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

Exam areas

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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C. Calculus

Calculus and Differential Equations

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

\Delta y / \Delta x

— 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

\dfrac{d}{dx}f(u) = f'(u)\,\dfrac{du}{dx}

, because composite functions like

e^{-kt}

and

\ln(C_0/C)

are everywhere. The derivative of

e^{u}

e^{u}\,du/dx

, which is exactly why first-order processes are self-similar: their rate is proportional to their state.

\dfrac{dy}{dx} = \lim_{\Delta x \to 0}\dfrac{f(x + \Delta x) - f(x)}{\Delta x}, \qquad \dfrac{d}{dx}e^{u} = e^{u}\dfrac{du}{dx}

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

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.

f'(x^\ast) = 0, \qquad \begin{cases} f''(x^\ast) < 0 & \text{maximum}\\ f''(x^\ast) > 0 & \text{minimum}\\ f''(x^\ast) = 0 & \text{inflection (if sign changes)}\end{cases}

The integral as accumulated mass

The definite integral is the limit of a sum of slivers,

\int_a^b f(x)\,dx = \lim \sum f(x_i)\,\Delta x_i

— the area under a curve. Physically it is an accumulation: the total mass discharged over a storm is

M = \int_0^T Q(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) = \int_a^x f(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.

\int_a^b f(x)\,dx = F(b) - F(a), \qquad \dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)

First-order linear ODEs: decay

The simplest and most-used model is

\dfrac{dC}{dt} + kC = 0

, whose solution is the exponential decay

C(t) = C_0 e^{-kt}

. This single equation governs radioactive decay, BOD exertion, first-order disinfection, and reaeration. The time constant is

\tau = 1/k

and the half-life is

\ln 2 / 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.

\dfrac{dC}{dt} = -kC \;\Longrightarrow\; C(t) = C_0\,e^{-kt}

First-order nonhomogeneous ODEs: the mixing tank

Add a source term and you get the workhorse completely mixed (CSTR) transient:

V\dfrac{dC}{dt} = Q C_{in} - Q C

, a first-order nonhomogeneous ODE with time constant

\tau = V/Q

. Its solution superposes a decaying memory of the initial state on an approach to the new steady state:

C(t) = C_{in}\big(1 - e^{-t/\tau}\big) + C_0\,e^{-t/\tau}

. After about

4\tau

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

\tau y' + y = K x(t)

in the handbook is this same equation with gain

K

and time constant

\tau

V\dfrac{dC}{dt} = Q C_{in} - Q C \;\Longrightarrow\; C(t) = C_{in}\big(1 - e^{-t/\tau}\big) + C_0\,e^{-t/\tau}, \;\; \tau = \dfrac{V}{Q}

Second-order linear ODEs and damping

A second-order equation

y'' + ay' + by = 0

is solved by seeking

y = Ce^{rx}

, which forces the characteristic equation

r^2 + ar + b = 0

. The discriminant

a^2 - 4b

decides the regime:

a^2 > 4b

gives two real roots and an overdamped response

y = C_1 e^{r_1 x} + C_2 e^{r_2 x}

;

a^2 = 4b

gives a repeated root and critical damping

y = (C_1 + C_2 x)e^{r_1 x}

; and

a^2 < 4b

gives complex roots and an underdamped, oscillatory response

y = e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)

with

\alpha = -a/2

. The same trichotomy describes coupled reactors and control loops, so reading damping from the roots is a transferable skill.

r = \dfrac{-a \pm \sqrt{a^2 - 4b}}{2}, \qquad a^2 - 4b \;\begin{cases} > 0 & \text{overdamped}\\ = 0 & \text{critically damped}\\ < 0 & \text{underdamped}\end{cases}

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

\int Q\,C\,dt

and watch the unit chain (m

^3

\times

1000 L/m

^3

\times

mg/L

\times

3600 s/h

\times

h gives milligrams; divide by

10^6

for kg). "How long until" a first-order quantity reaches a value is

t = \tfrac{1}{k}\ln(C_0/C)

. A filling or step-fed tank is the nonhomogeneous first-order solution; reach for

C_{in}(1 - e^{-t/\tau}) + C_0 e^{-t/\tau}

and identify

\tau = V/Q

first. For optimization, write one-variable, differentiate, set to zero, confirm with

f''

. Carry units through every substitution — most lost points here are unit-conversion slips, not calculus errors.

Key equations

Definition of the derivative

\dfrac{dy}{dx} = \lim_{\Delta x \to 0}\dfrac{f(x + \Delta x) - f(x)}{\Delta x}

Instantaneous rate of change / slope of $y = f(x)$ . In process terms, a flux or reaction rate.

