Rates of Change and Differential Equations — Topic Review — Solutions
This review covers all four lessons: Related Rates of Change, Separable Differential Equations, First-Order Linear Differential Equations, and Applications of Differential Equations. Questions are exam-style and increase in difficulty. Click each question to reveal the full worked solution.
Mixed Review Questions
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Fluency
Q1 — Related Rates
A spherical balloon is being inflated. When the radius is 6 cm, the radius is increasing at 0.5 cm/s. Find the rate of increase of the volume at that instant. (V = (4/3)πr³)
dV/dt = dV/dr × dr/dt = 4πr² × dr/dt.
At r = 6, dr/dt = 0.5:
dV/dt = 4π(6)² × 0.5 = 4π × 36 × 0.5 = 72π ≈ 226 cm³/s
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Fluency
Q2 — Related Rates
A 5-metre ladder leans against a vertical wall. The base slides away at 0.3 m/s. Find the rate at which the top slides down when the base is 3 m from the wall.
x² + y² = 25. Differentiate: 2x dx/dt + 2y dy/dt = 0.
At x = 3: y = √(25−9) = 4 m.
dy/dt = −x/y × dx/dt = −(3/4)(0.3) = −0.225 m/s.
The top slides down at 0.225 m/s.
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Fluency
Q3 — Separable DE
Solve the separable DE dy/dx = 2xy, given y(0) = 3.
Separate: dy/y = 2x dx.
Integrate: ln|y| = x² + C ⇒ y = Aex².
Apply y(0) = 3: A = 3.
y = 3ex²
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Fluency
Q4 — Separable DE
Solve dy/dx = y² sin x, given y(0) = 1.
Separate: dy/y² = sin x dx ⇒ −1/y = −cos x + C.
Apply y(0) = 1: −1 = −1 + C ⇒ C = 0.
−1/y = −cos x ⇒ y = 1/cos x.
y = sec x
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Fluency
Q5 — First-Order Linear DE
Solve dy/dx + 5y = 10, finding the general solution.
P = 5, μ = e5x. d/dx[e5xy] = 10e5x.
e5xy = 2e5x + C.
y = 2 + Ce−5x
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Understanding
Q6 — Related Rates
Water drains from a conical tank (apex down) of half-angle 30° at a rate of 2 m³/min. Find the rate at which the water depth decreases when the depth is 3 m. (V = (1/3)πr²h, and r = h tan 30° = h/√3.)
r = h/√3, so V = (1/3)π(h/√3)²h = (1/3)πh³/3 = πh³/9.
dV/dt = (πh²/3) dh/dt.
dV/dt = −2 (draining). At h = 3:
−2 = (π(9)/3) dh/dt = 3π dh/dt.
dh/dt = −2/(3π) ≈ −0.212 m/min.
The depth decreases at 2/(3π) ≈ 0.212 m/min.
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Understanding
Q7 — Separable DE with Initial Condition
Solve dP/dt = 0.04P(200 − P). Find P(t) given P(0) = 20. What is limt→∞ P(t)?
Write as logistic: dP/dt = (0.04 × 200)P(1 − P/200) = 8P(1 − P/200).
Here r = 8, K = 200. A = (200 − 20)/20 = 9.
P(t) = 200/(1 + 9e−8t).
As t → ∞: P → 200 (the carrying capacity).
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Understanding
Q8 — First-Order Linear DE
Solve dy/dx + (1/x)y = x² + 1, for x > 0. Give the general solution.
P(x) = 1/x, μ = eln x = x.
d/dx[xy] = x(x² + 1) = x³ + x.
xy = x4/4 + x²/2 + C.
y = x³/4 + x/2 + C/x
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Understanding
Q9 — Newton’s Law of Cooling
A metal bar at 800°C is plunged into water at 20°C. After 30 seconds the bar is at 400°C. How long until the bar reaches 50°C?
T(t) = 20 + 780e−kt. At t = 30: 400 = 20 + 780e−30k.
e−30k = 380/780 = 19/39. k = −(1/30)ln(19/39).
At T = 50: 50 = 20 + 780e−kt ⇒ e−kt = 30/780 = 1/26.
−kt = −ln26 ⇒ t = ln26/k = 30 ln26/ln(39/19).
ln26 ≈ 3.258, ln(39/19) ≈ 0.7178.
t ≈ 30 × 3.258/0.7178 ≈ 136 seconds
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Understanding
Q10 — Mixing Problem
A 100 L tank initially contains 20 g of salt dissolved in water. Brine with 0.3 g/L flows in at 5 L/min, and the well-mixed solution flows out at 5 L/min. Find Q(t), the amount of salt at time t.
dQ/dt = 0.3 × 5 − (Q/100) × 5 = 1.5 − Q/20.
