Modeling with First Order DE’s — Question 9

Question 9 – Solution

Strategy. Differentiate the geometric mass formula before applying the surface-loss law. A constant radius-loss rate need not imply exponential mass loss.

Step 1: Reduce the balance to radius. For $r>0$, the chain rule gives $$4\pi\rho r^2r'=-4\pi k r^2, \qquad r'=-k/\rho,\qquad r(0)=3\text{ mm}.$$ The observed radius loss is $1$ mm in $5$ min, hence $$\boxed{k/\rho=0.20\text{ mm/min}},\qquad \boxed{r(t)=3-0.20t\text{ mm}}.$$ If density is in mass/mm$^3$, $k$ has units mass/(mm$^2$ min). The observations identify only the ratio; a density value is needed to determine $k$ itself.

Step 2: Determine mass and extinction. The radius reaches zero at $\boxed{t_e=15\text{ min}}$. Since mass is proportional to $r^3$, $$\boxed{\frac{M(t)}{M(0)}=(1-t/15)^3},\qquad 0\le t\le 15.$$ The physical mass and radius remain zero afterward. Extending the linear radius formula past $15$ would produce negative radius and is not part of the droplet model. Differentiating the cubic mass formula recovers the original surface-loss balance while $r>0$.

Step 3: Compare fractional targets. Half mass occurs when $(1-t/15)^3=1/2$, giving $$\boxed{t_{1/2,M}=15(1-2^{-1/3})\approx 3.094\text{ min}}.$$ Half radius occurs at $t=7.5$ min, when only $(1/2)^3=1/8$ of the initial mass remains. Geometric scaling makes these distinct events.

Step 4: Test an exponential alternative. The initial relative mass slope from the cubic formula is $M'(0)/M(0)=-1/5$ min$^{-1}$. The matching exponential is therefore $$M_{\exp}(t)=M(0)e^{-t/5}.$$ At $10$ min the surface-loss model predicts $M/M(0)=1/27\approx 0.0370$, whereas the exponential predicts $e^{-2}\approx 0.1353$. The exponential never reaches zero at finite time; the geometric model empties at $15$ min. Agreement of initial value and slope does not imply agreement of the physical loss mechanism or its later predictions.