Solving D.E Word Problem: Radium Decompose, Half Life 1600, 200 Years

[bergausstein]

1. radium decompose at the rate proportional to the amount itself. if the half life is 1600, find the percentage remaining after at the end of 200 years.

can you me go about solving this. thanks!

 

[MarkFL]

Let $R(t)$ be the mass of radium in a given sample at time $t$. How can we mathematically state how this mass changes with time in general? Just look at the first sentence and use that to try to model this change with an initial value problem.

 

[bergausstein]

$R(t)=R_oe^{kt}$ where $k<0$

now the half-life is 1600 so,

$1600=3200e^{kt}$

now how wil I solve for the rate of decline here? please help!

 

[MarkFL]

$R(t)=R_oe^{kt}$ where $k<0$

now the half-life is 1600 so,

$1600=3200e^{kt}$

now how wil I solve for the rate of decline here? please help!

You have the correct equation for the mass of a sample, but the half-life being 1600 years means that at time t=1600 then $R(t)=\dfrac{1}{2}R_0$. That is (I like to always use a positive constant):

\(\displaystyle R(1600)=R_0e^{-1600k}=\frac{1}{2}R_0\)

Divide through by \(\displaystyle R_0\) and the convert from exponential to logarithmic form to solve for $k$. Another way to look at it is:

\(\displaystyle R(t)=R_02^{-\frac{t}{1600}}\)

And then to find the percentage remaining after $t$ years, use:

\(\displaystyle \frac{100R(t)}{R_0}=100\cdot2^{-\frac{t}{1600}}\)

 

[bergausstein]

from here solving for k

$\displaystyle R(1600)=R_0e^{-1600k}=\frac{1}{2}R_0$

dividing both sides by $R_o$

$\displaystyle e^{-1600k}=\frac{1}{2}$

taking the $\ln$ of both sides

$-1600k=\ln\frac{1}{2}$

dividing both sides by -1600

$\displaystyle k=-\frac{-\ln2}{1600}$ or $k=0.0004332$

now

$\displaystyle R(t)=R_0e^{-(0.0004332)t}$

how can I find the initial $R_o$?

 

[MarkFL]

You don't need to know the initial amount, you need to know what percentage of the initial amount remains after 200 years.

 

[bergausstein]

You don't need to know the initial amount, you need to know what percentage of the initial amount remains after 200 years.

how can I do that if there's no given initial amount? can you tell how to go about it?

 

[Prove It]

from here solving for k

$\displaystyle R(1600)=R_0e^{-1600k}=\frac{1}{2}R_0$

dividing both sides by $R_o$

$\displaystyle e^{-1600k}=\frac{1}{2}$

taking the $\ln$ of both sides

$-1600k=\ln\frac{1}{2}$

dividing both sides by -1600

$\displaystyle k=-\frac{-\ln2}{1600}$ or $k=0.0004332$

now

$\displaystyle R(t)=R_0e^{-(0.0004332)t}$

how can I find the initial $R_o$?

Do you think it's a good idea to give a decimal approximation when the function simplifies greatly if kept exact?

[tex]\displaystyle \begin{align*} k &= \frac{1}{1600} \ln{ (2) } \end{align*}[/tex]

and so

[tex]\displaystyle \begin{align*} R(t) &= R_0 \exp { \left[ \frac{t}{1600} \ln{(2)} \right] } \\ &= R_0 \exp{ \left[ \ln{ \left( 2 ^{ \frac{t}{1600} } \right) } \right] } \\ &= R_0 \cdot 2^{ \frac{t}{1600} } \end{align*}[/tex]

and so after 200 years, what proportion of the initial amount do you have?

 

[MarkFL]

how can I do that if there's no given initial amount? can you tell how to go about it?

This is what I posted before:

...to find the percentage remaining after $t$ years, use:

\(\displaystyle \frac{100R(t)}{R_0}\)

 

[bergausstein]

$\displaystyle \frac{100R(200)}{R_0}=100\cdot2^{\frac{200}{1600}}$

do you mean like this? how can I find the percentage of remaining here?

 

[MarkFL]

$\displaystyle \frac{100R(200)}{R_0}=100\cdot2^{\frac{200}{1600}}$

do you mean like this? how can I find the percentage of remaining here?

Reduce the exponent, and then that is the exact percentage remaining, and then you can use a calculator to obtain a decimal approximation if you like.

 

[Prove It]

It appears there was something wrong with your initial model.

You know that your model involves exponential decay, so after each unit of time passes, there is a certain constant amount multiplying through. In other words

[tex]\displaystyle \begin{align*} R_1 &= C\,R_0 \\ R_2 &= C^2\,R_0 \\ R_3 &= C^3\,R_0 \\ \vdots \\ R_t &= C^t\,R_0 \end{align*}[/tex]

Since you know that when [tex]\displaystyle \begin{align*} t = 1600, R_{1600} = \frac{1}{2}R_0 \end{align*}[/tex], that means

[tex]\displaystyle \begin{align*} \frac{1}{2}R_0 &= C^{1600}\,R_0 \\ \frac{1}{2} &= C^{1600} \\ \ln{ \left( \frac{1}{2} \right) } &= \ln{ \left( C^{1600} \right) } \\ \ln{ \left( \frac{1}{2} \right) } &= 1600 \ln{(C)} \\ \frac{1}{1600} \ln{ \left( \frac{1}{2} \right) } &= \ln{(C)} \\ \ln{ \left[ \left( \frac{1}{2} \right) ^{\frac{1}{1600}} \right] } &= \ln{(C)} \\ \left( \frac{1}{2} \right) ^{ \frac{1}{1600} } &= C \\ 2^{ -\frac{1}{1600}} &= C \end{align*}[/tex]

and thus your model is [tex]\displaystyle \begin{align*} R(t) = \left( 2^{-\frac{1}{1600}} \right) ^t \, R_0 = 2^{-\frac{t}{1600}}\,R_0 \end{align*}[/tex].

Now if [tex]\displaystyle \begin{align*} t = 200 \end{align*}[/tex], that gives

[tex]\displaystyle \begin{align*} R(200) &= 2^{-\frac{200}{1600}}\,R_0 \\ &= 2^{-\frac{1}{8}}\,R_0 \\ &\approx 0.917 \, R_0 \end{align*}[/tex]

So what percentage of the original amount do you have?

 

[MarkFL]

It appears there was something wrong with your initial model.

Yeah, I missed the dropped negative sign in the exponent.