FINAL EXAM SOLUTIONS

MATH 382

1. (a) The predator is represented by y and the prey by x. The prey grows exponentially (dy/dt = 2x) in the absence of the predator and is exterminated at a rate equal to the number of possible interactions of the two populations. The predator will die out exponentially (dy/dt = −y) in the absence of food (prey) and grows at a rate equal to 3 times the number of possible interactions.

(b) The equilibrium points are (0, 0), (1/3, 2).

(c) The orbits are given by the equations

\frac{xy^{2}}{e^{3x}e^{y} = C}

These are closed curves around the point (1/3, 2). They are \underline{not} ellipses.

2. (a) These are two competing populations which are trying to occupy the same niche. Population x has a slower natural growth rate than y, but has a larger stable population in the niche.

(b) The equilibrium points are (0, 0), (3, 0), (0, 2).

(c) If y = 0 initially, x goes to 3 (unless x = 0 initially, in which case x remains at 0). Iy x = 0 initially, y goes to 2 (unless y = 0 initially, in which case y remains at 0). In all other cases the population approaches (3, 0). This is proved by dividing the first quadrant into three regions and arguing that the population leaves the region where x′ > 0, y′ > 0, never leaves the region x′ > 0, y′ < 0 if it once enters this region. Also either the population either remains in region x′ < 0, y′ < 0 or enters region x′ > 0, y′ < 0. In all of these population must approach an equilibrium point and it cannot approach (0, 2). Hence it must approach (3, 0).

3. The logistic equation can be written in the form x′ = ax(b − x), where b = 3 × 10^{8}. If the population is small, the equation is approximately x′ = abx, with solution x = Ce^{abt}. If the population doubles in 1/2 hour, then 2 = e^{ab/2}, hence a = \frac{2 log(2)}{b} = \frac{2 log(2)}{3×10^{8}}. The logistic equation may be integrated to give

\frac{1}{ab} log(\frac{x}{b − x})|_{10^{8}}^{x} = 4

Thus

log \frac{x}{b − x} = 7 log 2,

and

x = \frac{128b}{129} = 2.9767 × 10^{8}.

4. a) The guerilla force is represented by y, since its operational losses are proportional to x and its combat losses are proportional to the product, xy (shooting at random, not concentrating fire at a target).

b) The orbits found by integrating are

\frac{x^{2}}{2a} − \frac{y}{c} + \frac{b}{c^{2}} log(cy + b) = C

Whether x or y reaches 0 first depends on initial conditions. If C = \frac{b}{c^{2}} log b there is a stalemate (both sides are eliminated).

5. Let dV/dt represent the rate of removal of snow. Let the snow be falling at rate rt, let the width of the sidewalk be W and let dx/dt be the forward rate of progress of the shoveler. Then dV/dt = W · rt · dx/dt = K (constant rate of snow removal). Integrating this equation gives

x_{2} − x_{1} = \frac{K}{rW} log(\frac{t_{2}}{t_{1}} )

Applying this equation to the two time intervals gives

2 = \frac{K}{rW} log(\frac{t + 2}{t}), 1 = \frac{K}{rW} log(\frac{t + 5}{t + 2})

where t is the number of hours before noon that the snow began to fall. We can eliminate \frac{K}{rW} from these two equations to get

(\frac{t + 2}{t}) = (\frac{t + 5}{t + 2})^{2}

The t^{3} term cancels in this equation and the resulting quadratic has .5292 ⋯ hr (= 31.75 ⋯ min) as solution. Thus the snowfall began at approximately at 11:28 am.

SOLUTIONS 3

6. The equation of motion of the baseball is

m\frac{dv}{dt} = mg − kv^{2}

where k = .00001. When the ball reaches its limiting velocity dv/dt = 0. Thus the limiting velocity is

v = \sqrt{\frac{mg}{k}} = \sqrt{\frac{5}{16} × 10^{5}} × \frac{3600}{5280} = 120.53 mph