# HELP WITH QUESTION PLEASEE (1 Viewer)

#### cossine

##### Member
//Hopefully it might be start cannot be bothered doing it all

Suppose we could model the amount of the chemical by

k * e ^ (-alpha*t) where alpha > 0 and t is the time in weeks.

Therefore k is the initial amount of the chemical.

The percentage of the chemical is given by

k * e ^ (-alpha*t) /k = e^-( alpha *t)

So the percentage of the chemical left is e^(-alpha*t)

For suspect A there is time of 1 week.

e^-alpha*(x) = 16% (1)

e^-alpha*(x+1) = 15% (2)

Where 0<x<1

(1) * (2) = e^-alpha*(x) * e^-alpha*(x+1)

=> e^-alpha(2x+1) = 2.4%

#### CM_Tutor

##### Moderator
Moderator
Let $\bg_white C(t)$ be the concentration of the chemical at time $\bg_white t$ weeks after the discovery of the dumping. Let $\bg_white C_0$ be the concentration at the time of the dumping, which occurred at some time $\bg_white t = T < 0$. We know that the concentration of the chemical decays according to the equation

$\bg_white \frac{dC}{dt} = -kC$

for some constant $\bg_white k>0$ and the solution of this differential equation is

$\bg_white C=Ae^{-kt}$

for some constant $\bg_white A$. We know that the concentration has fallen from $\bg_white C = C_0$ at time $\bg_white t = T$ to $\bg_white C = 0.16C_0$ at time $\bg_white t = 0$ and to $\bg_white C = 0.15C_0$ at time $\bg_white t = 1$. Using the latter two pieces of information, we get

$\bg_white 0.16C_0 = Ae^0 \qquad \qquad \text{. . . . . . . . . (1)$

and

$\bg_white 0.15C_0 = Ae^{-k} \qquad \qquad \text{. . . . . . . . . (2)$

Dividing equation (1) by equation (2) yields

\bg_white \begin{align*} \frac{0.16C_0}{0.15C_0} &= \frac{A \times 1}{Ae^{-k}} \\ \frac{16}{15} &= e^k \\ k &= \ln\left(\frac{16}{15}\right) \end{align*}

So, we can find $\bg_white T$, where $\bg_white C=C_0$:

\bg_white \begin{align*} C_0 &= Ae^{-kT} \\ C_0 &= 0.16C_0e^{-kT} \qquad \qquad \text{as A = 0.16C_0 from equation (1)} \\ \frac{C_0}{0.16C_0} &= e^{-kT} \\ 0.16 &= e^{kT} \\ \ln{0.16} &= kT \\ \ln\left(\frac{4}{25}\right) &= T\ln\left(\frac{16}{15}\right) \\ T &= \frac{\ln{4} - \ln{25}}{\ln{16}-\ln{15}} \\ &= \frac{2\ln{2} - 2\ln{5}}{4\ln{2}-\ln{15}} \\ &= -28.395... \\ &\approx\ \text{28 weeks and 3 days prior to t = 0} \end{align*}

So, based on decay rates, the contamination / dumping occurred 28 weeks and 3 days prior to it being discovered.

Suspect A was imprisoned for 26 weeks ending 1 week prior to the discovery, and so was imprisoned 27 weeks prior to the discovery of the contamination, and so was free at the time the dumping occurred.

Suspect B was imprisoned for 26 weeks ending 13 weeks prior to the discovery, and so was imprisoned at 39 weeks prior to the discovery and was still imprisoned when the dumping occurred.

The contamination decay rates are consistent with Suspect A being free at the time the crime was committed but Suspect B being incarcerated at that time and so unable to commit the crime. Based on these results, Suspect B is innocent and Suspect A remains under suspicion and is potentially the offender.

#### CM_Tutor

##### Moderator
Moderator
The calculation quoted below is correct for determining the time when the concentration falls to $\bg_white C=0.05C_0$.

However, it has been pointed out below that the question actually asked for the concentration to fall to 0.05% of its initial value, and thus to $\bg_white C=0.0005C_0$.

The correct calculation and answer is given in a post further below.

My mistake is preserved as a reminder that everyone can and does mistakes, that you should check independently, and that it is helpful to others to point out when mistakes are made. Thank you to @notme123 for pointing out this mistake.

The time when the concentration falls to $\bg_white C=0.05C_0$ can be calculated easily:

\bg_white \begin{align*} 0.05C_0 &= Ae^{-kt} \\ 0.05C_0 &= 0.16C_0e^{-kt} \qquad \qquad \text{as A = 0.16C_0 from equation (1)} \\ \frac{0.05C_0}{0.16C_0} &= e^{-kt} \\ \frac{16}{5} &= e^{kt} \\ \ln{\left(\frac{16}{5}\right)} &= t\ln\left(\frac{16}{15}\right) \\ t &= \frac{4\ln{2} - \ln{5}}{4\ln{2}-\ln{15}} \end{align*}
This is the time taken for the ground to become safe for replanting crops from the time the dumping occurred. Compensation happens from the time of the crime, so the period of compensation is