Standard form: dQ/dt + (1/20)Q = 1.5. μ = et/20.
d/dt[et/20Q] = 1.5et/20.
et/20Q = 30et/20 + C ⇒ Q = 30 + Ce−t/20.
Q(0) = 20: 20 = 30 + C ⇒ C = −10.
Q(t) = 30 − 10e−t/20 grams
As t → ∞: Q → 30 g = 0.3 g/L × 100 L. ✓
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Problem Solving
Q11 — Related Rates: Angle
A camera films a rocket launched vertically from a point 500 m away. When the rocket is 1200 m high and climbing at 80 m/s, find the rate at which the camera elevation angle θ is increasing.
tanθ = h/500, so θ = arctan(h/500). Differentiate:
sec²θ × dθ/dt = (1/500) dh/dt.
At h = 1200: tanθ = 1200/500 = 12/5, so sec²θ = 1 + (12/5)² = 1 + 144/25 = 169/25.
dθ/dt = (1/500) × 80 / (169/25) = (80/500) × (25/169) = 2000/84500 = 4/169 rad/s.
dθ/dt = 4/169 ≈ 0.0237 rad/s
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Problem Solving
Q12 — First-Order Linear DE (Variable Forcing)
Solve dy/dx − (2/x)y = x3ex, for x > 0, given y(1) = 0.
P(x) = −2/x, μ = e∫(−2/x) dx = e−2lnx = x−2 = 1/x².
d/dx[y/x²] = (1/x²) × x³ex = xex.
Integrate by parts: ∫xex dx = xex − ex + C = (x−1)ex + C.
y/x² = (x−1)ex + C ⇒ y = x²(x−1)ex + Cx².
Apply y(1) = 0: 0 = 1(0)e + C ⇒ C = 0.
y = x²(x − 1)ex
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Problem Solving
Q13 — Radioactive Decay Chain
Substance A decays into substance B at rate kA = 0.02 per year. Substance B decays at rate kB = 0.05 per year. Initially, A(0) = 100 g and B(0) = 0. Write the DE for B and solve to find B(t).
A(t) = 100e−0.02t. The DE for B is:
dB/dt = 0.02A − 0.05B = 2e−0.02t − 0.05B.
Standard form: dB/dt + 0.05B = 2e−0.02t. μ = e0.05t.
d/dt[e0.05tB] = 2e−0.02te0.05t = 2e0.03t.
e0.05tB = (2/0.03)e0.03t + C = (200/3)e0.03t + C.
B = (200/3)e−0.02t + Ce−0.05t.
Apply B(0) = 0: 0 = 200/3 + C ⇒ C = −200/3.
B(t) = (200/3)(e−0.02t − e−0.05t) grams
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Problem Solving
Q14 — RL Circuit with Sinusoidal Input
An RL circuit has L = 1 H, R = 1 Ω, and V(t) = sin t V. Find I(t) given I(0) = 0.
dI/dt + I = sin t. μ = et.
d/dt[etI] = et sin t.
∫et sin t dt = et(sin t − cos t)/2 + C (by integration by parts twice).
etI = et(sin t − cos t)/2 + C.
I = (sin t − cos t)/2 + Ce−t.
Apply I(0) = 0: 0 = (0 − 1)/2 + C ⇒ C = 1/2.
I(t) = (sin t − cos t + e−t)/2 amperes
The term e−t/2 is the transient; (sin t − cos t)/2 is the steady-state (sinusoidal) response.
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Problem Solving
Q15 — Logistic Growth: Maximum Rate
A population grows logistically: dP/dt = 0.1P(1 − P/400). Show that the growth rate dP/dt is maximised when P = 200, and find the maximum rate.
Let f(P) = 0.1P(1 − P/400) = 0.1P − P²/4000.
Differentiate with respect to P: df/dP = 0.1 − 2P/4000 = 0.1 − P/2000.
Set df/dP = 0: 0.1 = P/2000 ⇒ P = 200 = K/2. ✓
Second derivative: d²f/dP² = −1/2000 < 0, confirming maximum.
Maximum growth rate = f(200) = 0.1 × 200 × (1 − 200/400) = 20 × 0.5 = 10 individuals per unit time