For f(x) = x³ − 6x² + 9x, where does f have a local maximum?
x = 1 Set the first derivative to zero: f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3), so the critical points are x = 1 and x = 3. The second derivative is f″(x) = 6x − 12. At x = 1 it’s −6, which is negative, so that point is a local maximum.
What is the determinant of the matrix [[4, 2], [3, 5]]?
14 For a 2 × 2 matrix [[a, b], [c, d]], the determinant is ad − bc. Here that’s (4)(5) − (2)(3) = 20 − 6 = 14.
Evaluate the definite integral of 3x² dx from x = 0 to x = 2.
8 An antiderivative of 3x² is x³. Evaluate it at the limits: 2³ − 0³ = 8.
A simply supported beam is 10 m long, with supports at each end (A on the left, B on the right). A single 20 kN load acts 4 m from support A. Ignoring the beam’s weight, what is the vertical reaction at A?
12 kN Take moments about B. The load is 6 m from B, so R_A × 10 = 20 × 6, and R_A = 12 kN. Check: R_B = 20 − 12 = 8 kN, and moments about A give 8 × 10 = 20 × 4 = 80. It balances.
Two forces act on a particle: 30 N pointing east and 40 N pointing north. What is the size of the resultant force?
50 N The forces are at right angles, so the resultant is the hypotenuse: √(30² + 40²) = √(900 + 1,600) = √2,500 = 50 N.
You push with 50 N at the end of a wrench, at right angles to the handle. The distance from the bolt to where you push is 0.4 m. What moment do you apply to the bolt?
20 N·m Moment equals force times perpendicular distance: M = 50 N × 0.4 m = 20 N·m.
A car starts from rest and accelerates uniformly at 3 m/s² for 4 s. How far does it travel in that time?
24 m From rest with constant acceleration, s = ½at² = ½ × 3 × 4² = ½ × 3 × 16 = 24 m.
A 2 kg ball moves at 6 m/s. What is its kinetic energy?
36 J Kinetic energy is ½mv² = ½ × 2 × 6² = ½ × 2 × 36 = 36 J.
A steel rod with a cross-sectional area of 500 mm² carries an axial tensile load of 50 kN. What is the normal stress in the rod?
100 MPa Stress is force divided by area: σ = 50,000 N ÷ 500 mm² = 100 N/mm², and 1 N/mm² = 1 MPa. So σ = 100 MPa.
A steel bar 2 m long with a cross-sectional area of 1,000 mm² carries an axial load of 100 kN. Take E = 200 GPa. How much does the bar stretch?
1 mm Elongation is δ = PL/(AE). In N and mm: δ = (100,000 × 2,000) ÷ (1,000 × 200,000) = 200,000,000 ÷ 200,000,000 = 1 mm. (200 GPa = 200,000 N/mm².)
What is the gauge pressure at a depth of 10 m in fresh water? Use ρ = 1,000 kg/m³ and g = 9.81 m/s².
98.1 kPa Hydrostatic gauge pressure is p = ρgh = 1,000 × 9.81 × 10 = 98,100 Pa, which is 98.1 kPa.
Water flows at 2 m/s in a pipe with an inside diameter of 100 mm. The pipe narrows to 50 mm. Assuming incompressible flow, what is the velocity in the narrow section?
8 m/s Continuity says A₁V₁ = A₂V₂. Area goes with diameter squared, so halving the diameter cuts the area to a quarter. The velocity must go up four times: 2 × 4 = 8 m/s.
Water with a kinematic viscosity of 1.0 × 10⁻⁶ m²/s flows at 1 m/s through a pipe with a 0.1 m inside diameter. What is the Reynolds number, and what kind of flow is it?
100,000; turbulent Re = VD/ν = (1 × 0.1) ÷ (1.0 × 10⁻⁶) = 100,000. In a pipe, flow is laminar only at low Reynolds numbers, around 2,000 or below, so 100,000 is well into turbulent flow.
A heat engine works between a hot reservoir at 327°C and a cold reservoir at 27°C. What is the highest thermal efficiency it could possibly have?
50% The highest possible efficiency is the Carnot efficiency, 1 − T_cold/T_hot, using absolute temperatures. 327°C = 600 K and 27°C = 300 K, so the efficiency is 1 − 300/600 = 0.50, or 50%.
How much heat does it take to warm 2 kg of liquid water by 10°C? Use a specific heat of 4.18 kJ/(kg·K).
83.6 kJ Sensible heat is Q = mcΔT = 2 × 4.18 × 10 = 83.6 kJ. A change of 10°C is the same as a change of 10 K.
A 6 Ω resistor and a 3 Ω resistor are connected in parallel. What is their equivalent resistance?
2 Ω For two resistors in parallel, R = (R₁R₂)/(R₁ + R₂) = (6 × 3)/(6 + 3) = 18/9 = 2 Ω.
A 12 V source is connected across a 4 Ω resistor. How much power does the resistor dissipate?
36 W P = V²/R = 12² ÷ 4 = 144 ÷ 4 = 36 W. You can check with the current: I = 12 ÷ 4 = 3 A, and P = VI = 12 × 3 = 36 W.
What is the present worth of $10,000 to be received 5 years from now, at 6% interest compounded annually?
$7,473 Present worth is P = F ÷ (1 + i)ⁿ = 10,000 ÷ 1.06⁵ = 10,000 ÷ 1.3382 ≈ $7,473. This is the (P/F, 6%, 5) factor, about 0.7473.
You invest $5,000 today at 8% interest compounded annually. How much will you have after 3 years?
$6,299 Future worth is F = P(1 + i)ⁿ = 5,000 × 1.08³ = 5,000 × 1.259712 ≈ $6,299.
A licensed engineer finds a design flaw that could endanger the public. The engineer’s employer refuses to fix it. Under the NCEES Model Rules of Professional Conduct, what should the engineer do?
Tell the employer or client and the appropriate authority The Model Rules put the safety, health and welfare of the public first. If a licensee’s engineering judgment is overruled where public safety is at risk, they must inform their employer or client and the appropriate authority.