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Calculus & Differential Equations

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FE Environmental Study Guide Exam

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1 Introduction, Mathematics & Ethics

Exam Format, NCEES CBT Logistics & Blueprint Map Calculus & Differential Equations Probability, Statistics & Regression Numerical Methods & Engineering Math Mathematics Applications for Environmental FE

2 Ethics, Economics & Materials

NCEES Model Rules & Engineering Ethics Professional Practice, Liability & Public Safety Engineering Economics: Time Value of Money Benefit-Cost, Replacement & Inflation Analysis Materials: Concrete, Corrosion & Polymers

3 Environmental Science & Chemistry

Geosynthetics & Environmental Materials Environmental Chemistry & Stoichiometry Water Chemistry: pH, Alkalinity & Hardness Microbiology, BOD Kinetics & Monod Growth Henry's Law & Phase Equilibrium for Air Quality

4 Risk Assessment & Water Resources

Environmental Science Applications EPA Four-Step Risk Assessment Paradigm Exposure, Dose, Hazard Quotient & Cancer Risk Hydrology & Rainfall-Runoff Analysis Open-Channel Flow & Manning Equation

5 Groundwater & Water Treatment

Groundwater: Darcy, Theis & Cooper-Jacob Streeter-Phelps & Water-Quality Modeling Water Treatment: Sedimentation & Filtration Disinfection, CT Rule & Chlorination Wastewater: Activated Sludge & BOD Removal

6 Advanced Treatment & Air Quality

Advanced Treatment: MBR, GAC & RO Sludge Handling & Water Reuse NAAQS, Criteria Pollutants, HAPs & AQI Gaussian Plume, Pasquill Stability & Effective Stack Height AP-42 Factors & Air Pollution Control Technologies

7 Air Permitting & Solid/Hazardous Waste

PSD, NSR, BACT, LAER, Title V & Offsets RCRA Lists, Characteristics & Generator Categories Subtitle D Landfill Design, Liners & Leachate CERCLA, UST, TCLP & Land Disposal Restrictions MSW Collection, Transfer, Composting, WTE & Recycling

8 Sustainability, EHS & Exam Day

LCA Stages, Carbon Footprint & Scope 1/2/3 Emissions LEED Credits, ISO 14001 EMS & Triple Bottom Line PELs, TLVs, Hierarchy of Controls & Confined Spaces EPCRA Reporting, TRI Form R, DOT Hazmat & RCRA Manifests Exam Format, NCEES CBT Logistics & Blueprint Map

Calculus & Differential Equations

Key Takeaways Explain This Topic AI

Derivatives model rates of decay and reaction.

Integrals sum mass loading and runoff volume.

dC/dt = -kC gives C = C₀e^{-kt}.

Half-life t₁/₂ = ln(2)/k.

Optimization sets df/dx = 0 for minima.

Last updated: July 2026

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 d C d t \frac{dC}{dt} d t d C ​ is the instantaneous rate at which concentration changes. If C ( t ) = C 0 e − k t C(t) = C_0 e^{-kt} C ( t ) = C 0 ​ e − k t , then: d C d t = − k C 0 e − k t = − k C ( t ) \frac{dC}{dt} = -k C_0 e^{-kt} = -k C(t) d t d C ​ = − k C 0 ​ e − k t = − k C ( t )

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

  • Application Derivative meaning
  • Decay dC/dt negative when pollutant disappears
  • Growth dN/dt positive for bacterial population
  • Flow dV/dt = Q (flow rate)
  • Reaction d[A]/dt from rate law

Worked example: A lagoon concentration follows C = 100 e − 0.08 t C = 100 e^{-0.08t} C = 100 e − 0.08 t mg/L (t in days). Rate at t = 0: d C d t ∣ t = 0 = − 0.08 × 100 = − 8 mg/L ⋅ day \frac{dC}{dt}\bigg|_{t=0} = -0.08 \times 100 = -8 \text{ mg/L}\cdot \text{day} d t d C ​ ​ t = 0 ​ = − 0.08 × 100 = − 8 mg/L ⋅ day

