Sketching slope fields
A differential equation like dy/dx = x − y tells you the slope at any point (x, y). To sketch a slope field:
- Pick grid points, e.g. x and y from −2 to 2.
- Work out the slope at each point.
- Draw a short line with that slope.
Tips: slope 0 → flat line; equal slopes along a line are called isoclines (for x − y, every point with y = x has slope 0). If the rule uses only x, columns look the same; if only y, rows look the same.
Reasoning with slope fields
A solution curve through a start point (initial condition) follows the little lines without crossing them.
- Match a field to an equation: test easy points (where is slope 0? positive? does it depend on x only?).
- Equilibrium solutions: horizontal lines where dy/dx = 0 for all x (e.g. P = 0 and P = K in logistic).
- Long-run behaviour: follow the lines to see whether curves rise, fall, or approach a level.
- Concavity: find d²y/dx² by differentiating the rule (use the chain rule for y).
Approximating solutions using Euler's method
Start at (x₀, y₀). Choose a step size h. Repeat:
yₙ₊₁ = yₙ + h·f(xₙ, yₙ), xₙ₊₁ = xₙ + h.
Each step follows the tangent line, so it is a chain of linearizations.
- Smaller h → more steps → closer to the true solution.
- If the true solution is concave up, Euler underestimates; concave down → overestimates.
A table with columns x, y, slope, h × slope keeps the work tidy.
Logistic models
Exponential growth dP/dt = kP never stops. Real populations run out of food or space. The logistic model fixes this:
dP/dt = kP(1 − P/K)
- K = carrying capacity. If 0 < P₀ < K, P rises towards K; if P₀ > K, P falls towards K.
- Growth is fastest at P = K/2 (the inflection point of the S-curve).
- Solution: P(t) = K / (1 + A e^(−kt)), with A = (K − P₀)/P₀.
- lim P(t) = K as t → ∞.
Resistive (drag) forces and terminal velocity
A ball of mass m falls with air drag −bv (b in kg/s). Newton's second law: m dv/dt = mg − bv.
- At v = 0, acceleration is g.
- As v grows, drag grows, acceleration falls.
- When bv = mg, acceleration is 0: terminal velocity v_t = mg/b.
- Separating variables gives v(t) = v_t(1 − e^(−t/τ)), with time constant τ = m/b.
After one τ the speed is 63% of v_t; after 5τ it is over 99%. For fast objects drag can grow like v² instead; then v_t = √(mg/c).
Try it: a paper-cone race
Make two paper cupcake cases. Drop one, then two stacked together, from 2 m. The heavier stack reaches a higher terminal speed (v_t = mg/b grows with m) and lands first. Then draw a 5 × 5 slope field for dy/dx = y on paper and sketch the curve through (0, 1).
Key formulas and definitions
- Slope field: slope at (x, y) = f(x, y)
- Euler: yₙ₊₁ = yₙ + h·f(xₙ, yₙ)
- Logistic: dP/dt = kP(1 − P/K)
- P(t) = K / (1 + A e^(−kt)), A = (K − P₀)/P₀
- Fastest logistic growth at P = K/2
- Drag: m dv/dt = mg − bv
- v_t = mg/b, v(t) = v_t(1 − e^(−bt/m))
Worked examples
1. For dy/dx = x − y, find the slopes at (0, 0), (1, 0) and (0, 2).
Slope = x − y: (0,0) → 0, flat; (1,0) → 1; (0,2) → −2, steeply down.
2. Which equation fits a slope field where all lines in each row are the same, and lines are flat on y = 1: dy/dx = x − 1 or dy/dx = y − 1?
Same in each row means the rule uses only y. Flat at y = 1 means y − 1 = 0 there. So dy/dx = y − 1.
3. Use Euler's method with h = 0.5 for dy/dx = x − y, y(0) = 4, to estimate y(1).
Step 1: slope at (0, 4) = −4; y₁ = 4 + 0.5(−4) = 2 at x = 0.5. Step 2: slope at (0.5, 2) = −1.5; y₂ = 2 + 0.5(−1.5) = 1.25 at x = 1. Estimate y(1) ≈ 1.25.
4. The true solution in Example 3 is y = x − 1 + 5e^(−x). Is Euler's estimate too high or too low, and why?
True y(1) = 5/e ≈ 1.839. Euler gave 1.25, too low. y″ = 1 − y′ = 1 − x + y > 0 here, so the curve is concave up and tangent steps fall below it.
5. dP/dt = 0.8P(1 − P/10), P(0) = 1. Find the carrying capacity, the population when growth is fastest, and lim P(t).
K = 10. Fastest at P = K/2 = 5. As t → ∞, P → 10.
6. For the model in Example 5, find P(t) and P(2).
A = (10 − 1)/1 = 9. P(t) = 10/(1 + 9e^(−0.8t)). P(2) = 10/(1 + 9e^(−1.6)) = 10/(1 + 1.817) ≈ 3.55.
7. A 0.5 kg ball falls with drag constant b = 0.25 kg/s (g = 9.8 m/s²). Find v_t, τ and the speed after 2 s.
v_t = mg/b = 0.5 × 9.8 / 0.25 = 19.6 m/s. τ = m/b = 2 s. v(2) = 19.6(1 − e^(−1)) ≈ 19.6 × 0.632 ≈ 12.4 m/s.
Common mistakes
- Drawing solution curves that cross slope lines or each other. A solution follows the lines.
- In Euler's method, using the slope at the new point instead of the old one. Always use f(xₙ, yₙ) to step forward.
- Thinking logistic growth is fastest at P = K. At K growth stops; it is fastest at K/2.
- Saying terminal velocity means no forces act. Forces still act; they just balance (drag = weight), so acceleration is zero.