Modeling with First Order DE’s — Question 3

Question 3 – Solution

Strategy. Separate the unchanging baseline from the decaying component. Differences of equally spaced measurements remove the baseline.

Step 1: Write the component and observation models. The active mass satisfies $R'=-kR$, where $k>0$ has units day$^{-1}$. If $R(0)=R_0$, then $$R=R_0e^{-kt},\qquad M=B+R_0e^{-kt}.$$ The stable material contributes to the measurement but not to the loss rate.

Step 2: Identify the parameters. Let $r=e^{-k}$, where one day is the time step, so $0<r<1$. The measured successive differences satisfy $$-30=R_0(r-1),\qquad -15=R_0r(r-1).$$ Their ratio gives $r=1/2$. Therefore $$\boxed{R_0=60\text{ g},\quad B=40\text{ g},\quad k=\ln 2\text{ day}^{-1}},\qquad \boxed{M(t)=40+60\,2^{-t}}.$$ These values reproduce $100,70,55$ g. They are unique under the assumptions because the nonzero first difference fixes $r$, then $R_0$, then $B$. Also $R'=-kR$ verifies the rate law.

Step 3: Compute active-mass times. The active half-life is $\ln 2/k=\boxed{1\text{ day}}$. The condition $R=5$ gives $60\,2^{-t}=5$, hence $$\boxed{t=\frac{\ln 12}{\ln 2}\approx 3.585\text{ days}}.$$ The active mass is below $5$ g after that time. The total mass then is $45$ g, not $5$ g.

Step 4: Test the incorrect observation model. A single exponential through the first two total-mass measurements is $\widetilde M(t)=100(0.7)^t$. It predicts $\widetilde M(2)=49$ g, contradicting the measured $55$ g. It wrongly assigns the stable component the same disappearance law as the active material and predicts a zero limiting total mass instead of $40$ g.

The model is intended for $t\ge 0$. Its inference depends on exact data, constant $k$, and a truly constant baseline; noisy data would require an estimation method rather than exact identification from three values.