Ch 6.4 Exponential Growth & Decay Calculus Graphical, Numerical, Algebraic by Finney Demana, Waits,...
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Transcript of Ch 6.4 Exponential Growth & Decay Calculus Graphical, Numerical, Algebraic by Finney Demana, Waits,...
![Page 1: Ch 6.4 Exponential Growth & Decay Calculus Graphical, Numerical, Algebraic by Finney Demana, Waits, Kennedy.](https://reader033.fdocuments.in/reader033/viewer/2022050804/5518b54f550346881f8b50e8/html5/thumbnails/1.jpg)
Ch 6.4 Exponential Growth & DecayCalculus Graphical, Numerical, Algebraic by Finney Demana, Waits, Kennedy
![Page 2: Ch 6.4 Exponential Growth & Decay Calculus Graphical, Numerical, Algebraic by Finney Demana, Waits, Kennedy.](https://reader033.fdocuments.in/reader033/viewer/2022050804/5518b54f550346881f8b50e8/html5/thumbnails/2.jpg)
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![Page 4: Ch 6.4 Exponential Growth & Decay Calculus Graphical, Numerical, Algebraic by Finney Demana, Waits, Kennedy.](https://reader033.fdocuments.in/reader033/viewer/2022050804/5518b54f550346881f8b50e8/html5/thumbnails/4.jpg)
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Solving by Separation of Variables
2 2dySolve for y if = x y and y = 3 when x = 0.
dx
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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
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v
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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
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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
![Page 13: Ch 6.4 Exponential Growth & Decay Calculus Graphical, Numerical, Algebraic by Finney Demana, Waits, Kennedy.](https://reader033.fdocuments.in/reader033/viewer/2022050804/5518b54f550346881f8b50e8/html5/thumbnails/13.jpg)
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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?
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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
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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.
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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
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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
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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
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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?
![Page 23: Ch 6.4 Exponential Growth & Decay Calculus Graphical, Numerical, Algebraic by Finney Demana, Waits, Kennedy.](https://reader033.fdocuments.in/reader033/viewer/2022050804/5518b54f550346881f8b50e8/html5/thumbnails/23.jpg)
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