4.1 exponential functions - dr. travers page of...
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4.1 Exponential Functions
Linear v. Exponential
What is the defining characteristic of a linear function?
Constant rate of change.
The defining characteristic of an exponential function is thatthere is a constant ratio between consecutive y values.
x 1 2 3 4 5 6f (x) 4 8 12 16 20 24g(x) 4 5 6.25 7.8125 9.765625 12.20703125
Linear v. Exponential
What is the defining characteristic of a linear function?
Constant rate of change.
The defining characteristic of an exponential function is thatthere is a constant ratio between consecutive y values.
x 1 2 3 4 5 6f (x) 4 8 12 16 20 24g(x) 4 5 6.25 7.8125 9.765625 12.20703125
Linear v. Exponential
What is the defining characteristic of a linear function?
Constant rate of change.
The defining characteristic of an exponential function is thatthere is a constant ratio between consecutive y values.
x 1 2 3 4 5 6f (x) 4 8 12 16 20 24g(x) 4 5 6.25 7.8125 9.765625 12.20703125
Linear v. Exponential
What is the defining characteristic of a linear function?
Constant rate of change.
The defining characteristic of an exponential function is thatthere is a constant ratio between consecutive y values.
x 1 2 3 4 5 6f (x) 4 8 12 16 20 24g(x) 4 5 6.25 7.8125 9.765625 12.20703125
How Exponentials Work
Example
Suppose Sox tickets are $45 and increase at 4% per year. Howmuch are tickets after 1 year? 2 years? 3 years? t years?
1 year: 45(.04) + 45 = 45(1.04) = $46.802 years:
46.80 + 46.80(.04) = 46.80(1.04)= (45(1.04))(1.04)
= 45(1.04)2
= $48.67
How Exponentials Work
Example
Suppose Sox tickets are $45 and increase at 4% per year. Howmuch are tickets after 1 year? 2 years? 3 years? t years?
1 year:
45(.04) + 45 = 45(1.04) = $46.802 years:
46.80 + 46.80(.04) = 46.80(1.04)= (45(1.04))(1.04)
= 45(1.04)2
= $48.67
How Exponentials Work
Example
Suppose Sox tickets are $45 and increase at 4% per year. Howmuch are tickets after 1 year? 2 years? 3 years? t years?
1 year: 45(.04) + 45 = 45(1.04) = $46.80
2 years:
46.80 + 46.80(.04) = 46.80(1.04)= (45(1.04))(1.04)
= 45(1.04)2
= $48.67
How Exponentials Work
Example
Suppose Sox tickets are $45 and increase at 4% per year. Howmuch are tickets after 1 year? 2 years? 3 years? t years?
1 year: 45(.04) + 45 = 45(1.04) = $46.802 years:
46.80 + 46.80(.04) = 46.80(1.04)
= (45(1.04))(1.04)= 45(1.04)2
= $48.67
How Exponentials Work
Example
Suppose Sox tickets are $45 and increase at 4% per year. Howmuch are tickets after 1 year? 2 years? 3 years? t years?
1 year: 45(.04) + 45 = 45(1.04) = $46.802 years:
46.80 + 46.80(.04) = 46.80(1.04)= (45(1.04))(1.04)
= 45(1.04)2
= $48.67
How Exponentials Work
Example
Suppose Sox tickets are $45 and increase at 4% per year. Howmuch are tickets after 1 year? 2 years? 3 years? t years?
1 year: 45(.04) + 45 = 45(1.04) = $46.802 years:
46.80 + 46.80(.04) = 46.80(1.04)= (45(1.04))(1.04)
= 45(1.04)2
= $48.67
How Exponentials Work
Example
Suppose Sox tickets are $45 and increase at 4% per year. Howmuch are tickets after 1 year? 2 years? 3 years? t years?
1 year: 45(.04) + 45 = 45(1.04) = $46.802 years:
46.80 + 46.80(.04) = 46.80(1.04)= (45(1.04))(1.04)
= 45(1.04)2
= $48.67
How Exponentials Work
3 years:
48.67 + 48.67(.04) = 48.67(1.04)
(46.80(1.04))(1.04)= (45(1.04)(1.04)(1.04))
= 45(1.04)3
= $50.62
t years? P(t) = 45(1.04)t
How Exponentials Work
3 years:
48.67 + 48.67(.04) = 48.67(1.04)(46.80(1.04))(1.04)
= (45(1.04)(1.04)(1.04))= 45(1.04)3
= $50.62
t years? P(t) = 45(1.04)t
How Exponentials Work
3 years:
48.67 + 48.67(.04) = 48.67(1.04)(46.80(1.04))(1.04)
= (45(1.04)(1.04)(1.04))
= 45(1.04)3
= $50.62
t years? P(t) = 45(1.04)t
How Exponentials Work
3 years:
48.67 + 48.67(.04) = 48.67(1.04)(46.80(1.04))(1.04)
= (45(1.04)(1.04)(1.04))= 45(1.04)3
= $50.62
t years? P(t) = 45(1.04)t
How Exponentials Work
3 years:
48.67 + 48.67(.04) = 48.67(1.04)(46.80(1.04))(1.04)
= (45(1.04)(1.04)(1.04))= 45(1.04)3
= $50.62
t years? P(t) = 45(1.04)t
How Exponentials Work
3 years:
48.67 + 48.67(.04) = 48.67(1.04)(46.80(1.04))(1.04)
= (45(1.04)(1.04)(1.04))= 45(1.04)3
= $50.62
t years?
