model ptumbuhan dan preputan

8/11/2019 Model Ptumbuhan Dan Preputan

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Formula Eksponen

Pembolehubah X adalah eksponen. Oleh

itu, graf fungsi ini bukan garis lurus.Dalam satu barisan lurus, "kadar

perubahan" adalah sama di seluruh

graf.Dalam graf ini, "kadar perubahan"meningkat atau berkurangan merentasi

graf.


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Contoh:

Apabila> 0 dan b itu antara 0 dan 1,graf akan berkurangan (reput)

Untuk contoh ini, setiap masa x

meningkat sebanyak 1, y

berkurangan kepada separuh

daripada nilai sebelumnya.

Keadaan seperti itu dipanggil

Pereputan Eksponen.


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Model Pertumbuhan Populasi

Contoh: Populasi bakteria atau haiwan

Pembolehubah:t = masa (pembolehubah dimanipulasi)

P= bilangan dalam sesuatu populasi (pembolehubah

bergerakbalas)

Kadar pertumbuhan populasi:

k =pemalar perkadaran


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Pertumbuhan Populasi

Populasi = 0

P(t) > 0

Jadi, k > 0 Oleh itu P(t) > 0

Populasi akan sentiasa meningkat, maka

dP/dt akan bertambah banyak.

Kesimpulannya, kadar pertumbuhan

meningkat apabila populasi bertambah.


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and

For k > 0, kis the growth

constant or rate of growth

For k< 0, k is the decay

constant or rate of decline

Exponential

amount at time t amount initially

growth or decay constant

time ktoeAtA


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A0= kuantiti awal (pemalar arbitrari)t = masa

A (t) = kuantiti selepas masa t

k = pemalar (analogi kepada pereputan malar)ex = fungsi eksponen (e adalah asas logaritma asli)

A(t) = A0ekt(Pertumbuhan Eksponen)

A(t) = A0e-kt(Pereputan Eksponen)


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k > 0 (lengkungan pertumbuhan)

k < 0 (lengkungan pereputan)

Prinsip Pertumbuhan Semulajadi

Prinsip Pereputan Semulajadi


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Population Growth

kPdt

dP

Populasi bertumbuh pada kadar yang berkadaran dengan saiz

penduduk :

kdt

dP

P 1

atau

rategrowthrelativesizepopulation

rategrowthk


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Population Growth Formula

t=tahun, k = 0.02, Populasi semasa t(0)= P Populasi bertumbuh pada kadar relatif 2%

setahun.

P(t) = P e

eyty )(0

kt

0

00.02 t


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Example 1; Find the relative rate

Year Population (millions)

1900 1650

1910 1750

1920 1860

1930 2070

1940 2300

1950 2560

1960 3040

1970 3710

1980 4450

1990 5280

2000 6080


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kPdt

dP P (0) = 1650


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Pereputan Eksponen


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Radiocarbon Dating

All living things

contain Carbon 14.

When they die, theC-14 begins to

decay. We can

determine how

long somethinghas been dead by

the amount of C-

14 left.


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In 1940 a group of boys walking in the woods near the

village of Lascaux in France suddenly realized their dog

was missing. They soon found him in a hole too deep forhim to climb out. One of the boys was lowered into the

hole to rescue the dog and stumbled upon one of the

greatest archaeological discoveries of all time. What he

discovered was a cave whose walls were covered with

drawings of wild horses, cattle and a fierce-looking beast

resembling a modern bull. In addition, the cave contained

the charcoal remains of a small fire, andfrom these remains scientists were able to

determine that the cave was occupied 15,000

years ago.


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Charcoal left from the

logs contains Carbon-14


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So assuming the paintings were made at the time of the

fire in the cave, they are approximately 15,000 years old.

By chemical analysis it has been determined that

the amount of C-14 remaining in the samples of

the Lascaux charcoal was 15% of the amount

such trees would contain when living. The half-

life of C-14 is approximately 5600 years.

ktoeAtA

Divide both sides by Ao

000123776.0k

oA15.0

Take the ln of both sides

kteln15.0ln

Divide both sides by k327,15000123776.0

15.0ln

t


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By chemical analysis it has been determined that

the amount of C-14 remaining in the samples of

the Lascaux charcoal was 15% of the amount

such trees would contain when living. The half-

life of C-14 is approximately 5600 years.

The half-life of a radioactive element is the time it takes

for 1/2 the initial amount to decay.

kt

oeAtA

o

k

o AeAA 2

1

5600 5600

Solve this equation for k.

First divide both sides by A o

2

1lnln

5600 ke

Take the ln of both sidesDivide by 5600 to find k

000123776.05600

2

1ln

k


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ktoeAtA

Youll need to determine the k

value by using info that tells you

after a certain time it was a certain

amount and subbing these valuesin for t and A.

500800(1)

500 500

Now we know kwe are ready to answer questions

about the bacteria.

keln

5

8ln

47000363.0k

Notice thekis positive this time

since this is exponential growthinstead of decay.

A culture of bacteria obeys the law of uninhibited

growth. If 500 bacteria are present initially and there are

800 after 1 hour, how many will be present in the culture

after 5 hours? How long is it until there are 20,000bacteria?


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ktoeAtA Punch buttons in your

calculator using thek

value we already

found.

47000363.0k

500 (5)(5)

5243

ktoeAtA 50020,000

Solve for t

500 500

kteln40ln

hrs8.747000363.0

40lnt

A culture of bacteria obeys the law of uninhibited growth. If

500 bacteria are present initially and there are 800 after 1

hour, how many will be present in the culture after 5 hours?

How long is it until there are 20,000 bacteria?

model ptumbuhan dan preputan

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