9.3 Calculus, Change Rates, and Area-Volume Reasoning

Derivatives as Rates of Change

In surveying, the derivative is the slope of a curve at a point — most concretely, the grade of a profile. If elevation y is a function of horizontal distance x, then grade = dy/dx. A grade of −2% means dy/dx = −0.02, i.e., elevation drops 2 ft per 100 ft of run. The second derivative d²y/dx² measures how fast the grade itself is changing, which governs vertical curve design.

On an equal-tangent parabolic vertical curve of length L (in stations), the rate of grade change is constant:

r = (g₂ − g₁) / L (percent grade change per station)

where g₁ and g₂ are the entering and exiting grades. The elevation along the curve is y = y₀ + g₁x + (r/2)x², a quadratic whose first derivative dy/dx = g₁ + rx returns the grade at any station. Setting that derivative to zero locates the high or low point: x_turn = −g₁ / r stations from the curve start.

Worked turning point. A curve has g₁ = +3.00%, g₂ = −2.00%, length L = 6 stations (600 ft). Then r = (−2.00 − 3.00)/6 = −0.8333 % per station. The high point is x = −g₁/r = −3.00/(−0.8333) = 3.60 stations = 360 ft past the beginning of the vertical curve. Because g₁ is positive and g₂ negative, a crest (high point) is expected, which the calculation confirms.

Integrals as Accumulated Area and Volume

The integral accumulates a quantity. The area under a profile, the cross-sectional area of a cut, and the volume of earthwork are all integrals. When you have a continuous function, integrate directly; when you have tabulated field data, use a numerical rule.

The trapezoidal rule approximates the area under evenly spaced points (spacing h) as:

Area ≈ h · [ (y₀ + yₙ)/2 + y₁ + y₂ + ... + yₙ₋₁ ]

Worked trapezoidal area. Offsets to an irregular boundary at 50 ft spacing read 12.0, 18.0, 22.0, 19.0, 14.0 ft. With h = 50:

Area ≈ 50 · [ (12.0 + 14.0)/2 + 18.0 + 22.0 + 19.0 ] = 50 · [13.0 + 59.0] = 50 · 72.0 = 3,600 ft²

Simpson's one-third rule is more accurate when the number of intervals is even: Area ≈ (h/3)[y₀ + 4(odd ordinates) + 2(even interior ordinates) + yₙ]. For the same five readings (4 intervals): (50/3)[12.0 + 4(18.0+19.0) + 2(22.0) + 14.0] = (50/3)[12.0 + 148.0 + 44.0 + 14.0] = (50/3)(218.0) = 3,633 ft². The two methods agree closely; Simpson's gives the curved boundary slightly more area.

Earthwork Volumes

Between two surveyed cross-sections of areas A₁ and A₂ separated by distance L, the average end-area method estimates volume as a trapezoidal integral of area along the alignment:

V = L · (A₁ + A₂) / 2

To convert cubic feet to cubic yards, divide by 27.

Worked end-area volume. Two cross-sections 100 ft apart have cut areas A₁ = 220 ft² and A₂ = 280 ft². V = 100 · (220 + 280)/2 = 100 · 250 = 25,000 ft³ = 925.9 yd³.

The prismoidal formula refines this by including a middle section area Am (the area at the midpoint, found from interpolated dimensions — not the average of A₁ and A₂):

V = (L/6) · (A₁ + 4Am + A₂)

This is Simpson's rule applied to area along the alignment and is exact for a prismoid (a solid with plane ends and ruled sides). When the solid is warped or tapers nonlinearly, the prismoidal result is smaller and more accurate than end-area, and the difference (the prismoidal correction) can matter for pay quantities.

Sensitivity, Interpolation, and Average Value

The derivative also answers a practical FS question: how sensitive is a computed result to a small change in an input? If a quantity y depends on x, a small input change Δx produces an output change Δy ≈ (dy/dx)·Δx. This is the differential basis of error propagation. For example, horizontal distance HD = S·cos θ, so d(HD)/dθ = −S·sin θ; at a steep zenith angle the horizontal distance is far more sensitive to an angle error than at a near-level sight. That single derivative tells you where to spend measurement care.

Linear interpolation is the workhorse for reading intermediate values from a table or a profile. Between known points (x₁, y₁) and (x₂, y₂), the value at x is y = y₁ + (x − x₁)·(y₂ − y₁)/(x₂ − x₁). To interpolate an elevation at station 4+50 when station 4+00 reads 612.40 ft and station 5+00 reads 615.90 ft: y = 612.40 + (50/100)·(615.90 − 612.40) = 612.40 + 0.5·3.50 = 614.15 ft. This is just the discrete form of a derivative — a constant slope between the two stations.

The integral also gives an average value. The mean of a function over an interval is (1/(b−a))·∫f dx — concretely, the average depth of a channel from sounded cross-sections, or the average grade over a reach. Numerically, the average ordinate equals the trapezoidal area divided by the base length. These accumulation-and-average ideas are why FS calculus questions almost always wrap a real surveyed quantity (profile, area, volume, or rate) rather than testing differentiation as an abstract skill.