P(t) = 45(1.04)t
How Exponentials Work
3 years:
48.67 + 48.67(.04) = 48.67(1.04)(46.80(1.04))(1.04)
= (45(1.04)(1.04)(1.04))= 45(1.04)3
= $50.62
t years? P(t) = 45(1.04)t
What they look like ...
x
y
As x→ ∞, f (x)→ ∞As x→ −∞, f (x)→ 0y-intercept is (0, 1)
What they look like ...
x
y
As x→ ∞, f (x)→ ∞
As x→ −∞, f (x)→ 0y-intercept is (0, 1)
What they look like ...
x
y
As x→ ∞, f (x)→ ∞As x→ −∞, f (x)→ 0
y-intercept is (0, 1)
What they look like ...
x
y
As x→ ∞, f (x)→ ∞As x→ −∞, f (x)→ 0y-intercept is (0, 1)
Two Types of Exponential Functions
Annual growth
f (x) = a(1 + r)x or f (x) = abx
a is the initial valuer is the rater must be in decimal formb = 1 + rb > 0
Continuous growthf (x) = aekx
k is the rate in decimal formk > 0 for growth, k < 0 for decay
Two Types of Exponential Functions
Annual growthf (x) = a(1 + r)x or f (x) = abx
a is the initial valuer is the rater must be in decimal formb = 1 + rb > 0
Continuous growthf (x) = aekx
k is the rate in decimal formk > 0 for growth, k < 0 for decay
Two Types of Exponential Functions
Annual growthf (x) = a(1 + r)x or f (x) = abx
a is the initial valuer is the rate
r must be in decimal formb = 1 + rb > 0
Continuous growthf (x) = aekx
k is the rate in decimal formk > 0 for growth, k < 0 for decay
Two Types of Exponential Functions
Annual growthf (x) = a(1 + r)x or f (x) = abx
a is the initial valuer is the rater must be in decimal form
b = 1 + rb > 0
Continuous growthf (x) = aekx
k is the rate in decimal formk > 0 for growth, k < 0 for decay
Two Types of Exponential Functions
Annual growthf (x) = a(1 + r)x or f (x) = abx
a is the initial valuer is the rater must be in decimal formb = 1 + r
b > 0
Continuous growthf (x) = aekx
k is the rate in decimal formk > 0 for growth, k < 0 for decay
Two Types of Exponential Functions
Annual growthf (x) = a(1 + r)x or f (x) = abx
a is the initial valuer is the rater must be in decimal formb = 1 + rb > 0
Continuous growthf (x) = aekx
k is the rate in decimal formk > 0 for growth, k < 0 for decay
Two Types of Exponential Functions
Annual growthf (x) = a(1 + r)x or f (x) = abx
a is the initial valuer is the rater must be in decimal formb = 1 + rb > 0
Continuous growth
f (x) = aekx
k is the rate in decimal formk > 0 for growth, k < 0 for decay
Two Types of Exponential Functions
Annual growthf (x) = a(1 + r)x or f (x) = abx
a is the initial valuer is the rater must be in decimal formb = 1 + rb > 0
Continuous growthf (x) = aekx
k is the rate in decimal formk > 0 for growth, k < 0 for decay
Two Types of Exponential Functions
Annual growthf (x) = a(1 + r)x or f (x) = abx
a is the initial valuer is the rater must be in decimal formb = 1 + rb > 0
Continuous growthf (x) = aekx
k is the rate in decimal form
k > 0 for growth, k < 0 for decay
Two Types of Exponential Functions
Annual growthf (x) = a(1 + r)x or f (x) = abx
a is the initial valuer is the rater must be in decimal formb = 1 + rb > 0
Continuous growthf (x) = aekx
k is the rate in decimal formk > 0 for growth, k < 0 for decay
Writing Exponentials
Example
You invest $500 into a savings account at an interest rate of2.3% per year. Write an exponential function giving the amountin the account after t years.
Initial amount? $500Rate? .023
f (t) = 500(1.023)t
Writing Exponentials
Example
You invest $500 into a savings account at an interest rate of2.3% per year. Write an exponential function giving the amountin the account after t years.
