f(x)= ab^x is the basic function

How we model real life situations depends on what kind, or how much information we are given.

When you have:

Limited Information

A= A_0 (model)^t

Lots of Information

A= A_0 〖(m)〗^(t/p)

A = amount of the “substance” at the end of the time period.

A_0 = original amount of the “substance” at the beginning of the time period

Model = is our model for growth or decay of the “substance”, it is usually an exponential expression in base 10 or e although any base can be use.

t = is the amount of time that has passed.

m = is the multiplication factor

p = is the period; the amount of time required to multiply by “m” once

Okay, so having that information we can now solve the following problems.

The first problem has Limited Information.

The population of the earth was 5.3 billion in 1990. In 2000 it was 6.1 billion.

Model the population growth using an exponential function.

What was the population in 2008?

Before solving this problem, Mr. K showed us a really cool stuff in the internet. It is a world population clock. Here’s the link to the site World Population Growth .

Anyways here’s how you solve it:

There are two ways we solved this one with base 10 and the other with base e although any base can be use.

A= A_0 (model)^t

6.1=5.3(model)^10

6.1/5.3= 〖model〗^10

Base 10

log(6.1/5.3)〖= log〖(〖model)〗^10 〗 〗

(1/10) log(6.1/5.3)=log〖(model)〗

0.0061= log(model)

〖10〗^0.0061=model

P=5.3(〖10〗^(0.0061t) )

Base e

ln(6.1/5.3)=ln〖(model)^10 〗

(1/10) ln〖(6.1/5.3)=ln(model) 〗

0.0141= ln(model)

e^0.00141=model

P=5.3(e^(0.0141t) )

∴ 〖10〗^(0.0061)= e^(0.00141)=1.0142

e^x

*Solving this kind of problem with base e is preferable because the exponent in base e, when you rewrite it in percent, tells you the percent rate of growth or decay per time.

t=0 in 1990

t=18 in 2008

P=5.3(〖10〗^(0.0061t) )

P=5.3(〖10〗^((0.0061)(18) )

P=6.8261

P=5.3(e^(0.0141t) )

P=5.3(e^((0.0141)(18)) )

P=6.8261

Okay, so moving on to the next problem which we are given Lots of Information.

A colony of bacteria doubles every 6 days. If there were 3000 bacteria to begin with how many bacteria will there be in 15 days?

A= A_0 〖(m)〗^(t/p)

A=〖3000(2)〗^((15/6) )

A=16970.5628

A ≈16971

Okay so that it is for our morning class.

Now for our afternoon class, we watched a short movie clip about star trek and tribbles. Here’s the link for the clip mailto:http://youtube.com/watch?v=lZvmxmVVdk8.

Here is another example of exponential modeling given with Lots of Information.

The mass in (grams) of radioactive material in a sample is given by:

M(t)= 100e^(-0.0017t)

where t is measured in years.

Find the half-life of this radioactive substance.

Create a model using the half-life found in (a). How much of a 10 gram sample of the material will remain after 40 years?

M(t)= 100e^(-0.0017t)

50= 100e^(-0.0017t)

1/2= e^(-0.0017t)

ln〖(1/2)=ln〖e^(-0.0017t) 〗 〗

ln(1/2)= -0.0017t

(1/(-0.0017)) ln(1/2)=t

t=407.7336 years

The half-life of this radioactive substance is 407.7336 years approximately.

ALWAYS REMEMBER THAT A LOGARITHM IS AN EXPONENT!

A= A_0 〖(m)〗^(t/p)

A=10(1/2)^((t/407.7336) )

A=10(1/2)^((40/407.7336) )

A=9.3426 grams

OR

A= e^(-0.0017t)

A= e^((-0.0017)(40))

A=9.3426 grams

*Most of the class didn`t know the story about the life of ants and wasps. So Mr. K had to tell how they grow exponentially.*

So that’s it for Exponential Modeling. Let’s now go back to the Consumer Math stuff.

We’re given this problem to solve:

A $5000 investment earns interest at the annual rate of 8.4% compounded monthly.

What is the investment worth after one year?

What is it worth after 10 years?

How much interest is earned in 10 years?

We use this formula in compound interest.

A = The new Amount

P = Principal/original amount

R = The rate of which it will accumulate

N = Number of times per time amount given (rate will be divided by this value)

T = Amount of Time that it has been accumulating for (yearly, seconds, months, whatever it may be)

What is the investment worth after one year?

A=5000(1+0.084/12)^12

A=$5436.55

What is it worth after 10 years?

A=5000(1+0.084/12)^(12*10)

A=5000(1+0.084/12)^120

A=$11547.99

How much interest is earned in 10 years?

I_T=A-P

I_T= $11547.99-$5436.55

I_T=$6111.44

Okay so that’s pretty much all we did this day. I hope my scribe post helped you!

Ciao

NEXT SCRIBE is

CHRYCEL

the better version ( i think) of my post is

here....

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