\bg_white \begin{align*} \text{time being compensated} \qquad &= \qquad \frac{4\ln{2} - \ln{5}}{4\ln{2}-\ln{15}} + \left|\frac{2\ln{2} - 2\ln{5}}{4\ln{2}-\ln{15}}\right| \\ &= \qquad \frac{4\ln{2} - \ln{5} - 2\ln{2} + 2\ln{5}}{4\ln{2}-\ln{15}} \\ &= \qquad\frac{2\ln{2} + \ln{5}}{4\ln{2}-\ln{15}} \\ &= \qquad 46.41773...\ \text{weeks} \\ \\ \text{So, total compensation due} \qquad &= \qquad \text{ number of full or partial weeks affected} \times 425 \text{ dollars per affected week}\\ &= \qquad 425 \times 47 \\ &= \qquad 19975 \text{ dollars} \end{align*}

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#### notme123

##### Active Member
The time when the concentration falls to $\bg_white C=0.05C_0$ can be calculated easily:

\bg_white \begin{align*} 0.05C_0 &= Ae^{-kt} \\ 0.05C_0 &= 0.16C_0e^{-kt} \qquad \qquad \text{as A = 0.16C_0 from equation (1)} \\ \frac{0.05C_0}{0.16C_0} &= e^{-kt} \\ \frac{16}{5} &= e^{kt} \\ \ln{\left(\frac{16}{5}\right)} &= t\ln\left(\frac{16}{15}\right) \\ t &= \frac{4\ln{2} - \ln{5}}{4\ln{2}-\ln{15}} \end{align*}
This is the time taken for the ground to become safe for replanting crops from the time the dumping occurred. Compensation happens from the time of the crime, so the period of compensation is

\bg_white \begin{align*} \text{time being compensated} \qquad &= \qquad \frac{4\ln{2} - \ln{5}}{4\ln{2}-\ln{15}} + \left|\frac{2\ln{2} - 2\ln{5}}{4\ln{2}-\ln{15}}\right| \\ &= \qquad \frac{4\ln{2} - \ln{5} - 2\ln{2} + 2\ln{5}}{4\ln{2}-\ln{15}} \\ &= \qquad\frac{2\ln{2} + \ln{5}}{4\ln{2}-\ln{15}} \\ &= \qquad 46.41773...\ \text{weeks} \\ \\ \text{So, total compensation due} \qquad &= \qquad \text{ number of full or partial weeks affected} \times 425 \text{ dollars per affected week}\\ &= \qquad 425 \times 47 \\ &= \qquad 19975 \text{ dollars} \end{align*}
Isnt it 0.0005 C0, not 0.05 C0? Because 0.05% = 0.0005. If it was 0.05 C0 that would be 5% of the concentration.

#### CM_Tutor

##### Moderator
Moderator
Isnt it 0.0005 C0, not 0.05 C0? Because 0.05% = 0.0005. If it was 0.05 C0 that would be 5% of the concentration.
You are absolutely correct, I have made a mistake because the question did say 0.05%, which is a concentration of

$\bg_white \frac{0.05}{100} \times C_0 = 0.0005C_0$.​

I shall make a post correcting my answer. Thanks for pointing this out, @notme123.

#### CM_Tutor

##### Moderator
Moderator
The time when the concentration falls to 0.05% of its initial value, that is, to $\bg_white C=0.0005C_0$ can be calculated easily:
\bg_white \begin{align*} 0.0005C_0 &= Ae^{-kt} \\ 0.0005C_0 &= 0.16C_0e^{-kt} \qquad \qquad \text{as A = 0.16C_0 from equation (1)} \\ \frac{0.0005C_0}{0.16C_0} &= e^{-kt} \\ \frac{1600}{5} &= e^{kt} \\ \ln{320} &= t\ln\left(\frac{16}{15}\right) \qquad \qquad \text{as it was shown above that k=\ln\left(\frac{16}{15}\right)} \\ t &= \frac{\ln{320}}{\ln{16}-\ln{15}} \\ &= \frac{\ln{(64 \times 5)}}{\ln{16}-\ln{15}} \\ &= \frac{6\ln{2} + \ln{5}}{4\ln{2}-\ln{15}} \end{align*}
\bg_white \begin{align*} \text{time being compensated} \qquad &= \qquad \frac{6\ln{2} + \ln{5}}{4\ln{2}-\ln{15}} + \left|\frac{2\ln{2} - 2\ln{5}}{4\ln{2}-\ln{15}}\right| \\ &= \qquad \frac{6\ln{2} + \ln{5} - 2\ln{2} + 2\ln{5}}{4\ln{2}-\ln{15}} \\ &= \qquad\frac{4\ln{2} + 3\ln{5}}{4\ln{2}-\ln{15}} \\ &= \qquad 117.773...\ \text{weeks} \\ \\ \text{So, total compensation due} \qquad &= \qquad \text{number of full or partial weeks affected} \times 425 \text{ dollars per affected week}\\ &= \qquad 118 \times 425 \\ &= \qquad 50150 \text{ dollars} \end{align*}