Initial amount?
$500Rate? .023
f (t) = 500(1.023)t
Writing Exponentials
Example
You invest $500 into a savings account at an interest rate of2.3% per year. Write an exponential function giving the amountin the account after t years.
Initial amount? $500
Rate? .023
f (t) = 500(1.023)t
Writing Exponentials
Example
You invest $500 into a savings account at an interest rate of2.3% per year. Write an exponential function giving the amountin the account after t years.
Initial amount? $500Rate?
.023
f (t) = 500(1.023)t
Writing Exponentials
Example
You invest $500 into a savings account at an interest rate of2.3% per year. Write an exponential function giving the amountin the account after t years.
Initial amount? $500Rate? .023
f (t) = 500(1.023)t
Writing Exponentials
Example
You invest $500 into a savings account at an interest rate of2.3% per year. Write an exponential function giving the amountin the account after t years.
Initial amount? $500Rate? .023
f (t) = 500(1.023)t
Writing Exponentials
Example
Carbon-14 decays at a rate of 11.4% per 1000 years. Write aformula for the amount of carbon-14 left in an artifact after tthousand years if the artifact had 35 grams of carbon-14 whenit was made.
initial amount? 35Rate? -.114
f (t) = 35(.886)t
How much carbon-14 is left after 2500 years?
f (2.5) = 35(.886)2.5 = 25.86 g
Writing Exponentials
Example
Carbon-14 decays at a rate of 11.4% per 1000 years. Write aformula for the amount of carbon-14 left in an artifact after tthousand years if the artifact had 35 grams of carbon-14 whenit was made.
initial amount?
35Rate? -.114
f (t) = 35(.886)t
How much carbon-14 is left after 2500 years?
f (2.5) = 35(.886)2.5 = 25.86 g
Writing Exponentials
Example
Carbon-14 decays at a rate of 11.4% per 1000 years. Write aformula for the amount of carbon-14 left in an artifact after tthousand years if the artifact had 35 grams of carbon-14 whenit was made.
initial amount? 35
Rate? -.114
f (t) = 35(.886)t
How much carbon-14 is left after 2500 years?
f (2.5) = 35(.886)2.5 = 25.86 g
Writing Exponentials
Example
Carbon-14 decays at a rate of 11.4% per 1000 years. Write aformula for the amount of carbon-14 left in an artifact after tthousand years if the artifact had 35 grams of carbon-14 whenit was made.
initial amount? 35Rate?
-.114
f (t) = 35(.886)t
How much carbon-14 is left after 2500 years?
f (2.5) = 35(.886)2.5 = 25.86 g
Writing Exponentials
Example
Carbon-14 decays at a rate of 11.4% per 1000 years. Write aformula for the amount of carbon-14 left in an artifact after tthousand years if the artifact had 35 grams of carbon-14 whenit was made.
initial amount? 35Rate? -.114
f (t) = 35(.886)t
How much carbon-14 is left after 2500 years?
f (2.5) = 35(.886)2.5 = 25.86 g
Writing Exponentials
Example
Carbon-14 decays at a rate of 11.4% per 1000 years. Write aformula for the amount of carbon-14 left in an artifact after tthousand years if the artifact had 35 grams of carbon-14 whenit was made.
initial amount? 35Rate? -.114
f (t) = 35(.886)t
How much carbon-14 is left after 2500 years?
f (2.5) = 35(.886)2.5 = 25.86 g
Writing Exponentials
Example
Carbon-14 decays at a rate of 11.4% per 1000 years. Write aformula for the amount of carbon-14 left in an artifact after tthousand years if the artifact had 35 grams of carbon-14 whenit was made.
initial amount? 35Rate? -.114
f (t) = 35(.886)t
How much carbon-14 is left after 2500 years?
f (2.5) = 35(.886)2.5 = 25.86 g
Writing Exponentials
Example
Carbon-14 decays at a rate of 11.4% per 1000 years. Write aformula for the amount of carbon-14 left in an artifact after tthousand years if the artifact had 35 grams of carbon-14 whenit was made.
initial amount? 35Rate? -.114
f (t) = 35(.886)t
How much carbon-14 is left after 2500 years?
f (2.5) = 35(.886)2.5 = 25.86 g
Writing Exponentials
Example
Carbon-14 decays at a rate of 11.4% per 1000 years. Write aformula for the amount of carbon-14 left in an artifact after tthousand years if the artifact had 35 grams of carbon-14 whenit was made.
initial amount? 35Rate? -.114
f (t) = 35(.886)t
How much carbon-14 is left after 2500 years?
f (2.5) = 35(.886)2.5 = 25.86 g
Finding the Equation
Example
A census is taken every two years, beginning in 1990. In 1992,there were 321000 people in a certain city. In 1994, there were345000 people in that same city. If the population growsexponentially, find the formula for the population after t years.Express the population in thousands of people.