Chain and product rules

\dfrac{d}{dx}f(u) = f'(u)\dfrac{du}{dx}, \qquad \dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}

Differentiation of composites and products; the chain rule drives every $e^{-kt}$ derivative.

Max/min/inflection test

f'(x^\ast) = 0; \quad f''<0 \Rightarrow \max,\;\; f''>0 \Rightarrow \min,\;\; f''=0\text{ (sign change)} \Rightarrow \text{inflection}

First derivative locates stationary points; second derivative classifies them.

Fundamental theorem of calculus

\int_a^b f(x)\,dx = F(b) - F(a), \qquad \dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)

Integration and differentiation are inverse operations; $F$ is any antiderivative of $f$ .

Accumulated mass (loading)

M = \int_0^T Q(t)\,C(t)\,dt

Total mass discharged over time. $Q$ = flow, $C$ = concentration; mind unit conversion to kg or lb.

First-order decay

\dfrac{dC}{dt} = -kC \;\Rightarrow\; C(t) = C_0 e^{-kt}

Decay/BOD/disinfection model. $k$ = rate constant (time $^{-1}$ ); time constant $\tau = 1/k$ .

Half-life

t_{1/2} = \dfrac{\ln 2}{k}

Time for a first-order quantity to halve; independent of $C_0$ .

CSTR transient (first-order nonhomogeneous)

C(t) = C_{in}\big(1 - e^{-t/\tau}\big) + C_0\,e^{-t/\tau}, \qquad \tau = \dfrac{V}{Q}

Step response of a completely mixed tank; approaches $C_{in}$ as $t \to \infty$ . $\tau$ = hydraulic residence time.

Second-order characteristic equation

y'' + ay' + by = 0 \;\Rightarrow\; r^2 + ar + b = 0

Roots $r$ set the modes of the response; sign of $a^2 - 4b$ gives over/critical/underdamped.

Underdamped solution

y = e^{\alpha x}\big(C_1\cos\beta x + C_2\sin\beta x\big), \quad \alpha = -\tfrac{a}{2}

Complex-root case ( $a^2 < 4b$ ): decaying oscillation, with $\beta = \tfrac{1}{2}\sqrt{4b - a^2}$ .

Worked examples

Accumulated pollutant mass over a storm (integral)

Problem. During a $6\text{-h}$ storm an outfall discharges at a constant $Q = 0.60\ \text{m}^3/\text{s}$ . The pollutant concentration shows a first-flush decline $C(t) = 8 + 40e^{-0.5t}\ \text{mg/L}$ with $t$ in hours. Find the total mass of pollutant discharged.

Solution. Total mass

M = Q\displaystyle\int_0^6 C(t)\,dt

(with

Q

constant). Integrate the concentration over the

6\text{-h}

storm:

\displaystyle\int_0^6 \big(8 + 40e^{-0.5t}\big)\,dt = 8t\Big|_0^6 + \dfrac{40}{-0.5}e^{-0.5t}\Big|_0^6 = 48 + 80\big(1 - e^{-3}\big)

e^{-3} = 0.04979

, so the integral

= 48 + 80(0.9502) = 48 + 76.0 = 124.0\ \text{mg/L}\cdot\text{h}

. Now convert.

M = Q\,(124.0\ \text{mg/L}\cdot\text{h})

\;0.60\ \tfrac{\text{m}^3}{\text{s}} \times 3600\ \tfrac{\text{s}}{\text{h}} \times 1000\ \tfrac{\text{L}}{\text{m}^3} \times 124.0\ \tfrac{\text{mg}\cdot\text{h}}{\text{L}} = 2.678\times10^{8}\ \text{mg}

M = 2.678\times10^{8}\ \text{mg} = 268\ \text{kg}

. Sanity check: average concentration over the storm

\approx 124.0/6 = 20.7\ \text{mg/L}

; total volume

= 0.60 \times 6 \times 3600 = 12{,}960\ \text{m}^3 = 1.296\times10^{7}\ \text{L}

;

20.7\ \text{mg/L} \times 1.296\times10^{7}\ \text{L} = 2.68\times10^{8}\ \text{mg}

. Matches.

\boxed{M = 268\ \text{kg}}

M = Q\int_0^6\!\big(8 + 40e^{-0.5t}\big)dt = Q\,(124.0\ \text{mg/L}\cdot\text{h}) = 268\ \text{kg}

First-order decay and half-life

Problem. A reservoir contaminant decays first order with $k = 0.12\ \text{d}^{-1}$ . Starting at $C_0 = 2.5\ \text{mg/L}$ , find the time to reach the action level of $0.50\ \text{mg/L}$ , and the half-life.

