Calculus is fundamental in water engineering for analyzing rates of change, optimizing system designs, and predicting behaviour of hydraulic systems. Derivatives represent how quantities such as flow rate, pressure, or water level change with respect to time or space. Mastery of derivative operations allows water engineers to model dynamic systems and solve real-world problems efficiently in Kenya’s diverse water infrastructure projects.

1.1 Determination of derivatives Calculus functions Operations

This topic covers methods to find derivatives of functions commonly encountered in water engineering. Derivatives quantify the instantaneous rate of change of a variable, essential for understanding flow dynamics, sediment transport, and pressure variations in pipes and channels.

1.1.1 Derivative of Basic Polynomial Functions

The derivative of a polynomial function expresses how the function changes with respect to its independent variable, usually time \(t\) or distance \(x\). The power rule is the fundamental operation:

$$\frac{d}{dx} ( x^n ) = n x^{n-1}$$

Worked Examples

Example 1:
Given \(Q(x) = 5x^3\) representing flow rate in cubic meters per second along distance \(x\) in meters, find \(\frac{dQ}{dx}\).

$$\frac{dQ}{dx} = \frac{d}{dx} (5x^3)$$$$= 5 \times \frac{d}{dx} (x^3)$$$$= 5 \times 3x^{2}$$$$= 15x^{2}$$

Answer: \(\frac{dQ}{dx} = 15x^{2} \, \text{m}^3/\text{s}/\text{m}\)

Example 2:
Water depth \(h(t) = 2t^4 - 3t^2 + 7\) meters varies with time \(t\) in seconds. Find \(\frac{dh}{dt}\).

$$\frac{dh}{dt} = \frac{d}{dt} (2t^4 - 3t^2 + 7)$$$$= 2 \times 4t^{3} - 3 \times 2t^{1} + 0$$$$= 8t^{3} - 6t$$

Answer: \(\frac{dh}{dt} = 8t^{3} - 6t \, \text{m/s}\)

Example 3:
Discharge \(Q(x) = 4x^5 - 6x^3 + x\) m³/s along a pipe length \(x\) m. Find \(\frac{dQ}{dx}\).

$$\frac{dQ}{dx} = \frac{d}{dx} (4x^5 - 6x^3 + x)$$$$= 4 \times 5x^{4} - 6 \times 3x^{2} + 1$$$$= 20x^{4} - 18x^{2} + 1$$

Answer: \(\frac{dQ}{dx} = 20x^{4} - 18x^{2} + 1 \, \text{m}^3/\text{s}/\text{m}\)

1.1.2 Derivative of Exponential and Logarithmic Functions

Exponential and logarithmic functions model phenomena such as pollutant decay and infiltration rates. Their derivatives follow specific rules:

$$\frac{d}{dx} (e^{kx}) = k e^{kx}$$$$\frac{d}{dx} (\ln x) = \frac{1}{x}$$

Worked Examples

Example 1:
Pollutant concentration \(C(t) = 50 e^{-0.1 t}\) mg/L decreases over time \(t\) in hours. Find \(\frac{dC}{dt}\).

$$\frac{dC}{dt} = \frac{d}{dt} (50 e^{-0.1 t})$$$$= 50 \times (-0.1) e^{-0.1 t}$$$$= -5 e^{-0.1 t}$$

Answer: \(\frac{dC}{dt} = -5 e^{-0.1 t} \, \text{mg/L/hr}\)

Example 2:
Infiltration rate \(I(x) = 3 \ln x\) mm/hr depends on soil depth \(x\) in meters. Find \(\frac{dI}{dx}\).

$$\frac{dI}{dx} = 3 \times \frac{1}{x}$$$$= \frac{3}{x}$$

Answer: \(\frac{dI}{dx} = \frac{3}{x} \, \text{mm/hr/m}\)

Example 3:
Water pressure \(P(t) = 100 e^{0.05 t}\) kPa varies with time \(t\) in seconds. Find \(\frac{dP}{dt}\).

$$\frac{dP}{dt} = 100 \times 0.05 e^{0.05 t}$$$$= 5 e^{0.05 t}$$

Answer: \(\frac{dP}{dt} = 5 e^{0.05 t} \, \text{kPa/s}\)