Since we know this is an exponential model, we need only twopoints to find the formula. What are those points?
(2, 321), (4, 345)
Since we have an exponential growth function, we are lookingat something of the form P(t) = abt.
Finding the Equation
Example
A census is taken every two years, beginning in 1990. In 1992,there were 321000 people in a certain city. In 1994, there were345000 people in that same city. If the population growsexponentially, find the formula for the population after t years.Express the population in thousands of people.
Since we know this is an exponential model, we need only twopoints to find the formula. What are those points?
(2, 321), (4, 345)
Since we have an exponential growth function, we are lookingat something of the form P(t) = abt.
Finding the Equation
Example
A census is taken every two years, beginning in 1990. In 1992,there were 321000 people in a certain city. In 1994, there were345000 people in that same city. If the population growsexponentially, find the formula for the population after t years.Express the population in thousands of people.
Since we know this is an exponential model, we need only twopoints to find the formula. What are those points?
(2, 321), (4, 345)
Since we have an exponential growth function, we are lookingat something of the form P(t) = abt.
Finding the Equation
Example
A census is taken every two years, beginning in 1990. In 1992,there were 321000 people in a certain city. In 1994, there were345000 people in that same city. If the population growsexponentially, find the formula for the population after t years.Express the population in thousands of people.
Since we know this is an exponential model, we need only twopoints to find the formula. What are those points?
(2, 321), (4, 345)
Since we have an exponential growth function, we are lookingat something of the form P(t) = abt.
Finding the Equation
Next, we use the points to set up equations.
321 = ab2
345 = ab4
345321
=b4
b2
345321
= b2
b = 1.0367
So, we know that P(t) = a(1.0367)t.
Finding the Equation
Next, we use the points to set up equations.
321 = ab2
345 = ab4
345321
=b4
b2
345321
= b2
b = 1.0367
So, we know that P(t) = a(1.0367)t.
Finding the Equation
Next, we use the points to set up equations.
321 = ab2
345 = ab4
345321
=b4
b2
345321
= b2
b = 1.0367
So, we know that P(t) = a(1.0367)t.
Finding the Equation
Next, we use the points to set up equations.
321 = ab2
345 = ab4
345321
=b4
b2
345321
= b2
b = 1.0367
So, we know that P(t) = a(1.0367)t.
Finding the Equation
Next, we use the points to set up equations.
321 = ab2
345 = ab4
345321
=b4
b2
345321
= b2
b = 1.0367
So, we know that P(t) = a(1.0367)t.
Finding the Equation
Next, we use the points to set up equations.
321 = ab2
345 = ab4
345321
=b4
b2
345321
= b2
b = 1.0367
So, we know that P(t) = a(1.0367)t.
Finding the Equation
Now what?
P(t) = a(1.0367)t
321 = a(1.0367)2
a =321
1.0747a = 298.688
So, the function that gives the population after t years is
P(t) = 298.688(1.0367)t
Finding the Equation
Now what?
P(t) = a(1.0367)t
321 = a(1.0367)2
a =321
1.0747a = 298.688
So, the function that gives the population after t years is
P(t) = 298.688(1.0367)t
Finding the Equation
Now what?
P(t) = a(1.0367)t
321 = a(1.0367)2
a =321
1.0747a = 298.688
So, the function that gives the population after t years is
P(t) = 298.688(1.0367)t
Finding the Equation
Now what?
P(t) = a(1.0367)t
321 = a(1.0367)2
a =321
1.0747
a = 298.688
So, the function that gives the population after t years is
P(t) = 298.688(1.0367)t
Finding the Equation
Now what?
P(t) = a(1.0367)t
321 = a(1.0367)2
a =321
1.0747a = 298.688
So, the function that gives the population after t years is
P(t) = 298.688(1.0367)t
Finding the Equation
Now what?
P(t) = a(1.0367)t
321 = a(1.0367)2
a =321
1.0747a = 298.688
So, the function that gives the population after t years is
P(t) = 298.688(1.0367)t
Finding the Equation
Example
Find the equation of the exponential function that passesthrough (1, 3) and (5, 27).
3 = ab1
27 = ab5
273
=ab5
ab1
9 = b4
b = 1.732 =√
3
So, P(t) = a(√
3)t.
Finding the Equation
Example
Find the equation of the exponential function that passesthrough (1, 3) and (5, 27).
3 = ab1
27 = ab5
273
=ab5
ab1
9 = b4
b = 1.732 =√
3
So, P(t) = a(√
3)t.
Finding the Equation
Example
Find the equation of the exponential function that passesthrough (1, 3) and (5, 27).
3 = ab1
27 = ab5
273
=ab5
ab1
9 = b4
b = 1.732 =√
3
So, P(t) = a(√
3)t.