Solution. Model

C = C_0 e^{-kt}

, invert:

t = \dfrac{1}{k}\ln\!\left(\dfrac{C_0}{C}\right) = \dfrac{1}{0.12}\ln\!\left(\dfrac{2.5}{0.50}\right)

\ln 5 = 1.6094

, so

t = 1.6094 / 0.12 = 13.4\ \text{d}

. Half-life:

t_{1/2} = \ln 2 / k = 0.6931/0.12 = 5.78\ \text{d}

. Sanity check: dropping from

2.5

0.5

is a factor of

5

, between

2

and

3

halvings;

13.4/5.78 = 2.32

half-lives, and

2.5/2^{2.32} = 0.50

. Consistent.

\boxed{t = 13.4\ \text{d},\;\; t_{1/2} = 5.78\ \text{d}}

Mixing-tank step response (first-order nonhomogeneous ODE)

Problem. A completely mixed tank holds $V = 500\ \text{m}^3$ and is fed at $Q = 50\ \text{m}^3/\text{h}$ . It starts clean ( $C_0 = 0$ ). At $t = 0$ the influent steps to $C_{in} = 20\ \text{mg/L}$ . Find the tank concentration at $t = 15\ \text{h}$ , and the time to reach $18\ \text{mg/L}$ .

Solution. Mass balance

V\,dC/dt = Q(C_{in} - C)

has time constant

\tau = V/Q = 500/50 = 10\ \text{h}

. With

C_0 = 0

C(t) = C_{in}\big(1 - e^{-t/\tau}\big) = 20\big(1 - e^{-t/10}\big)

. At

t = 15\ \text{h}

C = 20\big(1 - e^{-1.5}\big) = 20(1 - 0.2231) = 20(0.7769) = 15.5\ \text{mg/L}

. Time to

18\ \text{mg/L}

18 = 20(1 - e^{-t/10}) \Rightarrow e^{-t/10} = 0.10 \Rightarrow t = -10\ln 0.10 = 10(2.3026) = 23.0\ \text{h}

. Sanity check:

t = 23.0\ \text{h}

2.3\tau

; a first-order approach reaches

90\%

of its final value at exactly

2.30\tau

, and

18/20 = 90\%

. Consistent.

\boxed{C(15\ \text{h}) = 15.5\ \text{mg/L};\;\; t_{18} = 23.0\ \text{h}}

C(t) = 20\big(1 - e^{-t/10}\big);\quad C(15) = 15.5\ \text{mg/L},\;\; t_{18} = 23.0\ \text{h}

Minimum-perimeter detention basin (optimization)

Problem. A rectangular detention basin must enclose a fixed plan area $A = 400\ \text{m}^2$ . To minimize the length of perimeter embankment $P$ , what dimensions should it have?

Solution. Let the basin be

x

A/x

. Perimeter

P(x) = 2x + 2\dfrac{A}{x} = 2x + \dfrac{800}{x}

. Differentiate and set to zero:

\dfrac{dP}{dx} = 2 - \dfrac{800}{x^2} = 0 \Rightarrow x^2 = 400 \Rightarrow x = 20\ \text{m}

. Second derivative

\dfrac{d^2P}{dx^2} = \dfrac{1600}{x^3} > 0

, confirming a minimum. Then

A/x = 400/20 = 20\ \text{m}

— a square — with

P = 2(20) + 2(20) = 80\ \text{m}

. Sanity check: a

25 \times 16

basin (same area) has

P = 2(25+16) = 82\ \text{m} > 80\ \text{m}

, so the square is indeed the minimum.

\boxed{20\ \text{m} \times 20\ \text{m},\;\; P = 80\ \text{m}}

Common pitfalls

•Unit chain in loading integrals: $\int Q\,C\,dt$ in m $^3$ /s, mg/L, and hours must be reconciled with $3600\ \text{s/h}$ and $1000\ \text{L/m}^3$ — which bring the result to milligrams — and then divided by $10^6$ before the mass comes out in kg. Most errors here are conversion slips, not calculus.
•Sign of the decay exponent: $C = C_0 e^{-kt}$ falls with time; writing $e^{+kt}$ (growth) flips the physics. Solving for time, use $t = \tfrac{1}{k}\ln(C_0/C)$ to keep $t$ positive.
•Confusing time constant with half-life: $\tau = 1/k$ but $t_{1/2} = \ln 2 / k = 0.693\,\tau$ . They differ by the factor $\ln 2$ — do not substitute one for the other.
•Forgetting the initial-condition term in the mixing tank: the full solution is $C_{in}(1-e^{-t/\tau}) + C_0 e^{-t/\tau}$ . Dropping $C_0 e^{-t/\tau}$ is only valid when the tank starts clean.
•Misreading $\tau$ for the CSTR: the time constant is $\tau = V/Q$ , the hydraulic residence time — not $Q/V$ . Compute and label $\tau$ 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.

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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