Ch 6.4 Exponential Growth & Decay Calculus Graphical, Numerical, Algebraic by Finney Demana, Waits, Kennedy.

Ch 6.4 Exponential Growth & DecayCalculus Graphical, Numerical, Algebraic by Finney Demana, Waits, Kennedy

Solving by Separation of Variables

2 2dySolve for y if = x y and y = 3 when x = 0.

dx

Law of Exponential Change

dy = ky

dtThe differential equation that describes growth is , where k is the growth constant (if positive) or decay constant (if negative). We can solve this equation by separating the variables.

kt + C

C kt

dy = ky

dtdy

= k dt separate the variablesy

dy = k dt integrate both sides

y

ln |y| = k t + C

| y | = e exponentiate both sides

| y | = e e propert

kt C

y of exponents

y = A e Let A = e

v

Continuously Compounded Interest

k t

0

rA(t) = A 1 +

k

If A0 dollars are invested at a fixed annual rate r (as a decimal) and compounded k times a year, the amount of money present after t years is:

If the interest is compounded continuously (or every instant) then the amount of money present after t years is:

k tr t

0 0k

rA(t) = lim A 1 + = A e

k

Finding Half Life

k t0

k t0 0

k t

Find the half-life of a radioactive substance with decay equation

y = y e

1The half-life is the solution to the equation y e = y .

21

Solvea lg ebraically : e = 2

1 - k t = ln

21 1 ln 2

Half life of an element:

ln 2

t = - ln

t

= k

=

2 k

k

Modeling Growth

At the beginning of summer, the population of a hive of hornets is growing at a rate proportional to the population. From a population of 10 on May 1, the number of hornets grows to 50 in thirty days. If the growth continues to follow the same model, how many days after May 1 will the population reach 100?

Modeling GrowthAt the beginning of summer, the population of a hive of hornets is growing at a rate proportional to the population. From a population of 10 on May 1, the number of hornets grows to 50 in thirty days. If the growth continues to follow the same model, how many days after May 1 will the population reach 100?

kt

kt

30kln5

t30

ln5t

0

30

1. y

y

50

ln

30

= y e

= 10 e

= 10 e

5

2.

y = 10 e

100 = 10 e

(0,10), (

=

30,5

0)

find t when y

=

10

0

k

10ln5

t30

30

ln

= 10 e

ln 10 t = = 42.920 days

5

Using Carbon-14 Dating

Scientists who use carbon-14 dating use 5700 years for its half-life. Find the age of a sample in which 10% of the radioactive nuclei originally present have decayed.

Using Carbon-14 Dating

Scientists who use carbon-14 dating use 5700 years for its half-life. Find the age of a sample in which 10% of the radioactive nuclei originally present have decayed.

t

5700

0

t

5700

0 0

t

5700

1A = A

2

1.9 A = A

2

1.9 =

2

t 1ln (.9) = ln

5700 2ln .9

t = 5700 = 866 yrs oldln .5

Newton’s Law of Cooling

Newton 's Law of Cooling: The rate at which an object's

temperature is changing at any time is proportional to the

difference between its temperature and the temperature of the

surrounding medium. If

s

s

s

s

k t + Cs

k ts

T = temperature at time t, and T is

the surrounding temperature, then:

dT = -k T - T

dtdT

= - k dtT - T

ln T - T = -k t + C

T - T = e

T - T = C e

k ts 0 s

k t0 s s

0 s

T - T = T - T e

at t = 0, C = T - T

T = T -

T

e

T

Newton’s Law of Cooling

0

0

Newton's Law of Cooling:

For T = initial temperature of object

T = temperature of surrounding environment

k = constant of variation, found by using given

Then,

T = - T e + kts

s

T T s

Using Newton’s Law of Cooling

A hard boiled egg at 98ºC is put in a pan under running 18ºC water to cool. After 5 minutes, the egg’s temperature is found to be 38ºC. How much longer will it take the egg to reach 20ºC?

Using Newton’s Law of Cooling

A hard boiled egg at 98ºC is put in a pan under running 18ºC water to cool. After 5 minutes, the egg’s temperature is found to be 38ºC. How much longer will it take the egg to reach 20ºC?

s 0

- k t

- k t

- 5k

- 5k

(-.2 ln 4

T = 18 and T = 98, then

T - 18 = (98 - 18) e

T = 18 + 80 e

Find k by using T = 38 when t = 5 to get:

38 = 18 + 80 e

1e =

41

5 k = ln = - ln 44

1k = ln 4

5

so now, T = 18 + 80 e

) t

(-.2 ln 4) t Solve 20 = 18 + 80 e to get t = 13.3 min

13.3 - 5 = 8.3 seconds longer