Finding the Equation
Example
Find the equation of the exponential function that passesthrough (1, 3) and (5, 27).
3 = ab1
27 = ab5
273
=ab5
ab1
9 = b4
b = 1.732 =√
3
So, P(t) = a(√
3)t.
Finding the Equation
Example
Find the equation of the exponential function that passesthrough (1, 3) and (5, 27).
3 = ab1
27 = ab5
273
=ab5
ab1
9 = b4
b = 1.732 =√
3
So, P(t) = a(√
3)t.
Finding the Equation
Example
Find the equation of the exponential function that passesthrough (1, 3) and (5, 27).
3 = ab1
27 = ab5
273
=ab5
ab1
9 = b4
b = 1.732 =√
3
So, P(t) = a(√
3)t.
Finding the Equation
Now, we need to find a.
P(t) = a(√
3)t
3 = a(√
3)1
a =3√3
Therefore, P(t) = 3√3(√
3)t.
Can we simplify this?
P(t) = 3(√
3)t−1 or P(t) = (√
3)t+1
Finding the Equation
Now, we need to find a.
P(t) = a(√
3)t
3 = a(√
3)1
a =3√3
Therefore, P(t) = 3√3(√
3)t.
Can we simplify this?
P(t) = 3(√
3)t−1 or P(t) = (√
3)t+1
Finding the Equation
Now, we need to find a.
P(t) = a(√
3)t
3 = a(√
3)1
a =3√3
Therefore, P(t) = 3√3(√
3)t.
Can we simplify this?
P(t) = 3(√
3)t−1 or P(t) = (√
3)t+1
Finding the Equation
Now, we need to find a.
P(t) = a(√
3)t
3 = a(√
3)1
a =3√3
Therefore, P(t) = 3√3(√
3)t.
Can we simplify this?
P(t) = 3(√
3)t−1 or P(t) = (√
3)t+1
Finding the Equation
Now, we need to find a.
P(t) = a(√
3)t
3 = a(√
3)1
a =3√3
Therefore, P(t) = 3√3(√
3)t.
Can we simplify this?
P(t) = 3(√
3)t−1 or P(t) = (√
3)t+1
Finding an Equation
Example
Find the equation of the exponential function that passesthrough (−4, 3) and (3, 21).
213
=ab3
ab−4 ⇒ 7 = b7 ⇒ b ≈ 1.32
21 = a(1.32)3 ⇒ a =21
(1.32)3 ≈ 9.131
So,f (x) = 9.131(1.32)x
Finding an Equation
Example
Find the equation of the exponential function that passesthrough (−4, 3) and (3, 21).
213
=ab3
ab−4 ⇒ 7 = b7 ⇒ b ≈ 1.32
21 = a(1.32)3 ⇒ a =21
(1.32)3 ≈ 9.131
So,f (x) = 9.131(1.32)x
Finding an Equation
Example
Find the equation of the exponential function that passesthrough (−4, 3) and (3, 21).
213
=ab3
ab−4 ⇒ 7 = b7 ⇒ b ≈ 1.32
21 = a(1.32)3 ⇒ a =21
(1.32)3 ≈ 9.131
So,f (x) = 9.131(1.32)x
Finding an Equation
Example
Find the equation of the exponential function that passesthrough (−4, 3) and (3, 21).
213
=ab3
ab−4 ⇒ 7 = b7 ⇒ b ≈ 1.32
21 = a(1.32)3 ⇒ a =21
(1.32)3 ≈ 9.131
So,f (x) = 9.131(1.32)x
Finding the Equation
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
5
4
3
2
1
-2
-1
Finding the Equation
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
5
4
3
2
1
-2
-1
Finding the Equation
The points we have are (1, 2) and (2.5, 4).
2 = ab1
4 = ab52
2 = b32
b = 223 ≈ 1.587
2 = a(1.587)1
a ≈ 1.26
So,
f (x) = 1.26(1.587)x
Finding the Equation
The points we have are (1, 2) and (2.5, 4).
2 = ab1
4 = ab52
2 = b32
b = 223 ≈ 1.587
2 = a(1.587)1
a ≈ 1.26
So,
f (x) = 1.26(1.587)x
Finding the Equation
The points we have are (1, 2) and (2.5, 4).
2 = ab1
4 = ab52
2 = b32
b = 223 ≈ 1.587
2 = a(1.587)1
a ≈ 1.26
So,
f (x) = 1.26(1.587)x
Finding the Equation
The points we have are (1, 2) and (2.5, 4).
2 = ab1
4 = ab52
2 = b32
b = 223 ≈ 1.587
2 = a(1.587)1
a ≈ 1.26
So,
f (x) = 1.26(1.587)x
Finding the Equation
The points we have are (1, 2) and (2.5, 4).
