bcb12e ppt 6 3

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Chapter 6

Integration

Section 3

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Learning Objectives for Section 6.3

Differential Equations, Growth, Decay

■

The student will be able to identify differential equations and their

slope fields.

■

The student will be able to solve continuous compound interest problems using differential equations.

■

The student will be able to solve problems involving exponential growth and decay using

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Barnett/Ziegler/Byleen Business Calculus 12e 3

Intro to Differential Equations

dy

dx

= 3

x

2

−

5

x

+ 3

′

p

(

x

) = 300

e

−0.05x

′′

y

= 3

x

2

−

5

x

+ 4

We previously studied equations like

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Differential Equations

and Slope Fields

Slope fields will be introduced through an example.

Let dy/dx = y + 1.

Remember the geometric interpretation of the derivative is the slope of the line at the given point. For the given function, the slope at the point (0, 2) would be 3. At the points (–2, 1) and (2, 3), the slopes would be 2 and 4, respectively.

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Differential Equations

and Slope Fields (continued)

continued

Point (–2,1) (0,2) (2,3)

Slope 2 3 4

The graph of the slope field for all points.

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Differential Equations

1 1 ) 1 ( ) 1 ( + = + − = = − y e C Ce e C dx

d _x _x _x

We claim that the solution of dy/dx = y + 1 is y = Cex – 1.

Let’s check to see if it works. Substitute in the original equation for y:

We have confirmed that y = Cex – 1 is a solution to the

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Differential Equations

y = Cex – 1 is the general solution to the differential equation

dy/dx = y + 1. A particular solution going through the point (0, 0) would be 0 = Ce0 – 1 or C = 1. This yields the equation

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Slope Fields in General

A slope field for a first-order differential equation is obtained by drawing tangent line segments determined by the equation at each point in a grid. In general, this is done by computers, but to obtain a few points by hand we can do the following:

1. Draw tangent lines for a solution curve of the differential equation that passes through a few points.

2. Sketch an approximate graph of the solution curve that passes through these points.

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Continuous Compound Interest

Revisited

To find the function A = A(t) that satisfies the above conditions we divide both sides of the above equation by A and integrate with respect to time t.

rA dt

dA

=

Let P be the initial amount of money deposited in an account. Let

A be the amount at any time t. Continuous compound interest means that the rate of growth of the money in the account at any time t is proportional to the amount present at that time. Since

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Continuous Compound Interest

Revisited - continued

and A(0) = P

rA

dt

dA

=

C

rt

A

C

rt

C

A

dt

r

dA

A

dt

r

dt

dA

A

+

=

+

=

+

=

∫

|

ln

|

ln

1

2 1

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Continuous Compound Interest

Revisited - continued

rt rt rt rt C C rt

Pe

De

A

De

e

A

C

rt

A

=

±

=

+

=

+

|

ln

If C is an arbitrary real number, D = eC is an arbitrary positive

number. (Remember that an exponential is always positive).

P = ±D is an arbitrary nonzero number, but we can verify that

P = 0 also works, so P is finally an arbitrary real number.

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Exponential Growth Law

In general, the rate of growth of money in the previous case may be extended to any quantity that grows proportionally to the amount present with respect to time.

If and

r

Q

Q(0) = Q₀, then Q = Q₀ ert.

dt

dQ

=

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Relative Growth Rate

The constant r in the exponential growth law is called the

relative growth rate. If the relative growth rate is r = 0.02, then the quantity Q is growing at a rate dQ/dt = 0.02Q (that is 2% of the quantity Q per unit of time t). Note the distinction between the relative growth rate r and the rate of growth dQ/dt

of the quantity Q. Relative growth rate is 0.02 and the rate of growth is 0.02Q. Once we know that the rate of growth of

something is proportional to the amount present, we know that it has exponential growth and we can use the exponential

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Example 1

China had a population of 1.32 billion in 2007

(t = 0). Let P represent the population (in billions)

t years after 2007, and assume a continuous growth rate of 0.6%. Find the estimated population for

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Example 1

China had a population of 1.32 billion in 2007

(t = 0). Let P represent the population (in billions)

t years after 2007, and assume a continuous growth rate of 0.6%. Find the estimated population for

China in the year 2025.

The exponential growth/decay law applies, so that

P = 1.32 e 0.006 t_.

Substituting 18 = 25 – 7 for t yields

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Example 2

A bone from an ancient tomb was discovered and was

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Example 2

A bone from an ancient tomb was discovered and was

found to have 5% of the original radioactive carbon present. Estimate the age of the bone.

Solution: The exponential growth/decay law that applies is

The solution is

Q(t) = Q₀ e-0.0001238 t_{= 0.05}_Q

0, so

0.05 = e-0.0001238 t

0

) 0 ( 0001238

.

0 Q Q Q

dt

dQ ₌

−

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Example 2

(continued)

0.05 = e – 0.0001238 t

ln 0.05 = ln (e – 0.0001238 t) = – 0.0001238 t

0001238

.

0

05

.

0

ln

−

=

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Comparison of Growth

and Decay Behaviors

The graphs and equations in the following table compare several growth models. These are divided into two groups: unlimited growth (first model), and limited growth or decay (other three).

Description Model Solution Graph Unlimited

Growth Rate of growth is propor-tional to amount present

dy/dt = ky y(0) = C k, t > 0

y = C e kt

Exponential Decay

Rate of decay is propor-tional to amount present

dy/dt = –ky y(0) = C k, t > 0

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Comparison of Growth and Decay

(continued)

Description Model Solution Graph Limited Growth

Rate of growth is proportional to the difference between the amount present and a fixed limit

dy/dt = k (M – y)

y(0) = 0

k, t > 0

y = M (1 – e –kt₎

Logistic Growth

Rate of growth is proportional to the amount present and to the difference between the amount present and a fixed amount.

dy/dt = ky (M – y) y(0) = M/(1 + C)

k, t > 0

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Uses of Exponential Growth Models

 Unlimited Growth

• Short term population growth

• Growth of money at continuous compound interest

• Price-supply curves

 Exponential Decay

• Depletion of natural resources

• Radioactive decay

• Light absorption in water

• Price-demand curves

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Uses of Exponential Growth Models

(continued)

 Limited Growth:

• Sales fads (for example, skateboards)

• Depreciation of equipment

• Company growth

• Learning

 _{Logistic Growth}

• Long-term population growth

• Epidemics

• Sales of new products

• Rumor spread