Derivatives & Integrals in Physical Models

A rate answers “how fast is Q changing right now?”, and “right now” needs a limit. Take the chord slope ΔQ/Δt across an interval, then shrink the interval: the chord slopes settle on a single number, dQ/dt, the slope of the tangent at that instant. A leaking bucket holding m(t) = (12.0 kg)exp(−t/τ), with τ = 40 s, shows the limit arriving. From t = 20 s, chords 10 s, 1.0 s and 0.10 s wide give −0.161, −0.180 and −0.1817 kg s⁻¹, closing on the exact slope −(12.0 kg ÷ 40 s)exp(−0.5) = −0.182 kg s⁻¹. The widest chord is about 12% too shallow, because the leak slows across those 10 s. Two checks come free with every derivative. Its unit is the unit of Q divided by the unit of t, and its sign says whether Q is rising or falling at that instant, independent of how large Q is.

Physical models are usually written as rates. Newton's second law fixes dv/dt, a leak fixes dm/dt, a motor fixes dE/dt, because what nature settles locally is how a quantity changes in the next instant, not what its value is. The derivative dQ/dt is that local statement: the limit of ΔQ/Δt as the interval shrinks, drawn as the slope of the Q–t graph and carrying the unit of Q per unit of time. The definite integral runs the argument backwards, adding r dt over thin slices until the net change appears as the signed area under the rate graph. The two operations undo each other with one gap: a rate fixes the shape of Q(t) but not its level, so an absolute prediction needs one measured value, the initial condition, to pin that level down. With data instead of formulas, the limits become finite differences and sums whose truncation errors typically shrink as h or h² for sufficiently smooth functions. At an idealised jump in a finite rate, the accumulated quantity may have a corner: its derivative need not exist there, but piecewise integration still works. And measured values carry noise, which a difference quotient divides by h: shrink the step far enough and the derivative gets worse, not better.

A derivative dQ/dt is how fast a quantity changes at one instant, the slope of its graph; a definite integral of that rate over an interval is the net change it builds up, the signed area under the rate graph. A motor feeds a flywheel power P(t) = (40 W s⁻¹)t. At t = 5.0 s the stored energy is rising at dE/dt = 200 W, and over those 5.0 s it has gained ΔE = ∫ (40 W s⁻¹)t dt, taken from t = 0 to 5.0 s, which is (20 W s⁻¹)(5.0 s)² = 500 J.

Run the argument backwards. Across a thin slice dt the quantity changes by r dt, so across a whole interval the net change is the sum of every slice, which in the limit of thin slices is the definite integral ΔQ = ∫ r dt. On the rate graph each slice is a strip of height r and width dt, so ΔQ is the area under the curve, carrying the unit of r × t: L s⁻¹ × s = L, W × s = J, N × s = kg m s⁻¹. The area is signed. A net force F(t) = (6.0 N) − (1.2 N s⁻¹)t on a trolley delivers +15 N s of momentum in the first 5.0 s and −15 N s in the next 5.0 s, so over 10 s the integral is zero: the trolley leaves with the momentum it arrived with, though it was pushed hard both ways. ∫F dt answers “how much did p change?” The total push, ∫|F| dt = 30 N s, is a different question with a different integral.

Differentiation throws information away. Q(t) and Q(t) + C have the same slope at every instant, so a rate on its own pins Q down only up to a constant. One measured value restores it. Write Q(t) = Q₀ + ∫ r(t′) dt′ from t₀ to t, with t′ a dummy variable and Q₀ = Q(t₀): differentiating returns r(t), and setting t = t₀ returns Q₀, so both halves of the model hold. The measurement need not be at t = 0. A tank filling at r(t) = (0.60 L s⁻¹) − (0.020 L s⁻²)t has V(t) = C + 0.60t − 0.010t², and a reading of 30.0 L at t = 10 s gives 30.0 = C + 6.0 − 1.0, so C = 25.0 L. Each integration costs one condition. Newton's second law fixes acceleration, so position needs two: a = dv/dt leaves one constant and v = dx/dt leaves another, which is why a launch needs both a starting position and a starting velocity.

The pairing works on graphs before any algebra. The height of the rate graph at a time is the slope of the Q graph at that time, and the area under the rate graph between two times is the rise of the Q graph between them. Three readings follow: Q has a peak or a trough wherever r crosses zero, Q rises most steeply wherever r is largest, and Q curves upward wherever r is increasing. Grain pours into a hopper at a rate that climbs steadily from zero to 2.0 kg s⁻¹ at t = 4.0 s, then falls steadily back to zero at 10 s. The mass never decreases, because the rate is never negative. It gains the triangle's area, ½ × 10 s × 2.0 kg s⁻¹ = 10 kg. It grows fastest at 4.0 s, yet only 4.0 kg has arrived by then, and the halfway mass of 5.0 kg arrives at 10 − √30 = 4.5 s. The fastest instant and the halfway instant are different, and only the area locates the second.