Decay and Growth

takuto · #1 $Old$ November 4th, 2009, 04:15 PM

Im having problems with decay and growth.
The problems states:

Radioactive Decay: They half-life of phosphorous-32 is 14 Days, if there are 5 grams presently, when will there be 1 gram left?

Doubling your Money: How much time is required for a $1000 deposit to double at an interest rate of 2.5% compounded annually.

I have no idea what to do. Do I have to find an equation?
Please help. Any help would be greatly appreciated.

skeeter · #2 $Old$ November 4th, 2009, 04:37 PM

Quote:

Originally Posted by takuto $View Post$

Im having problems with decay and growth.
The problems states:

Radioactive Decay: They half-life of phosphorous-32 is 14 Days, if there are 5 grams presently, when will there be 1 gram left?

Doubling your Money: How much time is required for a $1000 deposit to double at an interest rate of 2.5% compounded annually.

I have no idea what to do. Do I have to find an equation?
Please help. Any help would be greatly appreciated.

you seem to state this on most every post you've made.

are you not familiar with the equations

$y = y_0 e^{kt}$

and

$A = P\left(1 + \frac{r}{n}\right)^{nt}$

???

TKHunny · #3 $Old$ November 4th, 2009, 04:46 PM

Quote:

I have no idea what to do.

That's ridiculous. NEVER say that again.

Quote:

Do I have to find an equation?

It's not like looking under rocks. Did you study the section? Have you been to class?

Quote:

Originally Posted by takuto $View Post$

Radioactive Decay: They half-life of phosphorous-32 is 14 Days, if there are 5 grams presently, when will there be 1 gram left?

Doubling your Money: How much time is required for a $1000 deposit to double at an interest rate of 2.5% compounded annually.

You need a model. Since we have a post entitled "Decay and Growth", and you seem to be providing exponential examples, might I sugggest:

$f(t) = A_{0}e^{k \cdot t}$

There are many ways to write it. This should look familiar. It MUST be in your book.

Quote:

Radioactive Decay: The half-life of phosphorous-32 is 14 Days, if there are 5 grams presently, when will there be 1 gram left?

"there are 5 grams presently"

$f(0) = A_{0}e^{k \cdot (0)} = A_{0} = 5$

Now the model looks like this:

$f(t) = 5 \cdot e^{k \cdot (t)}$

"The half-life of phosphorous-32 is 14 Days"

$f(14) = 5 \cdot e^{k \cdot (14)} = 2.5$

See how that 2.5 is 1/2 of 5? That's what a half-life is all about.

Solve for 'k'

$5 \cdot e^{k \cdot (14)} = 2.5$

$e^{k \cdot (14)} = 0.5$

$k \cdot (14) = ln(0.5) = -ln(2)$

$k = \frac{-ln(2)}{14}$

Now the model looks like this:

$f(t) = 5 \cdot e^{\left[\frac{-ln(2)}{14}\right] \cdot (t)}$

"when will there be 1 gram left?"

$1 = 5 \cdot e^{\left[\frac{-ln(2)}{14}\right] \cdot (t_{0})}$

Solve for $t_{0}$ .

As you can see, it is more algebra than anything else. Once you have the model, you are on your way.

Finish this one up and let's see the next one.