§7.4 Separable differential equations (continued)

Exercises (continued)

Summary

A radioactive sample has half-life $5730$ years, with mass satisfying $\frac{dM}{dt} = kM$. The decay constant equals

For the initial value problem $\frac{dy}{dt} = \frac{-2ty}{t^2+1}$ with $y(0) = 4$, separating variables gives

Torricelli's Law implies the water height $h(t)$ in a leaking tank satisfies $\frac{dh}{dt} = k\sqrt{h}$ with $k \lt 0$. The square root appears because

The initial value problem $\frac{dy}{dt} = -\frac{t}{y}$, $y(0) = 8$ has solution

For $\frac{dy}{dt} = -\frac{t}{y}$ with $y(0) = 8$, the solution cannot be continued past $t = 8$ because

Using the half-life $5730$ years, compute the time for a carbon sample to decay to one quarter of its original mass.

A carbon sample retains $30\%$ of its original mass. Using $k = -\frac{\ln 2}{5730}$, the age is $t = -\frac{5730\ln(0.3)}{\ln 2}$. Report the value in years, rounded to one decimal.

With $k = -2$ and initial height $h(0) = 100$, the solution is $h(t) = (10-t)^2$. Find the time when the tank empties.

A tank's height drops at $20$ inches per minute when filled to $100$ inches. Using $\frac{dh}{dt} = k\sqrt{h}$, find $k$.

For the Gompertz model $\frac{dP}{dt} = -P\ln\left(\frac{P}{3}\right)$, the equilibrium $P = 3$ is stable, since the rate is positive below $3$ and negative above it.

In the Gompertz model, both the solution with $P(0) = 1$, namely $P = 3e^{\ln(1/3)e^{-t}}$, and the one with $P(0) = 6$ tend to $3$ as $t \to \infty$.