  • Integrals —
  • Accumulated
  • Quantities

The definite integral sums continuous contributions: Total mass loaded = ∫ 0 T Q ( t ) C ( t ) d t \text{Total mass loaded} = \int_0^T Q(t) \, C(t) \, dt Total mass loaded = ∫ 0 T ​ Q ( t ) C ( t ) d t

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

  • First-Order
  • Linear
  • ODEs

The separable equation d C d t = − k C \frac{dC}{dt} = -kC d t d C ​ = − k C integrates to C = C 0 e − k t C = C_0 e^{-kt} C = C 0 ​ e − k t . Half-life: t 1 / 2 = ln ⁡ ( 2 ) / k t_{1/2} = \ln(2)/k 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 − 0.693 t ⇒ t = ln ⁡ ( 8 ) 0.693 = 3 days 0.125 = e^{-0.693t} \Rightarrow t = \frac{\ln(8)}{0.693} = 3 \text{ days} 0.125 = e − 0.693 t ⇒ t = 0.693 ln ( 8 ) ​ = 3 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 − 4 a c b^2 - 4ac b 2 − 4 a c .

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 ∂ C ∂ x v \frac{\partial C}{\partial x} v ∂ x ∂ C ​ appears in transport equations. FE items may ask conceptually which term dominates — advection downstream versus dispersion spreading.

  • Optimization with
  • Derivatives

Set d f d x = 0 \frac{d f}{dx} = 0 d x df ​ = 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 − k t C = C_0 - kt C = C 0 ​ − k t (zero-order) when the problem states first-order .

Forgetting initial condition when integrating — always apply C ( 0 ) = C 0 C(0) = C_0 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 t_{1/2} t 1/2 ​ , compute k = 0.693 / t 1 / 2 k = 0.693/t_{1/2} 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 o u t = C i n 1 + k τ where τ = V / Q = 2500 s C_{out} = \frac{C_{in}}{1 + k\tau} \quad \text{where } \tau = V/Q = 2500\text{ s} C o u t ​ = 1 + k τ C in ​ ​ where τ = V / Q = 2500 s C o u t = 40 1 + 25 = 1.54 mg/L C_{out} = \frac{40}{1 + 25} = 1.54\text{ mg/L} C o u t ​ = 1 + 25 40 ​ = 1.54 mg/L

Compare to plug-flow C o u t = C i n e − k τ = 40 e − 25 ≈ 0 C_{out} = C_{in} e^{-k\tau} = 40 e^{-25} \approx 0 C o u t ​ = C in ​ e − k τ = 40 e − 25 ≈ 0 mg/L — same k and τ, different hydraulics.

Integration by Substitution (Mass Loading)

If hourly concentration varies as C ( t ) = 15 + 5 sin ⁡ ( π t / 12 ) C(t) = 15 + 5\sin(\pi t/12) C ( t ) = 15 + 5 sin ( π 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 C ˉ T M = Q \bar{C} T M = Q C ˉ T .

Worked example: Average C = 15 mg/L, Q = 2 m³/s, T = 86400 s: M = 2 × 15 × 86400 = 2,592,000 mg = 2.59 kg/day M = 2 \times 15 \times 86400 = 2{,}592{,}000\text{ mg} = 2.59\text{ kg/day} M = 2 × 15 × 86400 = 2 , 592 , 000 mg = 2.59 kg/day

Logarithmic and Exponential Forms

pH problems use [ H + ] = 10 − pH [H^+] = 10^{-\text{pH}} [ H + ] = 1 0 − pH . Sound intensity and some attenuation models use decibels — know I = I 0 × 10 d B / 10 I = I_0 \times 10^{dB/10} I = I 0 ​ × 1 0 d B /10 conceptually.

Worked example: pH drops from 7.0 to 5.0. Hydrogen ion activity ratio: [ H + ] 2 [ H + ] 1 = 10 − ( 5 − 7 ) = 10 2 = 100 \frac{[H^+]_2}{[H^+]_1} = 10^{-(5-7)} = 10^{2} = 100 [ H + ] 1 ​ [ H + ] 2 ​ ​ = 1 0 − ( 5 − 7 ) = 1 0 2 = 100

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

Taylor-Series Approximation (Small Changes)

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

ODE Summary