Calculus & Differential Equations

Quick Answer: FE Environmental calculus is applied, not theoretical. Master derivatives as rates, integrals as accumulated totals, and separable ODEs for first-order decay. The Handbook supplies standard forms — you supply correct initial conditions and units.

Environmental systems change with time and space. Concentrations decay downstream, BOD is consumed in rivers, and reactor concentrations approach steady state. Calculus provides the language for these dynamics on the FE Environmental exam (~10% math block).

Derivatives —

Rates of

Change

The derivative ( \frac{dC}{dt} ) is the instantaneous rate at which concentration changes. If ( C(t) = C_0 e^{-kt} ), then:

[ \frac{dC}{dt} = -k C_0 e^{-kt} = -k C(t) ]

The rate is proportional to current concentration — the defining property of first-order kinetics.

Worked example: A lagoon concentration follows ( C = 100 e^{-0.08t} ) mg/L (t in days). Rate at t = 0:

[ \frac{dC}{dt}\bigg|_{t=0} = -0.08 \times 100 = -8 \text{ mg/L·day} ]

Integrals —

Accumulated

Quantities

The definite integral sums continuous contributions:

[ \text{Total mass loaded} = \int_0^T Q(t) , C(t) , dt ]

When data are tabulated, use the trapezoidal rule (covered in numerical methods) instead of analytic integration.

Worked example: Constant flow Q = 0.5 m³/s for 3600 s with C = 20 mg/L:

[ M = Q \cdot C \cdot t = 0.5 \times 20 \times 3600 = 36{,}000 \text{ mg} = 36 \text{ g} ]

First-Order

Linear

ODEs

The separable equation ( \frac{dC}{dt} = -kC ) integrates to ( C = C_0 e^{-kt} ). Half-life: ( t_{1/2} = \ln(2)/k ).

Worked example: k = 0.693/day (implying t₁/₂ = 1 day). How long to reach 12.5% of C₀?

[ 0.125 = e^{-0.693t} \Rightarrow t = \frac{\ln(8)}{0.693} = 3 \text{ days} ]

Higher-Order and

Coupled

Streeter-Phelps BOD-DO sag couples two first-order relationships — solved analytically in the Handbook. Second-order ODEs ( ay'' + by' + cy = 0 ) yield exponential or oscillatory solutions depending on the discriminant ( b^2 - 4ac ).

For environmental FE, know the form of the solution; full derivation is not required.

Partial

Derivatives (Light)

Concentration in a river may vary with distance x and time t: ( C(x,t) ). The advection term ( v \frac{\partial C}{\partial x} ) appears in transport equations. FE items may ask conceptually which term dominates — advection downstream versus dispersion spreading.

Optimization with

Derivatives

Set ( \frac{d f}{dx} = 0 ) to find maxima/minima — e.g., optimal recycle ratio or minimum cost tank volume. Verify with second derivative or endpoints if the domain is bounded.

Exam

Traps

• Mixing days and seconds in k and t invalidates decay answers.
• Using ( C = C_0 - kt ) (zero-order) when the problem states first-order.
• Forgetting initial condition when integrating — always apply ( C(0) = C_0 ).

Handbook

Connections

Search for differential equations, calculus, and first-order decay. Pair with environmental chemistry and microbiology chapters for BOD and reactor problems.

Tip: If a problem gives ( t_{1/2} ), compute ( k = 0.693/t_{1/2} ) before solving — faster than guessing k from tables.

FE Exam

Integration

Environmental FE items on this topic often combine regulatory classification with a quantitative step. Read the stem for the governing law (CWA, CAA, RCRA, OSHA) before selecting equations. Flag multi-step problems and return after your first pass — average time is under three minutes per question across 110 items.

Practice locating handbook relationships by keyword during timed drills. Confirm units on every constant: mg/L versus μg/m³, ft³/s versus MGD, and days versus seconds in decay and pumping problems are frequent error sources.

FE Exam Integration

Chain Rule in Environmental Rate Laws

When concentration depends on temperature through (k(T)), use the chain rule: (dC/dt = (dC/dk)(dk/dT)(dT/dt)). FE items rarely require full derivation but may ask whether a rate increases when temperature rises — for most biochemical rates, yes, via Arrhenius or θ factors.

Worked example — reactor mass balance: A CSTR with volume V = 500 m³ receives flow Q = 0.2 m³/s at C_in = 40 mg/L. First-order decay (k = 0.01) s⁻¹ inside the tank. Steady-state outlet:

[ C_{out} = \frac{C_{in}}{1 + k\tau} \quad \text{where } \tau = V/Q = 2500\text{ s} ]

[ C_{out} = \frac{40}{1 + 25} = 1.54\text{ mg/L} ]

Compare to plug-flow (C_{out} = C_{in} e^{-k\tau} = 40 e^{-25} \approx 0) mg/L — same k and τ, different hydraulics.

Integration by Substitution (Mass Loading)

If hourly concentration varies as (C(t) = 15 + 5\sin(\pi t/12)) mg/L over 24 h with constant Q = 2 m³/s, daily mass requires integrating (QC(t)). When only average C is given, use (M = Q \bar{C} T).

Worked example: Average C = 15 mg/L, Q = 2 m³/s, T = 86400 s:

[ M = 2 \times 15 \times 86400 = 2{,}592{,}000\text{ mg} = 2.59\text{ kg/day} ]

Logarithmic and Exponential Forms

pH problems use ( [H^+] = 10^{-\text{pH}} ). Sound intensity and some attenuation models use decibels — know (I = I_0 \times 10^{dB/10}) conceptually.

Worked example: pH drops from 7.0 to 5.0. Hydrogen ion activity ratio:

[ \frac{[H^+]_2}{[H^+]_1} = 10^{-(5-7)} = 10^{2} = 100 ]

Acidity increases 100-fold — critical for metal solubility and disinfection speciation.

Taylor-Series Approximation (Small Changes)

For small (\Delta x), (f(x+\Delta x) \approx f(x) + f'(x)\Delta x). Useful when an exam asks for sensitivity of effluent concentration to a 5% flow increase without full re-solution.

ODE Summary