Let's work through each part step-by-step: --- ## **Question 4** Given: \[ \frac{dP}{dt} = 7P + 14P^2 \] ### **(4.1) Is this a logistic model? Justify your answer!** **Logistic model:** A standard logistic model has the form: \[ \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) \] where \( r \) is the growth rate and \( K \) is the carrying capacity. **Given model:** \[ \frac{dP}{dt} = 7P + 14P^2 = 7P(1 + 2P) \] - This is NOT a logistic model because: - The logistic model has a **negative quadratic term** (\(-P^2\)), which leads to a carrying capacity. - Here, the quadratic term is **positive** (\(+14P^2\)), which means the population will grow **faster** as \(P\) increases, and there is **no limiting value or carrying capacity**. - Thus, the population grows without bound (super-exponential growth). **Final answer:** > **No, this is not a logistic model. In a logistic model, the quadratic term is negative to represent a carrying capacity that limits growth, but here the quadratic term is positive, leading to unbounded growth.** --- ### **(4.2) Draw a phase line of the model** The phase line is based on the sign of \( \frac{dP}{dt} \): \[ \frac{dP}{dt} = 7P + 14P^2 = 7P(1 + 2P) \] - Equilibrium points: Set \( \frac{dP}{dt} = 0 \): - \( 7P = 0 \implies P = 0 \) - \( 1 + 2P = 0 \implies P = -\frac{1}{2} \) - Test intervals: - For \( P < -\frac{1}{2} \): both factors negative and positive → \( \frac{dP}{dt} > 0 \) - For \( -\frac{1}{2} < P < 0 \): first factor negative, second positive → \( \frac{dP}{dt} < 0 \) - For \( P > 0 \): both factors positive → \( \frac{dP}{dt} > 0 \) **Phase line:** ``` P: -∞ -----> -1/2 <----- 0 -----> ∞ (↑) (↓) (↑) ``` --- ### **(4.3) Draw a sketch of the solution curve \( P(t) \), if \( P(0) = 7 \).** - Since \( P(0) = 7 > 0 \): - From the phase line, for \( P > 0 \), \( \frac{dP}{dt} > 0 \) so \( P(t) \) increases. - The growth is faster than exponential (super-exponential, "blow-up" in finite time). - The curve will rise steeply and go to infinity in finite time. **Sketch:** - Start at \( P = 7 \) at \( t = 0 \). - Rapidly rises, curving upwards more and more steeply. --- ## **Question 5** Given the phase line for \( \frac{dx}{dt} = G(x) \): Arrows: \( x < -1 \): right \( -1 < x < 0.5 \): right \( 0.5 < x < 2 \): left \( x > 2 \): left Equilibrium points: at \( x = -1, 0.5, 2 \) ### **(5.1) Draw a rough graph of what the function \( G(x) \) could look like.** - \( G(x) = 0 \) at \( x = -1, 0.5, 2 \) - \( G(x) > 0 \) where arrows are right (i.e., \( x < 0.5 \)) - \( G(x) < 0 \) where arrows are left (i.e., \( x > 0.5 \)) **Rough graph:** ``` G(x) ↑ | | o o o | / \ / \ / \ |-------|---|--|---|--|---|----> x -1 0.5 2 ``` - A possible sketch: G(x) positive between \( -\infty \) and \( 0.5 \), negative between \( 0.5 \) and \( \infty \), touching zero at the equilibrium points. --- ### **(5.2) Draw a rough sketch of the solution curves \( x(t) \) for:** - \( x(0) = 1 \) - \( x(0) = 0 \) **For \( x(0) = 1 \):** - \( 0.5 < 1 < 2 \), so arrows point left, \( x \) decreases toward \( x = 0.5 \). **For \( x(0) = 0 \):** - \( -1 < 0 < 0.5 \), so arrows point right, \( x \) increases toward \( x = 0.5 \). **Sketch:** - Both solutions approach \( x = 0.5 \) as \( t \to \infty \). - \( x(t) \) increases from 0 to 0.5. - \( x(t) \) decreases from 1 to 0.5. --- ## **Summary Table** | Question | Step | Answer/Explanation | |----------|------|-------------------| | 4.1 | | No, not logistic; positive quadratic term. | | 4.2 | | Phase line: Equilibria at \( P = 0, -0.5 \), arrows →, ←, →. | | 4.3 | | \( P(t) \) for \( P(0)=7 \) grows rapidly, blows up in finite time. | | 5.1 | | \( G(x) \) positive below 0.5 (zero at -1, 0.5), negative above 0.5 (zero at 2) | | 5.2 | | Both \( x(0)=1 \) and \( x(0)=0 \) approach \( x=0.5 \) as \( t \to \infty \) | --- Do you want the sketches/graphs drawn out? Let me know!

