§7.5 Modeling with differential equations (continued)

Summary

Lottery payments of one million dollars per year flow into an account earning $5\%$, with $A$ in millions. The differential equation is

Over $25$ years the annuity account $A = 20(e^{0.05t}-1)$ reaches about $49.81$ million, while a lump sum of $10$ million at the same rate reaches $10e^{1.25} \approx 34.90$ million. Which option wins?

Brine with $20$ grams per gallon enters a $100$ gallon tank at $3$ gallons per minute, with equal outflow of the mixed solution. If $S(t)$ is the salt total, then

A baby gains weight at a rate proportional to the reciprocal of its weight $w$, so the differential equation is

A skydiver accelerates under gravity at $9.8$ m/s$^2$ while drag removes a rate proportional to velocity $v$. The differential equation is

For the skydiver model $\frac{dv}{dt} = 9.8-kv$ with $v(0) = 0$, the solution is

The lottery annuity account follows $A(t) = 20(e^{0.05t}-1)$ in millions. Compute $A(25)$, rounded to three decimals.

A skydiver's terminal velocity is $54$ meters per second. Using $\frac{9.8}{k} = 54$, find $k$, rounded to three decimals.

With $k \approx 0.181481$, the skydiver reaches half the terminal velocity when $e^{-kt} = 0.5$. Find that time in seconds, rounded to three decimals.

For the brine tank $\frac{dS}{dt} = 60-\frac{3}{100}S$, find the stable equilibrium amount of salt in grams.

The baby weighing $8$ lb at birth and $9$ lb at one month has model $w(t) = \sqrt{17t+64}$. Compute $w(12)$, rounded to two decimals.

Because the brine enters and leaves at the same $3$ gallons per minute, the tank's volume stays at $100$ gallons for all time.

The model $w(t) = \sqrt{17t+64}$ predicts only about $30$ pounds at age ten and $42$ pounds at age twenty, so it is unrealistic over a lifetime.