1.1.3 Derivative of Trigonometric Functions

Trigonometric functions describe oscillatory behaviours such as tidal fluctuations and wave motion. Their derivatives are:

$$\frac{d}{dx} (\sin x) = \cos x$$$$\frac{d}{dx} (\cos x) = -\sin x$$

Worked Examples

Example 1:
Tidal height \(H(t) = 2 \sin (\frac{\pi t}{6})\) meters, where \(t\) is time in hours. Find \(\frac{dH}{dt}\).

$$\frac{dH}{dt} = 2 \times \cos (\frac{\pi t}{6}) \times \frac{\pi}{6}$$$$= \frac{\pi}{3} \cos (\frac{\pi t}{6})$$

Answer: \(\frac{dH}{dt} = \frac{\pi}{3} \cos (\frac{\pi t}{6}) \, \text{m/hr}\)

Example 2:
Flow velocity \(v(x) = 4 \cos (2x)\) m/s varies with distance \(x\) m. Find \(\frac{dv}{dx}\).

$$\frac{dv}{dx} = 4 \times (-\sin (2x)) \times 2$$$$= -8 \sin (2x)$$

Answer: \(\frac{dv}{dx} = -8 \sin (2x) \, \text{m/s/m}\)

Example 3:
Water level oscillation \(L(t) = 3 \sin (5t) + 2 \cos (3t)\) meters, \(t\) in seconds. Find \(\frac{dL}{dt}\).

$$\frac{dL}{dt} = 3 \times 5 \cos (5t) - 2 \times 3 \sin (3t)$$$$= 15 \cos (5t) - 6 \sin (3t)$$

Answer: \(\frac{dL}{dt} = 15 \cos (5t) - 6 \sin (3t) \, \text{m/s}\)

1.1.4 Rules for Derivatives of Composite Functions

Water engineering often involves composite functions such as flow rate depending on depth and velocity. The chain rule is essential:

$$\frac{d}{dx} f(g(x)) = f'(g(x)) \times g'(x)$$

Worked Examples

Example 1:
Discharge \(Q(t) = (3t^2 + 1)^4\) m³/s, find \(\frac{dQ}{dt}\).

$$\frac{dQ}{dt} = 4 (3t^2 + 1)^3 \times \frac{d}{dt} (3t^2 + 1)$$$$= 4 (3t^2 + 1)^3 \times 6t$$$$= 24 t (3t^2 + 1)^3$$

Answer: \(\frac{dQ}{dt} = 24 t (3t^2 + 1)^3 \, \text{m}^3/\text{s}^2\)

Example 2:
Water level \(h(x) = \sin (2x^2)\) meters along distance \(x\) in meters. Find \(\frac{dh}{dx}\).

$$\frac{dh}{dx} = \cos (2x^2) \times \frac{d}{dx} (2x^2)$$$$= \cos (2x^2) \times 4x$$$$= 4x \cos (2x^2)$$

Answer: \(\frac{dh}{dx} = 4x \cos (2x^2) \, \text{m/m}\)

Example 3:
Pressure \(P(t) = e^{3t^2 + 2t}\) kPa, find \(\frac{dP}{dt}\).

$$\frac{dP}{dt} = e^{3t^2 + 2t} \times \frac{d}{dt} (3t^2 + 2t)$$$$= e^{3t^2 + 2t} \times (6t + 2)$$

Answer: \(\frac{dP}{dt} = (6t + 2) e^{3t^2 + 2t} \, \text{kPa/s}\)

Chapter Summary

This chapter covered the determination of derivatives for various calculus functions and the fundamental operations involved in differentiation. It explored derivative functions in detail, including the differentiation of trigonometric and hyperbolic functions, as well as inverse trigonometric functions through targeted exercises. The concepts of rate of change and small change were examined to understand their practical applications. Techniques for calculating stationary points involving two variables were also performed, providing insight into critical points analysis. The chapter then shifted focus to the determination of integral functions, emphasizing the integration process. Finally, it addressed the integration of algebraic functions, with specific attention to trigonometric, logarithmic, hyperbolic, and inverse functions, completing a comprehensive overview of calculus operations.