Certainly! Let's proceed with each part step-by-step, providing detailed calculations, reasoning, and final answers. --- ## **Question 4** Given differential equation: \[ \frac{dP}{dt} = 7P + 14P^2 \] --- ### **(4.1) Is this a logistic model? Justify your answer** **Step 1:** Recall the standard logistic model: \[ \frac{dP}{dt} = r P \left( 1 - \frac{P}{K} \right) \] which expands to: \[ \frac{dP}{dt} = r P - \frac{r P^2}{K} \] with: - **Linear term:** \( r P \) - **Quadratic term:** \( - \frac{r}{K} P^2 \) (negative quadratic) **Step 2:** Compare with the given model: \[ \frac{dP}{dt} = 7 P + 14 P^2 \] - The quadratic term here is **positive** (\(+14 P^2\)). **Step 3:** **Implication of quadratic term sign:** - In logistic growth, the quadratic term is negative, which models a **limiting effect** (carrying capacity). - Here, the quadratic term is positive, implying **accelerated growth** with no limiting population—super-exponential growth. **Final conclusion:** \[ \boxed{ \text{No, this is not a logistic model because the quadratic term is positive, indicating unbounded growth.} } \] --- ### **(4.2) Draw a phase line for the model** **Step 1:** Find equilibrium points where \(\frac{dP}{dt} = 0\): \[ 7P + 14P^2 = 0 \] Factor out \(7P\): \[ 7P (1 + 2P) = 0 \] So, \[ P = 0 \quad \text{or} \quad 1 + 2P = 0 \implies P = -\frac{1}{2} \] **Step 2:** Sign analysis of \(\frac{dP}{dt}\): - For \(P < -\frac{1}{2}\): - \(P\) negative, \(1 + 2P\) negative \(\Rightarrow\) \(7P\) negative, \(1+2P\) negative - Product: negative \(\times\) negative = **positive**, so \(\frac{dP}{dt} > 0\). - For \(-\frac{1}{2} < P < 0\): - \(P\) negative, \(1 + 2P\) positive (since \(2P > -1\)) - Product: negative \(\times\) positive = **negative**, so \(\frac{dP}{dt} < 0\). - For \(P > 0\): - \(P\) positive, \(1 + 2P\) positive - Product: positive \(\times\) positive = **positive**, so \(\frac{dP}{dt} > 0\). **Step 3:** Sketch the phase line: ``` P: --------------|--------------|---------------- -∞ -1/2 0 +∞ Direction of \(\frac{dP}{dt}\): - For \(P < -1/2\): + (arrow to the right) - For \(-1/2 < P < 0\): - (arrow to the left) - For \(P > 0\): + (arrow to the right) Equilibria at \(P = -\frac{1}{2}\) and \(P= 0\). ``` **Optional:** Indicate stability: - Near \(P = -\frac{1}{2}\): - For \(P\) just less than \(-\frac{1}{2}\), \(\frac{dP}{dt} > 0\): solutions move away from \(-\frac{1}{2}\) to the left. - For \(P\) just greater than \(-\frac{1}{2}\), \(\frac{dP}{dt} < 0\): solutions move toward \(-\frac{1}{2}\). **Thus:** \(P = -\frac{1}{2}\) is **stable**. - Near \(P=0\): - For \(P\) just less than 0, \(\frac{dP}{dt} < 0\), solutions move toward 0. - For \(P\) just greater than 0, \(\frac{dP}{dt} > 0\), solutions move away from 0. **Thus:** \(P=0\) is **unstable**. --- ### **(4.3) Sketch the solution curve \( P(t) \) with \( P(0) = 7 \).** **Step 1:** Initial condition: \[ P(0) = 7 > 0 \] - From the phase line, for \(P > 0\), \(\frac{dP}{dt} > 0\), **population increases**. **Step 2:** Behavior: - Since \(\frac{dP}{dt} = 7 P + 14 P^2\), for large \(P\), the quadratic term dominates, and the growth accelerates rapidly. - The solution "blows-up" (becomes infinite) in **finite time** because the differential equation resembles a super-exponential growth. **Step 3:** Qualitative sketch: - The curve starts at \(P=7\) at \(t=0\). - As \(t\) increases, \(P(t)\) increases rapidly, becoming steeper. - Approaching some finite time \(T\), \(P(t) \to +\infty\). **Final note:** - The solution is **not bounded** and exhibits **finite-time blow-up**. --- ## **Summary for Question 4:** | Part | Explanation | Final