Mathematics · Study · FE Environmental · FE → PE Prep
Mathematics
5% of exam
Analytic geometry and trigonometry, algebraic equations and roots, differential and integral calculus, differential equations, and numerical methods including error propagation.
4 concepts
A. Analytic geometry and trigonometry
Analytic Geometry and Trigonometry
Lines and conics, the unit circle and trig identities, the laws of sines and cosines, and polar-rectangular conversion as the geometry engine behind environmental areas, volumes, and slopes.
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B. Algebraic equations and roots
Algebraic Equations and Roots
Linear, quadratic, and polynomial roots, the discriminant, logarithms and exponentials, partial fractions, and the characteristic equation that governs reactor and decay models.
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Derivatives as rates and integrals as accumulated mass, maxima and minima, the fundamental theorem, and first- and second-order linear ODEs for decay, reactor, and mixing models.
Calculus is the language environmental engineering uses to describe change: a derivative is a rate (a mass flux, a slope, a velocity), and an integral is an accumulation (the total mass that passed an outfall, the volume under a hydrograph). Every reactor model, every decay curve, and every mixing-tank transient is a differential equation whose solution is an exponential. The FE Reference Handbook Mathematics chapter supplies the derivative and integral tables, the maximum/minimum tests, and the standard ODE solution forms; the exam rewards engineers who can translate a worded process — "a tank is filling," "a pollutant is decaying" — into the right equation and read its behavior. This concept is the bridge between the formulas you carry and the processes you model.
The derivative as a rate of change
The derivative dy/dx is the limit of Δy/Δx — the instantaneous slope of a curve, or in process terms the rate at which a quantity changes. In environmental work dC/dt is a reaction rate, dV/dt is a flow, and dh/dx is a hydraulic gradient. You will lean on the product rule, the quotient rule, and above all the chain rule dxdf(u)=f′(u)dxdu, because composite functions like e−kt and ln(C0/C) are everywhere. The derivative of eu is eudu/dx, which is exactly why first-order processes are self-similar: their rate is proportional to their state.
dxdy=Δx→0limΔxf(x+Δx)−f(x),dxdeu=eudxdu
Maxima, minima, and optimization
A function is stationary where its first derivative vanishes, f′(x)=0; the second derivative then classifies the point — f′′(x)<0 is a maximum, f′′(x)>0
The integral as accumulated mass
The definite integral is the limit of a sum of slivers, ∫abf(x)dx=lim∑f(xi)Δxi
First-order linear ODEs: decay
The simplest and most-used model is dtdC+kC=0, whose solution is the exponential decay C(t)=C0e−kt
First-order nonhomogeneous ODEs: the mixing tank
Add a source term and you get the workhorse completely mixed (CSTR) transient: VdtdC=QCin−QC, a first-order nonhomogeneous ODE with time constant
Second-order linear ODEs and damping
A second-order equation y′′+ay′+by=0 is solved by seeking y=Cerx
Exam strategy
First decide whether the problem is asking for a rate (differentiate), an accumulation (integrate), or a time evolution (solve an ODE). "Total mass/volume over a period" is an integral — set up ∫QCdt and watch the unit chain (m3/s × mg/L × s, then convert to kg). "How long until" a first-order quantity reaches a value is
Key equations
Definition of the derivativedxdy=limΔx→0Δxf(x+Δx)−f(x)
Worked examples
Accumulated pollutant mass over a storm (integral)
Problem. During a 6-h storm an outfall discharges at a constant Q=0.60m3/s. The pollutant concentration shows a first-flush decline C(t)=8+40e−0.5tmg/L
Common pitfalls
•Unit chain in loading integrals: ∫QCdt in m3/s, mg/L, and hours must be reconciled with 3600s/h and 1000L/m3
References
NCEES FE Reference Handbook — Mathematics (Differential and Integral Calculus, Differential Equations)
Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery — reactor transient and CSTR models
Davis & Cornwell, Introduction to Environmental Engineering — first-order kinetics and mass balances
D. Numerical methods
Numerical Methods and Error Propagation
Newton-Raphson root finding, trapezoidal and Simpson integration of tabulated hydrograph and loading data, finite differences, and Kline-McClintock propagation of measurement error.
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is a minimum, and
f′′(x)=0
with a sign change is an inflection point. This is the engine of design optimization: the least-cost basin proportions, the depth that maximizes channel velocity, the detention time that minimizes residual. The procedure is always the same — write the quantity as a function of one variable, differentiate, set to zero, and confirm with the second derivative.
— the area under a curve. Physically it is an accumulation: the total mass discharged over a storm is
M=∫0TQ(t)C(t)dt
, the area under the load-versus-time curve. The fundamental theorem of calculus ties accumulation back to rate: differentiation and integration are inverse operations, so the running total
F(x)=∫axf(t)dt
has derivative
F′(x)=f(x)
. This is why a flow-weighted mass and a concentration profile are two views of the same process.
