s customary to express the relative rate of decay in the units of
half-life: the amount of time it takes a pure sample to decay to one
which is only half pure.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
43. Computing the amount remaining of a decaying
sample
Example
The half-life of polonium-210 is about 138 days. How much of a 100 g
sample remains after t years?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 23 / 40
44. Computing the amount remaining of a decaying
sample
Example
The half-life of polonium-210 is about 138 days. How much of a 100 g
sample remains after t years?
Solution
We have y = y0 ekt , where y0 = y(0) = 100 grams. Then
365 · ln 2
50 = 100ek·138/365 =⇒ k = − .
138
Therefore
= 100 · 2−365t/138
365·ln 2
y(t) = 100e− 138
t
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 23 / 40
45. Computing the amount remaining of a decaying
sample
Example
The half-life of polonium-210 is about 138 days. How much of a 100 g
sample remains after t years?
Solution
We have y = y0 ekt , where y0 = y(0) = 100 grams. Then
365 · ln 2
50 = 100ek·138/365 =⇒ k = − .
138
Therefore
= 100 · 2−365t/138
365·ln 2
y(t) = 100e− 138
t
Notice y(t) = y0 · 2−t/t1/2 , where t1/2 is the half-life.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 23 / 40
46. Carbon-14 Dating
The ratio of carbon-14 to
carbon-12 in an organism
decays exponentially:
p(t) = p0 e−kt .
The half-life of carbon-14 is
about 5700 years. So the
equation for p(t) is
ln2
p(t) = p0 e− 5700 t
Another way to write this would
be
p(t) = p0 2−t/5700
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 24 / 40
47. Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
48. Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
Solution
p(t)
We are looking for the value of t for which = 0.1
p0
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
49. Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
Solution
p(t)
We are looking for the value of t for which = 0.1 From the
p0
equation we have
2−t/5700 = 0.1
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
50. Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
Solution
p(t)
We are looking for the value of t for which = 0.1 From the
p0
equation we have
t
2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1
5700
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
51. Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
Solution
p(t)
We are looking for the value of t for which = 0.1 From the
p0
equation we have
t ln 0.1
2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700
5700 ln 2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
52. Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
Solution
p(t)
We are looking for the value of t for which = 0.1 From the
p0
equation we have
t ln 0.1
2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700 ≈ 18, 940
5700 ln 2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
53. Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
Solution
p(t)
We are looking for the value of t for which = 0.1 From the
p0
equation we have
t ln 0.1
2−t/5700 = 0.1 =⇒ − ln 2 = ln 0.1 =⇒ t = · 5700 ≈ 18, 940
5700 ln 2
So the fossil is almost 19,000 years old.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
54. Outline
Recall
The differential equation y′ = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 26 / 40
55. Newton's Law of Cooling
Newton’s Law of Cooling
states that the rate of
cooling of an object is
proportional to the
temperature difference
between the object and its
surroundings.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 27 / 40
56. Newton's Law of Cooling
Newton’s Law of Cooling
states that the rate of
cooling of an object is
proportional to the
temperature difference
between the object and its
surroundings.
This gives us a differential
equation of the form
dT
= k(T − Ts )
dt
(where k < 0 again).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 27 / 40
57. General Solution to NLC problems
To solve this, change the variable y(t) = T(t) − Ts . Then y′ = T′ and
k(T − Ts ) = ky. The equation now looks like
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
58. General Solution to NLC problems
To solve this, change the variable y(t) = T(t) − Ts . Then y′ = T′ and
k(T − Ts ) = ky. The equation now looks like
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt
Now we can solve!
y′ = ky
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
59. General Solution to NLC problems
To solve this, change the variable y(t) = T(t) − Ts . Then y′ = T′ and
k(T − Ts ) = ky. The equation now looks like
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt
Now we can solve!
y′ = ky =⇒ y = Cekt
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
60. General Solution to NLC problems
To solve this, change the variable y(t) = T(t) − Ts . Then y′ = T′ and
k(T − Ts ) = ky. The equation now looks like
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt
Now we can solve!
