Source: ENGR 216 Comprehensive Exam Practice Bank | Texas A&M University
Tags: finite differences, forward difference, numerical derivative, Fourier's law, momentum, normal distribution, z-score, probability, standard deviation, ENGR 216
When you have discrete data points instead of a continuous function, finite difference methods approximate derivatives numerically. Forward difference uses the current point and the next one. Separately, the normal distribution is the backbone of engineering statistics: z-scores convert any normal variable into a standard form so you can look up probabilities in a table.
Finite difference method
A numerical technique for approximating derivatives from tabulated data. Three main types: forward, backward, and central.
Forward difference
Uses the current data point and the next one: df/dx ≈ (f(x_{i+1}) − f(x_i)) / (x_{i+1} − x_i). Lower accuracy than central difference, but only needs data ahead of the point.
Backward difference
Uses the current data point and the previous one: df/dx ≈ (f(x_i) − f(x_{i−1})) / (x_i − x_{i−1}).
Central difference
Uses the points on either side: df/dx ≈ (f(x_{i+1}) − f(x_{i−1})) / (x_{i+1} − x_{i−1}). More accurate than forward or backward alone.
Fourier's Law of Heat Conduction
q = k · A · (dT/dx), where k is thermal conductivity, A is cross-sectional area, and dT/dx is the temperature gradient.
Normal distribution (Gaussian distribution)
A symmetric, bell-shaped probability distribution fully described by its mean (μ) and standard deviation (σ). About 68% of values fall within ±1σ, 95% within ±2σ, and 99.7% within ±3σ.
Z-score (standard score)
z = (x − μ) / σ. Converts any normally distributed variable into the standard normal distribution (μ = 0, σ = 1) so you can use a single probability table.
Given data:
Point 4: x = 0.04 m, T = 82.6°C
Point 5: x = 0.05 m, T = 94.5°C
The forward difference approximation for dT/dx at Point 4:
dT/dx ≈ (T₅ − T₄) / (x₅ − x₄) = (94.5 − 82.6) / (0.05 − 0.04) = 11.9 / 0.01 = 1190 °C/m
Applying Fourier's Law:
q = k · A · (dT/dx) = 0.053 × 0.73 × 1190 ≈ 46.0 kW
A 2.0 kg ball at x = 9.1 cm when t = 0.3 s and x = 15.8 cm when t = 0.4 s.
Velocity at t = 0.3 s (forward difference):
v = (x₂ − x₁) / (t₂ − t₁) = (15.8 − 9.1) / (0.4 − 0.3) = 6.7 / 0.1 = 67 cm/s = 0.67 m/s
Momentum:
p = m · v = 2.0 × 0.67 = 1.34 kg·m/s
For any normal distribution problem, the workflow is:
Identify μ, σ, and the boundary value x.
Compute z = (x − μ) / σ.
Look up the cumulative probability from a z-table or use the complement as needed.
μ = 200 m, σ = 30 m. Damage occurs if the parachute opens below 100 m.
z = (100 − 200) / 30 = −100/30 = −3.33
P(Z < −3.33) ≈ 0.0004 (from z-table)
The probability of damage is roughly 0.04%, which is very small but not zero.
The filter catches 90% of particles, meaning 10% pass through. Particles are normally distributed with μ = 0.5 microns, σ = 0.2 microns.
The filter catches the largest 90%, so the 10% that pass through are the smallest particles. We need the diameter below which 10% of particles fall.
z for P(Z < z) = 0.10 is z ≈ −1.28
x = μ + z · σ = 0.5 + (−1.28)(0.2) = 0.5 − 0.256 = 0.244 microns
The largest particle that passes through the filter is approximately 0.24 microns in diameter.
Forward difference: df/dx ≈ (f_{i+1} − f_i) / (x_{i+1} − x_i)
Backward difference: df/dx ≈ (f_i − f_{i−1}) / (x_i − x_{i−1})
Central difference: df/dx ≈ (f_{i+1} − f_{i−1}) / (x_{i+1} − x_{i−1})
Fourier's Law: q = k · A · (dT/dx)
Momentum: p = m · v
Z-score: z = (x − μ) / σ
⚠️ The exam will specify which difference method to use (forward, backward, or central). Read the question carefully and do not default to central difference unless instructed.
⚠️ Watch your units in Fourier's Law problems. If k is in kW/(m·°C), your answer will be in kW. Convert if the question asks for watts.
⚠️ For normal distribution problems, draw a quick sketch of the bell curve and shade the region you need. This prevents sign errors and helps you decide whether to use P(Z < z) or 1 − P(Z < z).
⚠️ The air filtration problem requires careful reading: "catches 90%" means the 10th percentile passes through, not the 90th.
Q: When would you choose forward difference over central difference?
A: When you only have data at the current point and the next point (no previous point available), or when the problem specifically requires forward difference.
Q: A temperature reading at x = 0.02 m is 60°C and at x = 0.03 m is 71°C. What is the forward-difference estimate of dT/dx at x = 0.02 m?
A: (71 − 60) / (0.03 − 0.02) = 11 / 0.01 = 1100 °C/m.
Q: If a z-score is −2.5, what does that mean physically?
A: The value is 2.5 standard deviations below the mean. In a standard normal table, P(Z < −2.5) ≈ 0.0062, so about 0.62% of values fall below this point.
Q: In the parachute problem, what would the probability of damage be if σ increased to 50 m?
A: z = (100 − 200)/50 = −2.0, giving P(Z < −2.0) ≈ 0.0228, or about 2.3%. A larger standard deviation means less predictable opening altitude and higher risk.
finite difference, forward difference, backward difference, central difference, numerical derivative, Fourier's law, thermal conductivity, heat conduction, temperature gradient, momentum, velocity, normal distribution, Gaussian, bell curve, z-score, standard score, cumulative probability, z-table, percentile, ENGR 216, Texas A&M