∫abf(x)dx=F(b)−F(a),dxd∫axf(t)dt=f(x)
. This single equation governs radioactive decay, BOD exertion, first-order disinfection, and reaeration. The time constant is
τ=1/k
and the half-life is
ln2/k
. The defining feature is that the rate of loss is proportional to how much is present, so the curve never quite reaches zero but falls by a fixed fraction each time constant.
dtdC=−kC⟹C(t)=C0e−kt
τ=V/Q
. Its solution superposes a decaying memory of the initial state on an approach to the new steady state:
C(t)=Cin(1−e−t/τ)+C0e−t/τ
. After about
4τ
the tank has effectively forgotten its starting condition — the basis for the "three-to-five residence times" rule of thumb for reaching steady operation. The general form
τy′+y=Kx(t)
in the handbook is this same equation with gain
K
and time constant
τ
.
VdtdC=QCin−QC⟹C(t)=Cin(1−e−t/τ)+C0e−t/τ,τ=QV
, which forces the characteristic equation
r2+ar+b=0
. The discriminant
a2−4b
decides the regime:
a2>4b
gives two real roots and an overdamped response
y=C1er1x+C2er2x
;
a2=4b
gives a repeated root and critical damping
y=(C1+C2x)er1x
; and
a2<4b
gives complex roots and an underdamped, oscillatory response
y=eαx(C1cosβx+C2sinβx)
with
α=−a/2
. The same trichotomy describes coupled reactors and control loops, so reading damping from the roots is a transferable skill.
Complex-root case (a2<4b): decaying oscillation, with β=214b−a2
with
t
in hours. Find the total mass of pollutant discharged.
Solution. Total mass M=Q∫06C(t)dt (with Q constant). Integrate the concentration over the 6-h storm:
∫06(8+40e−0.5t)dt=8t06+−0.540e−0.5t06=48+80(1−e−3).
e−3=0.04979, so the integral =48+80(0.9502)=48+76.0=124.0mg/L⋅h.
Now convert. M=Q(124.0mg/L⋅h): 0.60sm3×3600hs×1000m3L×124.0Lmg⋅h=2.678×108mg.
M=2.678×108mg=268kg.
Sanity check: average concentration over the storm ≈124.0/6=20.7mg/L; total volume =0.60×6×3600=12,960m3=1.296×107L; 20.7mg/L×1.296×107L=2.68×108mg. Matches.
M=268kg
M=Q∫06(8+40e−0.5t)dt=Q(124.0mg/L⋅h)=268kg
First-order decay and half-life
Problem. A reservoir contaminant decays first order with k=0.12d−1. Starting at C0=2.5mg/L, find the time to reach the action level of 0.50mg/L, and the half-life.
Solution. Model C=C0e−kt, invert: t=k1ln(CC0)=0.121ln(0.502.5)
Problem. A completely mixed tank holds V=500m3 and is fed at Q=50m3/h. It starts clean (C0=0). At t=0 the influent steps to Cin=20mg/L. Find the tank concentration at t=15h, and the time to reach 18mg/L.
Solution. Mass balance VdC/dt=Q(Cin−C) has time constant τ=V/Q=500/50=10h
C(t)=20(1−e−t/10);C(15)=15.5mg/L,t18=23.0h
Minimum-perimeter detention basin (optimization)
Problem. A rectangular detention basin must enclose a fixed plan area A=400m2. To minimize the length of perimeter embankment P, what dimensions should it have?
Solution. Let the basin be x by A/x. Perimeter P(x)=2x+2xA=2x+x800.
Differentiate and set to zero: dxdP=2−x2800=0⇒x2=400⇒x=20m.
Second derivative dx2d2P=x31600>0, confirming a minimum.
Then A/x=400/20=20m — a square — with P=2(20)+2(20)=80m.
Sanity check: a 25×16 basin (same area) has P=2(25+16)=82m>80m, so the square is indeed the minimum.
20m×20m,P=80m
before the mass comes out in kg. Most errors here are conversion slips, not calculus.
•Sign of the decay exponent: C=C0e−kt falls with time; writing e+kt (growth) flips the physics. Solving for time, use t=k1ln(C0/C) to keep t positive.
•Confusing time constant with half-life: τ=1/k but t1/2=ln2/k=0.693τ. They differ by the factor ln2 — do not substitute one for the other.
•Forgetting the initial-condition term in the mixing tank: the full solution is Cin(1−e−t/τ)+C0e−t/τ. Dropping C0e−t/τ is only valid when the tank starts clean.
•Misreading τ for the CSTR: the time constant is τ=V/Q, the hydraulic residence time — not Q/V. Compute and label τ first; everything else follows.
•Skipping the second-derivative test: f′(x)=0 alone does not tell you maximum from minimum. A design "optimum" that is actually a maximum cost (or an inflection) is a classic trap — confirm the sign of f′′.
•Radian vs. degree mode in calculus: derivatives and integrals of trig functions assume radians. A sin or cos evaluated in degrees inside an integral gives a wrong area.