y′ = ky =⇒ y = Cekt =⇒ T − Ts = Cekt
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
61. General Solution to NLC problems
To solve this, change the variable y(t) = T(t) − Ts . Then y′ = T′ and
k(T − Ts ) = ky. The equation now looks like
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt
Now we can solve!
y′ = ky =⇒ y = Cekt =⇒ T − Ts = Cekt =⇒ T = Cekt + Ts
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
62. General Solution to NLC problems
To solve this, change the variable y(t) = T(t) − Ts . Then y′ = T′ and
k(T − Ts ) = ky. The equation now looks like
dT dy
= k(T − Ts ) ⇐⇒ = ky
dt dt
Now we can solve!
y′ = ky =⇒ y = Cekt =⇒ T − Ts = Cekt =⇒ T = Cekt + Ts
Plugging in t = 0, we see C = y0 = T0 − Ts . So
Theorem
The solution to the equation T′ (t) = k(T(t) − Ts ), T(0) = T0 is
T(t) = (T0 − Ts )ekt + Ts
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
63. Computing cooling time with NLC
Example
A hard-boiled egg at 98 ◦ C is put in a sink of 18 ◦ C water. After 5
minutes, the egg’s temperature is 38 ◦ C. Assuming the water has not
warmed appreciably, how much longer will it take the egg to reach
20 ◦ C?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 29 / 40
64. Computing cooling time with NLC
Example
A hard-boiled egg at 98 ◦ C is put in a sink of 18 ◦ C water. After 5
minutes, the egg’s temperature is 38 ◦ C. Assuming the water has not
warmed appreciably, how much longer will it take the egg to reach
20 ◦ C?
Solution
We know that the temperature function takes the form
T(t) = (T0 − Ts )ekt + Ts = 80ekt + 18
To find k, plug in t = 5:
38 = T(5) = 80e5k + 18
and solve for k.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 29 / 40
65. Finding k
Solution (Continued)
38 = T(5) = 80e5k + 18
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
66. Finding k
Solution (Continued)
38 = T(5) = 80e5k + 18
20 = 80e5k
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
70. Finding k
Solution (Continued)
38 = T(5) = 80e5k + 18
20 = 80e5k
1
= e5k
( )4
1
ln = 5k
4
1
=⇒ k = − ln 4.
5
Now we need to solve for t:
t
20 = T(t) = 80e− 5 ln 4 + 18
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
71. Finding t
Solution (Continued)
t
20 = 80e− 5 ln 4 + 18
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
72. Finding t
Solution (Continued)
t
20 = 80e− 5 ln 4 + 18
t
2 = 80e− 5 ln 4
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
73. Finding t
Solution (Continued)
t
20 = 80e− 5 ln 4 + 18
t
2 = 80e− 5 ln 4
1 t
= e− 5 ln 4
40
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
74. Finding t
Solution (Continued)
t
20 = 80e− 5 ln 4 + 18
t
2 = 80e− 5 ln 4
1 t
= e− 5 ln 4
40
t
− ln 40 = − ln 4
5
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
75. Finding t
Solution (Continued)
t
20 = 80e− 5 ln 4 + 18
t
2 = 80e− 5 ln 4
1 t
= e− 5 ln 4
40
t
− ln 40 = − ln 4
5
ln 40 5 ln 40
=⇒ t = 1
= ≈ 13 min
5 ln 4 ln 4
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
76. Computing time of death with NLC
Example
A murder victim is discovered at
midnight and the temperature of
the body is recorded as 31 ◦ C.
One hour later, the temperature
of the body is 29 ◦ C. Assume
that the surrounding air
temperature remains constant
at 21 ◦ C. Calculate the victim’s
time of death. (The “normal”
temperature of a living human
being is approximately 37 ◦ C.)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 32 / 40
77. Solution
Let time 0 be midnight. We know T0 = 31, Ts = 21, and
T(1) = 29. We want to know the t for which T(t) = 37.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 33 / 40
78. Solution
Let time 0 be midnight. We know T0 = 31, Ts = 21, and
T(1) = 29. We want to know the t for which T(t) = 37.