2 = ab1
4 = ab52
2 = b32
b = 223 ≈ 1.587
2 = a(1.587)1
a ≈ 1.26
So,
f (x) = 1.26(1.587)x
Finding the Equation
The points we have are (1, 2) and (2.5, 4).
2 = ab1
4 = ab52
2 = b32
b = 223 ≈ 1.587
2 = a(1.587)1
a ≈ 1.26
So,
f (x) = 1.26(1.587)x
Finding the Equation
The points we have are (1, 2) and (2.5, 4).
2 = ab1
4 = ab52
2 = b32
b = 223 ≈ 1.587
2 = a(1.587)1
a ≈ 1.26
So,
f (x) = 1.26(1.587)x
Interest
Example
Suppose you invest $1000 in an account that accumulates 5%interest annually. How much would you have in the account atthe end of the year? 2 years?
Simple Interest
A(t) = a (1 + r)t
whereA(t) is the balance at time ta is the principalr is the ratet is the time in years
Interest
Example
Suppose you invest $1000 in an account that accumulates 5%interest annually. How much would you have in the account atthe end of the year? 2 years?
Simple Interest
A(t) = a (1 + r)t
whereA(t) is the balance at time t
a is the principalr is the ratet is the time in years
Interest
Example
Suppose you invest $1000 in an account that accumulates 5%interest annually. How much would you have in the account atthe end of the year? 2 years?
Simple Interest
A(t) = a (1 + r)t
whereA(t) is the balance at time ta is the principal
r is the ratet is the time in years
Interest
Example
Suppose you invest $1000 in an account that accumulates 5%interest annually. How much would you have in the account atthe end of the year? 2 years?
Simple Interest
A(t) = a (1 + r)t
whereA(t) is the balance at time ta is the principalr is the rate
t is the time in years
Interest
Example
Suppose you invest $1000 in an account that accumulates 5%interest annually. How much would you have in the account atthe end of the year? 2 years?
Simple Interest
A(t) = a (1 + r)t
whereA(t) is the balance at time ta is the principalr is the ratet is the time in years
Interest
A(1) =
1000 (1 + .05)1
A(1) = $1050
A(2) = 1000 (1 + .05)2
A(2) = $1102.50
Interest
A(1) = 1000
(1 + .05)1
A(1) = $1050
A(2) = 1000 (1 + .05)2
A(2) = $1102.50
Interest
A(1) = 1000 (1 + .05)1
A(1) =
$1050
A(2) = 1000 (1 + .05)2
A(2) = $1102.50
Interest
A(1) = 1000 (1 + .05)1
A(1) = $1050
A(2) = 1000 (1 + .05)2
A(2) = $1102.50
Interest
A(1) = 1000 (1 + .05)1
A(1) = $1050
A(2) = 1000 (1 + .05)2
A(2) = $1102.50
Compound Interest
Example
What if we were to invest that same $1000 in to an account thataccumulates 5% interest, but instead it is compoundedquarterly. How much would be in the account after 1 year? 2years?
Compound Interest
A(t) = a(
1 +rk
)kt
wheret is measured in yearsk is the number of compounding periods per year
Compound Interest
Example
What if we were to invest that same $1000 in to an account thataccumulates 5% interest, but instead it is compoundedquarterly. How much would be in the account after 1 year? 2years?
Compound Interest
A(t) = a(
1 +rk
)kt
wheret is measured in yearsk is the number of compounding periods per year
Compound Interest
So we would have
A(1) = 1000(
1 +.054
)4(1)
A(1) =
$1050.95
A(2) = 1000(
1 +.054
)4(2)
A(2) = $1104.49
Compound Interest
So we would have
A(1) = 1000(
1 +.054
)4(1)
A(1) = $1050.95
A(2) = 1000(
1 +.054
)4(2)
A(2) = $1104.49
Compound Interest
So we would have
A(1) = 1000(
1 +.054
)4(1)
A(1) = $1050.95
A(2) =
1000(
1 +.054
)4(2)
A(2) = $1104.49
Compound Interest
So we would have
A(1) = 1000(
1 +.054
)4(1)
A(1) = $1050.95
A(2) = 1000(
1 +.054
)4(2)
A(2) = $1104.49
Compound Interest
Example
What if we compounded monthly? How much after 2 years?
A(2) = 1000(
1 +.0512
)12(2)
A(2) = $1104.94
Compound Interest
Example
What if we compounded monthly? How much after 2 years?
A(2) =
1000(
1 +.0512
)12(2)
A(2) = $1104.94
Compound Interest
Example
What if we compounded monthly? How much after 2 years?
A(2) = 1000
(1 +
.0512
)12(2)
A(2) = $1104.94
Compound Interest
Example
What if we compounded monthly? How much after 2 years?
A(2) = 1000(
1 +
.0512
)12(2)
A(2) = $1104.94
Compound Interest
Example
What if we compounded monthly? How much after 2 years?
