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D.E application

bergausstein

Active member
Jul 30, 2013
191
1. Radium decomposes at a rate proportional to the amount at any instant. In 100 years, 100mg of radium decomposes to 96mg. How many mg will be left after 200 years?

2. if a population of a town doubled in the past 25 years and the present population is 300,000 when will the town have a population of 800,000?


prob 1.

since 3mg of 100mg radium have decomposed over a period of 100 years this amount is 3% of the original amount.


$\frac{R_0-0.03R_0}{R_0}=\frac{R_0\,e^{k100}}{R_0}$

$\ln(1-0.03)=\ln(e^{k100})$

$\ln(1-0.03)=100k$

$k=\frac{\ln(1-0.03)}{100}$

when t=200

$R(200)=R_0\,e^{\ln(1-0.03)^2}$

$R(200)=100(0.9409)$

$R=94.09mg$ is this correct?

prob 2

$\frac{dP}{dt}=kP$

$P(t)=P_0\,e^{kt}$

when t=25; $P_0=2P_0$

$\frac{2P_0}{P_0}=\frac{P_0\,e^{25k}}{P_0}$

$2=e^{25k}$

$\ln2=25k$

$k=\frac{\ln2}{25}$


$\frac{dP}{dt}=kR$

$P(t)=P_0\,e^{kt}$

$P(t)=P_0\,e^{\frac{\ln2}{25}t}$

$800,000=300,000(2^{\frac{t}{25}}$

$\ln2.67=\frac{t}{25}\ln2$

$t= 35.42$years

is this correct?
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
1. Radium decomposes at a rate proportional to the amount at any instant. In 100 years, 100mg of radium decomposes to 96mg. How many mg will be left after 200 years?

2. if a population of a town doubled in the past 25 years and the present population is 300,000 when will the town have a population of 800,000?


prob 1.

since 3mg of 100mg radium have decomposed over a period of 100 years this amount is 3% of the original amount.


$\frac{R_0-0.03R_0}{R_0}=\frac{R_0\,e^{k100}}{R_0}$

Where did ".03" come from? The problem said that amount reduced from 100 to 96 mg. Where did the "3 mg" you refer to come from?
100-96= 4.

$\ln(1-0.03)=\ln(e^{k100})$

$\ln(1-0.03)=100k$

$k=\frac{\ln(1-0.03)}{100}$

when t=200

$R(200)=R_0\,e^{\ln(1-0.03)^2}$

$R(200)=100(0.9409)$

$R=94.09mg$ is this correct?

prob 2

$\frac{dP}{dt}=kP$

$P(t)=P_0\,e^{kt}$

when t=25; $P_0=2P_0$

$\frac{2P_0}{P_0}=\frac{P_0\,e^{25k}}{P_0}$

$2=e^{25k}$

$\ln2=25k$

$k=\frac{\ln2}{25}$


$\frac{dP}{dt}=kR$

$P(t)=P_0\,e^{kt}$

$P(t)=P_0\,e^{\frac{\ln2}{25}t}$

$800,000=300,000(2^{\frac{t}{25}}$

$\ln2.67=\frac{t}{25}\ln2$

$t= 35.42$years

is this correct?
When I do it, using 8/3= 2.66666666666666667, I get 35.38 years. Don't round off until you have to! But the data given: 800,000, 300,000, and 25 are at most to 2 significant figures so I would say answer years anyway.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
If we're going to actually solve an initial value problem, we could develop the general formula as follows:

\(\displaystyle \frac{dy}{dt}=ky\) where \(\displaystyle y(0)=y_0\)

Separate variables, integrate with respect to $t$, switch dummy variables for clarity, and use the given boundaries as limits:

\(\displaystyle \int_{y_0}^{y(t)}\frac{1}{u}\,du=k\int_0^t\,dv\)

Apply the FTOC:

\(\displaystyle \ln\left|\frac{y(t)}{y_0} \right|=kt\)

Now, if we are given another point on the solution $\left(a,y_a \right)$ then we have:

\(\displaystyle \ln\left|\frac{y_a}{y_0} \right|=ka\implies k=\frac{1}{a}\ln\left|\frac{y_a}{y_0} \right|\)

And so we find:

\(\displaystyle \ln\left|\frac{y(t)}{y_0} \right|=\frac{1}{a}\ln\left|\frac{y_a}{y_0} \right|t\)

Hence:

(1) \(\displaystyle y(t)=y_0\left(\frac{y_a}{y_0} \right)^{\frac{t}{a}}\)

(2) \(\displaystyle t=\frac{a\ln\left|\frac{y(t)}{y_0} \right|}{\ln\left|\frac{y_a}{y_0} \right|}\)

Now we have formulas to answer this type of problem.

Problem 1: We want to use (1).

We identify:

\(\displaystyle y_0=100\text{ mg},\,y_a=96\text{ mg},\,a=100\text{ yr},\,t=200\text{ yr}\)

Plugging this into (1), we obtain:

\(\displaystyle y(t)=100\left(\frac{96}{100} \right)^{\frac{200}{100}}\text{ mg}=100\left(\frac{24}{25} \right)^2\text{ mg}=\frac{2304}{25}\text{ mg}=92.16\text{ mg}\)

Your error was, as HallsofIvy pointed out, one of subtraction.

Problem 2: We want to use (2).

We identify:

\(\displaystyle a=25\text{ yr},\,y_0=150000\,y_a=300000,\,y(t)=800000\)

Plugging this into (2), we obtain:

\(\displaystyle t=\frac{25\ln\left|\frac{800000}{150000} \right|}{\ln\left|\frac{300000}{150000} \right|}\text{ yr}=\frac{25\ln\left(\frac{16}{3} \right)}{\ln\left(2 \right)}\text{ yr}\)

Now, we want to subtract 25 from this because the starting point is 25 years ago. So we find the population will be 800,000 approximately 35.3759374819711 years from now.