To find k:
29 = 10ek·1 + 21 =⇒ k = ln 0.8
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 33 / 40
79. Solution
Let time 0 be midnight. We know T0 = 31, Ts = 21, and
T(1) = 29. We want to know the t for which T(t) = 37.
To find k:
29 = 10ek·1 + 21 =⇒ k = ln 0.8
To find t:
37 = 10et·ln(0.8) + 21
1.6 = et·ln(0.8)
ln(1.6)
t= ≈ −2.10 hr
ln(0.8)
So the time of death was just before 10:00pm.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 33 / 40
80. Outline
Recall
The differential equation y′ = ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 34 / 40
81. Interest
If an account has an compound interest rate of r per year
compounded n times, then an initial deposit of A0 dollars becomes
( r )nt
A0 1 +
n
after t years.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 35 / 40
82. Interest
If an account has an compound interest rate of r per year
compounded n times, then an initial deposit of A0 dollars becomes
( r )nt
A0 1 +
n
after t years.
For different amounts of compounding, this will change. As
n → ∞, we get continously compounded interest
( r )nt
A(t) = lim A0 1 + = A0 ert .
n→∞ n
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 35 / 40
83. Interest
If an account has an compound interest rate of r per year
compounded n times, then an initial deposit of A0 dollars becomes
( r )nt
A0 1 +
n
after t years.
For different amounts of compounding, this will change. As
n → ∞, we get continously compounded interest
( r )nt
A(t) = lim A0 1 + = A0 ert .
n→∞ n
Thus dollars are like bacteria.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 35 / 40
84. Continuous vs. Discrete Compounding of interest
Example
Consider two bank accounts: one with 10% annual interested
compounded quarterly and one with annual interest rate r compunded
continuously. If they produce the same balance after every year, what
is r?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 36 / 40
85. Continuous vs. Discrete Compounding of interest
Example
Consider two bank accounts: one with 10% annual interested
compounded quarterly and one with annual interest rate r compunded
continuously. If they produce the same balance after every year, what
is r?
Solution
The balance for the 10% compounded quarterly account after t years is
A1 (t) = A0 (1.025)4t = P((1.025)4 )t
The balance for the interest rate r compounded continuously account
after t years is
A2 (t) = A0 ert
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 36 / 40
86. Solving
Solution (Continued)
A1 (t) = A0 ((1.025)4 )t
A2 (t) = A0 (er )t
For those to be the same, er = (1.025)4 , so
r = ln((1.025)4 ) = 4 ln 1.025 ≈ 0.0988
So 10% annual interest compounded quarterly is basically equivalent
to 9.88% compounded continuously.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 37 / 40
87. Computing doubling time with exponential growth
Example
How long does it take an initial deposit of $100, compounded
continuously, to double?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 38 / 40
88. Computing doubling time with exponential growth
Example
How long does it take an initial deposit of $100, compounded
continuously, to double?
Solution
We need t such that A(t) = 200. In other words
ln 2
200 = 100ert =⇒ 2 = ert =⇒ ln 2 = rt =⇒ t = .
r
For instance, if r = 6% = 0.06, we have
ln 2 0.69 69
t= ≈ = = 11.5 years.
0.06 0.06 6
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 38 / 40
89. I-banking interview tip of the day
ln 2
The fraction can also
r
be approximated as either
70 or 72 divided by the
percentage rate (as a
number between 0 and
100, not a fraction between
0 and 1.)
This is sometimes called
the rule of 70 or rule of 72.
72 has lots of factors so it’s
used more often.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 39 / 40
90. Summary
When something grows or decays at a constant relative rate, the
growth or decay is exponential.
Equations with unknowns in an exponent can be solved with
logarithms.
Your friend list is like culture of bacteria (no offense).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 40 / 40