Unit 7 practice test: Differential Equations

$kP(K - P) = 0$ at $P = 0$ and $P = K$; the graph returns to 0 at 800.

• 0: $P = 0$ is the other equilibrium: no animals, no growth. The carrying capacity is the positive one.
• Correct: 800: Right: the rate returns to 0 at $P = 800$, the positive equilibrium the population approaches.
• 400: 400 is where the rate is largest, half of $K$.
• 32: 32 animals per year is the largest rate, not a population.

From the zeros, $K = 800$. From the top point, $k \cdot 400 \cdot 400 = 32$, so $k$$= \frac{32}{160{,}000}$$= 0.0002$.

• $\frac{dP}{dt} = 32P(800 - P)$: 32 is the largest rate, not $k$. At $P = 400$ this formula gives over 5 million.
• $\frac{dP}{dt} = 0.0002P(P - 800)$: This rate is negative for $0 < P < 800$, but the graph is above the axis there.
• Correct: $\frac{dP}{dt} = 0.0002P(800 - P)$: Right: $K = 800$, and at $P = 400$: $k(400)(400) = 32$ gives $k = 0.0002$.
• $\frac{dP}{dt} = 0.0002P(400 - P)$: This rate is 0 at $P = 400$, but the graph is highest there and returns to 0 at 800.

The rate $kP(K - P)$ is symmetric about $\frac{K}{2} = 400$, so $P = 200$ and $P = 600$ give the same rate, 24 animals per year.

• The rate is larger at $P = 600$, since there are more animals: More animals means more births, but less room: $800 - 600 = 200$. The two effects balance exactly.
• The rate is larger at $P = 200$, since there is more room to grow: More room, but fewer animals to grow. The graph has the same height at 200 and 600.
• The graph alone does not allow a comparison: Read the heights of the graph, or use the model: both rates are 24.
• Correct: They are equal: 24 animals per year each: Right: the parabola is symmetric about $P = 400$; $0.0002(200)(600)$$= 0.0002(600)(200)$$= 24$.

$\frac{d^2P}{dt^2}$$= \frac{d}{dP}\left(kP(K - P)\right) \cdot \frac{dP}{dt}$$= k(K - 2P)\frac{dP}{dt}$. With $\frac{dP}{dt} > 0$, the sign is that of $K - 2P$: positive for $P < 400$, negative above. $P = 400$ is a point of inflection.

• Correct: Concave up while $P < 400$, concave down while $400 < P < 800$: Right: $\frac{dP}{dt}$ increases as $P$ grows toward 400 and decreases after, so $P(t)$ bends up and then down.
• Concave down while $P < 400$, concave up while $400 < P < 800$: Reversed: below 400 the rate is rising as $P$ rises, so the graph of $P$ gets steeper (concave up).
• Concave up throughout $0 < P < 800$: Above 400 the rate falls as $P$ rises, so the graph of $P$ flattens out: concave down.
• Neither: the graph of $P(t)$ is a straight line: A straight line has a constant rate. Here the rate changes with $P$, from 0 up to 32 and back to 0.

$y^{-2}\,dy = 2\,dx$, so $-\frac{1}{y} = 2x + C$. At $(0, \frac{1}{2})$: $-2 = C$. Then $\frac{1}{y} = 2 - 2x$.

• $y = \frac{1}{2 + 2x}$: $\int y^{-2}\,dy = -\frac{1}{y}$: the minus sign flips the $2x$.
• $y = -\frac{1}{2x} + \frac{1}{2}$: The constant was added after solving for $y$.
• $y = \frac{2}{1 - x}$: At $x = 0$ this is 2, not $\frac{1}{2}$.
• Correct: $y = \frac{1}{2 - 2x}$: $-\frac{1}{y} = 2x + C$; at $(0, \frac{1}{2})$, $C = -2$, so $y = \frac{1}{2 - 2x}$.

$y = \frac{1}{2 - 2x}$ is undefined at $x = 1$. The largest open interval containing $x = 0$ is $x < 1$.

• $x \ne 1$: A solution's domain is one interval; the piece with $x > 1$ is not connected to $(0, \frac{1}{2})$.
• $x > 1$: The initial point $x = 0$ must be in the domain.
• Correct: $x < 1$: The formula breaks at $x = 1$; the solution lives on the interval containing $x = 0$.
• $(-\infty, \infty)$: At $x = 1$ the formula divides by 0.

As $x \to 1^-$, $2 - 2x \to 0$ through positive values, so $y \to \infty$: a vertical asymptote ends the solution.

• Correct: $y$ increases without bound.: $2 - 2x \to 0^+$, so $\frac{1}{2 - 2x} \to \infty$.
• $y$ approaches $\frac{1}{2}$.: $\frac{1}{2}$ is the value at $x = 0$, not near $x = 1$.
• $y$ approaches 0.: The denominator shrinks to 0, so the fraction grows.
• $y$ decreases without bound.: For $x < 1$ the denominator is positive.