A(2) = 1000(
1 +.0512
)
12(2)
A(2) = $1104.94
Compound Interest
Example
What if we compounded monthly? How much after 2 years?
A(2) = 1000(
1 +.0512
)12(2)
A(2) =
$1104.94
Compound Interest
Example
What if we compounded monthly? How much after 2 years?
A(2) = 1000(
1 +.0512
)12(2)
A(2) = $1104.94
Compound Interest
Example
What if we could compound continuously?
Continuously Compounded Interest
A(t) = limx→∞
a(
1 +rx
)xt
Let’s simplify to an interest rate of 100% and an initial amountof 1.
Continuously Compounded Interest
A(t) = limx→∞
(1 +
1x
)xt
When we do, this is equal to the number e.
Compound Interest
Example
What if we could compound continuously?
Continuously Compounded Interest
A(t) = limx→∞
a(
1 +rx
)xt
Let’s simplify to an interest rate of 100% and an initial amountof 1.
Continuously Compounded Interest
A(t) = limx→∞
(1 +
1x
)xt
When we do, this is equal to the number e.
Compound Interest
Example
What if we could compound continuously?
Continuously Compounded Interest
A(t) = limx→∞
a(
1 +rx
)xt
Let’s simplify to an interest rate of 100% and an initial amountof 1.
Continuously Compounded Interest
A(t) = limx→∞
(1 +
1x
)xt
When we do, this is equal to the number e.
Compound Interest
Example
What if we could compound continuously?
Continuously Compounded Interest
A(t) = limx→∞
a(
1 +rx
)xt
Let’s simplify to an interest rate of 100% and an initial amountof 1.
Continuously Compounded Interest
A(t) = limx→∞
(1 +
1x
)xt
When we do, this is equal to the number e.
Compound Interest
Example
What if we could compound continuously?
Continuously Compounded Interest
A(t) = limx→∞
a(
1 +rx
)xt
Let’s simplify to an interest rate of 100% and an initial amountof 1.
Continuously Compounded Interest
A(t) = limx→∞
(1 +
1x
)xt
When we do, this is equal to the number e.
Continuous Growth
Example
Same example, compounded continuously for 2 years.
A(2) = 1000e(.05)2
A(2) = $1105.17
Continuous Growth
Example
Same example, compounded continuously for 2 years.
A(2) = 1000e(.05)2
A(2) =
$1105.17
Continuous Growth
Example
Same example, compounded continuously for 2 years.
A(2) = 1000e(.05)2
A(2) = $1105.17
Continuous Growth and Decay
Example
A radioactive element decays at a rate of 2.5% per year. Howmuch of the initial 50 grams will be left after 21 years?
P(t) = P0ekt
P(21) = 50e−.025(21)
P(21) = 29.58 grams
Continuous Growth and Decay
Example
A radioactive element decays at a rate of 2.5% per year. Howmuch of the initial 50 grams will be left after 21 years?
P(t) = P0ekt
P(21) = 50e−.025(21)
P(21) = 29.58 grams
Continuous Growth and Decay
Example
A radioactive element decays at a rate of 2.5% per year. Howmuch of the initial 50 grams will be left after 21 years?
P(t) = P0ekt
P(21) = 50e−.025(21)
P(21) = 29.58 grams
Continuous Growth and Decay
Example
A radioactive element decays at a rate of 2.5% per year. Howmuch of the initial 50 grams will be left after 21 years?
P(t) = P0ekt
P(21) = 50e−.025(21)
P(21) = 29.58 grams
Continuous Growth and Decay
Example
You take cold medicine at 100 grams per dose. The medicineleaves the blood stream at a rate of 22% per hour. How much isleft in your blood stream after 4 hours?
Continuous or annual decay?
Q(t) = 100ekt
Q(t) = 100e−.22(t)
Q(4) = 100e−.88
Q(4) = 41.48 grams
Continuous Growth and Decay
Example
You take cold medicine at 100 grams per dose. The medicineleaves the blood stream at a rate of 22% per hour. How much isleft in your blood stream after 4 hours?
Continuous or annual decay?
Q(t) = 100ekt
Q(t) = 100e−.22(t)
Q(4) = 100e−.88
Q(4) = 41.48 grams
Continuous Growth and Decay
Example
You take cold medicine at 100 grams per dose. The medicineleaves the blood stream at a rate of 22% per hour. How much isleft in your blood stream after 4 hours?
Continuous or annual decay?
Q(t) = 100ekt
Q(t) = 100e−.22(t)
Q(4) = 100e−.88
Q(4) = 41.48 grams
Continuous Growth and Decay
Example
You take cold medicine at 100 grams per dose. The medicineleaves the blood stream at a rate of 22% per hour. How much isleft in your blood stream after 4 hours?
