Calculus is fundamental in chemical engineering for analysing rates of reaction, heat and mass transfer, and process dynamics. Derivatives allow engineers to understand how variables change with respect to one another, crucial in reaction kinetics and process control. Integration helps determine quantities such as total accumulation or work done over time or space. This chapter focuses on applying calculus concepts to chemical engineering problems encountered in Kenya’s industries.

6.1 Definition of derivatives of a function

The derivative of a function measures the rate at which the function value changes as its input changes, essential for modelling dynamic chemical processes. It quantifies instantaneous rates such as reaction rate or concentration change. The formal definition uses limits to express the derivative from first principles.

$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$

Worked Examples

Example 1: A concentration of a reactant in a batch reactor is \(C(t) = 5t^2 + 3t\) mol/L, where \(t\) is time in minutes. Find the rate of change of concentration at \(t=2\) min using first principles.

Given: \(C(t) = 5t^2 + 3t\), \(t=2\) min$$ f'(2) = \lim_{h \to 0} \frac{C(2+h) - C(2)}{h} $$$$ = \lim_{h \to 0} \frac{5(2+h)^2 + 3(2+h) - [5(2)^2 + 3(2)]}{h} $$$$ = \lim_{h \to 0} \frac{5(4 + 4h + h^2) + 6 + 3h - (20 + 6)}{h} $$$$ = \lim_{h \to 0} \frac{20 + 20h + 5h^2 + 6 + 3h - 26}{h} $$$$ = \lim_{h \to 0} \frac{20h + 5h^2 + 3h}{h} = \lim_{h \to 0} (20 + 5h + 3) = 23 $$

Answer: 23 mol/L·min

Example 2: For \(f(x) = \sqrt{x}\), find the derivative at \(x=4\) using first principles.

Given: \(f(x) = \sqrt{x}\), \(x=4\)$$ f'(4) = \lim_{h \to 0} \frac{\sqrt{4+h} - \sqrt{4}}{h} $$ Multiply numerator and denominator by conjugate:$$ = \lim_{h \to 0} \frac{(\sqrt{4+h} - 2)(\sqrt{4+h} + 2)}{h(\sqrt{4+h} + 2)} = \lim_{h \to 0} \frac{(4+h) - 4}{h(\sqrt{4+h} + 2)} $$$$ = \lim_{h \to 0} \frac{h}{h(\sqrt{4+h} + 2)} = \lim_{h \to 0} \frac{1}{\sqrt{4+h} + 2} = \frac{1}{2 + 2} = \frac{1}{4} $$

Answer: \(\frac{1}{4}\) units of \(f'(4)\)

Example 3: Given \(f(x) = \frac{1}{x}\), find \(f'(1)\) using first principles.

Given: \(f(x) = \frac{1}{x}\), \(x=1\)$$ f'(1) = \lim_{h \to 0} \frac{\frac{1}{1+h} - 1}{h} = \lim_{h \to 0} \frac{\frac{1 - (1+h)}{1+h}}{h} = \lim_{h \to 0} \frac{-h}{h(1+h)} = \lim_{h \to 0} \frac{-1}{1+h} = -1 $$

Answer: \(-1\) units of \(f'(1)\)

Chapter Summary

This chapter introduces the concept of derivatives as the fundamental measure of how a function changes, starting with differentiation from first principles. It includes tables of common derivatives and explains the essential rules of differentiation that simplify the process. The chapter explores the practical significance of derivatives in calculating rates of change and small changes in quantities. It also covers the identification of turning and stationary points for functions of two variables, which are critical in optimizing engineering designs. Integration is defined as the reverse process of differentiation, with a focus on both indefinite and definite integrals. Various methods of integration are presented, including those involving hyperbolic and inverse functions. Finally, the chapter demonstrates the application of integration techniques to solve real-world engineering problems.