Answer | |--------|--------------|--------------| | 4.1 | Comparing with logistic form; quadratic term positive | **No**, not logistic | | 4.2 | Sign analysis of \(\frac{dP}{dt}\); equilibria at \(P=-1/2, 0\) | **Phase line**: stable at \(-1/2\), unstable at 0 | | 4.3 | Behavior of \(P(t)\) for \(P(0)=7\) | Rapid, unbounded growth, blow-up in finite time | --- ## **Question 5** Given the phase line arrows: - \(x < -1\): right - \(-1 < x < 0.5\): right - \(0.5 < x < 2\): left - \(x > 2\): left Equilibria at \(x = -1, 0.5, 2\). --- ### **(5.1) Draw a rough graph of \( G(x) \).** **Step 1:** Zeroes at \(x = -1, 0.5, 2\). **Step 2:** Sign of \(G(x)\): - To the **left of \(-1\)** (\(x < -1\)): \(G(x) > 0\), solutions move right. - Between \(-1\) and \(0.5\): \(G(x) > 0\), solutions move right. - Between \(0.5\) and \(2\): \(G(x) < 0\), solutions move left. - To the **right of \(2\)**: \(G(x) < 0\), solutions move left. **Step 3:** Sketch: - \(G(x)\) crosses zero at the equilibrium points. - Positive between \(-1\) and \(0.5\), negative between \(0.5\) and \(2\), zero at the equilibrium points. **Final sketch:** ``` G(x) ↑ | | / \ / \ |-------|---|------|---|------> x -1 0.5 2 ``` - The graph dips below zero between \(0.5\) and \(2\), crosses zero at the equilibrium points, and remains positive elsewhere. --- ### **(5.2) Sketch of solution curves \(x(t)\) for** - \(x(0) = 1\) - \(x(0) = 0\) **Behavior analysis:** - At \(x=1\): \(0.5 < 1 < 2\), so \(G(x) < 0\), solutions **decrease** toward \(x=0.5\). - At \(x=0\): \(-1 < 0 < 0.5\), so \(G(x) > 0\), solutions **increase** toward \(x=0.5\). **Final solution behavior:** - Both initial points tend toward the stable equilibrium at \(x=0.5\). **Sketch:** - \(x(t)\) starting at 1 decreases toward 0.5. - \(x(t)\) starting at 0 increases toward 0.5. --- ## **Summary for Question 5:** | Part | Explanation | Final Answer | |--------|--------------|--------------| | 5.1 | \(G(x)\) crosses zero at \(-1, 0.5, 2\), positive and negative regions as per phase line | Graph crosses at these points, positive before \(-1\) and between \(-1, 0.5\), negative between \(0.5, 2\), negative after 2 | | 5.2 | \(x(0)=1\) decreases toward 0.5; \(x(0)=0\) increases toward 0.5 | Both approach \(x=0.5\) as \(t \to \infty\) | --- **Note:** For actual sketches, plotting with approximate curves showing zero crossings and direction arrows would help visualize the phase line and solution trajectories. --- **If you'd like, I can generate the approximate sketches or further elaborate on the solution behavior.**

Certainly! Here are the final concise answers for each part: --- ### **Question 4** **(4.1)** **Answer:** No, this is not a logistic model because the quadratic term is positive, leading to unbounded (super-exponential) growth. **(4.2)** **Answer:** Equilibria at \( P = -\frac{1}{2} \) (stable) and \( P=0 \) (unstable). Phase line: - \( P < -\frac{1}{2} \): \( \frac{dP}{dt} > 0 \) (arrow right) - \(-\frac{1}{2} < P < 0 \): \( \frac{dP}{dt} < 0 \) (arrow left) - \( P > 0 \): \( \frac{dP}{dt} > 0 \) (arrow right) **(4.3)** **Answer:** For initial \( P(0)=7 \), the solution \( P(t) \) increases rapidly and blows up to infinity in finite time. --- ### **Question 5** **(5.1)** **Answer:** \( G(x) \) has zeros at \( x = -1, 0.5, 2 \). It is positive on \((-\infty, -1)\) and \((-1, 0.5)\), negative on \((0.5, 2)\), and negative beyond \( 2 \). **(5.2)** **Answer:** The solutions starting at \( x=1 \) (between 0.5 and 2) decrease toward \( x=0.5 \); those starting at \( x=0 \) increase toward \( x=0.5 \). Both tend toward the stable equilibrium at \( x=0.5 \). ---