Continuous or annual decay?
Q(t) = 100ekt
Q(t) =
100e−.22(t)
Q(4) = 100e−.88
Q(4) = 41.48 grams
Continuous Growth and Decay
Example
You take cold medicine at 100 grams per dose. The medicineleaves the blood stream at a rate of 22% per hour. How much isleft in your blood stream after 4 hours?
Continuous or annual decay?
Q(t) = 100ekt
Q(t) = 100e−.22(t)
Q(4) = 100e−.88
Q(4) = 41.48 grams
Continuous Growth and Decay
Example
You take cold medicine at 100 grams per dose. The medicineleaves the blood stream at a rate of 22% per hour. How much isleft in your blood stream after 4 hours?
Continuous or annual decay?
Q(t) = 100ekt
Q(t) = 100e−.22(t)
Q(4) =
100e−.88
Q(4) = 41.48 grams
Continuous Growth and Decay
Example
You take cold medicine at 100 grams per dose. The medicineleaves the blood stream at a rate of 22% per hour. How much isleft in your blood stream after 4 hours?
Continuous or annual decay?
Q(t) = 100ekt
Q(t) = 100e−.22(t)
Q(4) = 100e−.88
Q(4) = 41.48 grams
Continuous Growth and Decay
Example
You take cold medicine at 100 grams per dose. The medicineleaves the blood stream at a rate of 22% per hour. How much isleft in your blood stream after 4 hours?
Continuous or annual decay?
Q(t) = 100ekt
Q(t) = 100e−.22(t)
Q(4) = 100e−.88
Q(4) = 41.48 grams
Annual↔ Continuous
Example
Suppose you invest $1000 at a continuous rate of 7%. What isthe equivalent annual interest rate?
P(t) = 1000e.07t
P(1) = 1000e.07(1)
P(1) = 1072.51
1072.51 = 1000 (1 + r)1
1.07251 = (1 + r)r = .07251
Annual↔ Continuous
Example
Suppose you invest $1000 at a continuous rate of 7%. What isthe equivalent annual interest rate?
P(t) =
1000e.07t
P(1) = 1000e.07(1)
P(1) = 1072.51
1072.51 = 1000 (1 + r)1
1.07251 = (1 + r)r = .07251
Annual↔ Continuous
Example
Suppose you invest $1000 at a continuous rate of 7%. What isthe equivalent annual interest rate?
P(t) = 1000e.07t
P(1) = 1000e.07(1)
P(1) = 1072.51
1072.51 = 1000 (1 + r)1
1.07251 = (1 + r)r = .07251
Annual↔ Continuous
Example
Suppose you invest $1000 at a continuous rate of 7%. What isthe equivalent annual interest rate?
P(t) = 1000e.07t
P(1) = 1000e.07(1)
P(1) = 1072.51
1072.51 = 1000 (1 + r)1
1.07251 = (1 + r)r = .07251
Annual↔ Continuous
Example
Suppose you invest $1000 at a continuous rate of 7%. What isthe equivalent annual interest rate?
P(t) = 1000e.07t
P(1) = 1000e.07(1)
P(1) = 1072.51
1072.51 = 1000 (1 + r)1
1.07251 = (1 + r)r = .07251
Annual↔ Continuous
Example
Suppose you invest $1000 at a continuous rate of 7%. What isthe equivalent annual interest rate?
P(t) = 1000e.07t
P(1) = 1000e.07(1)
P(1) = 1072.51
1072.51 =
1000 (1 + r)1
1.07251 = (1 + r)r = .07251
Annual↔ Continuous
Example
Suppose you invest $1000 at a continuous rate of 7%. What isthe equivalent annual interest rate?
P(t) = 1000e.07t
P(1) = 1000e.07(1)
P(1) = 1072.51
1072.51 = 1000
(1 + r)1
1.07251 = (1 + r)r = .07251
Annual↔ Continuous
Example
Suppose you invest $1000 at a continuous rate of 7%. What isthe equivalent annual interest rate?
P(t) = 1000e.07t
P(1) = 1000e.07(1)
P(1) = 1072.51
1072.51 = 1000 (1 + r)1
1.07251 = (1 + r)r = .07251
Annual↔ Continuous
Example
Suppose you invest $1000 at a continuous rate of 7%. What isthe equivalent annual interest rate?
P(t) = 1000e.07t
P(1) = 1000e.07(1)
P(1) = 1072.51
1072.51 = 1000 (1 + r)1
1.07251 = (1 + r)
r = .07251
Annual↔ Continuous
Example
Suppose you invest $1000 at a continuous rate of 7%. What isthe equivalent annual interest rate?
P(t) = 1000e.07t
P(1) = 1000e.07(1)
P(1) = 1072.51
1072.51 = 1000 (1 + r)1
1.07251 = (1 + r)r = .07251