Differential calculus is essential for Science Laboratory Technology professionals in Kenya, enabling precise analysis of changing quantities in experiments and processes. Mastery of differentiation allows laboratory technologists to model rates of reaction, concentration changes, and physical phenomena accurately. This chapter equips students with the skills to apply differential calculus to typical laboratory problems, enhancing data interpretation and experimental design.
Differential calculus studies how functions change by examining their derivatives. In Science Laboratory Technology, this is critical for understanding reaction rates, sensor readings, and dynamic system behavior. This section covers functional notation, standard differentiation rules, and differential equations relevant to laboratory contexts.
Functional notation provides a concise way to represent variables dependent on one another, which is fundamental to describing laboratory measurements varying with time or other parameters.
The function of a variable \(x\) is denoted as \(f(x)\), representing the output value for each input \(x\).
$$f(x) = \text{expression involving } x$$
Example 1: A laboratory sensor measures temperature \(T\) varying with time \(t\) as \(T(t) = 5t + 20\). Find the temperature at \(t = 4\) seconds.
Given: \(T(t) = 5t + 20\), \(t = 4\) s
$$T(4) = 5 \times 4 + 20$$
$$= 20 + 20$$
Answer: 40^\circ C
Example 2: The concentration \(C\) of a solution changes with volume \(V\) as \(C(V) = \frac{100}{V}\). Find \(C\) when \(V = 5\) mL.
Given: \(C(V) = \frac{100}{V}\), \(V = 5\) mL
$$C(5) = \frac{100}{5}$$
$$= 20 \text{ units}$$
Answer: 20 units
Example 3: A chemical reaction rate \(r\) depends on temperature \(T\) as \(r(T) = 3T^2 + 2T + 1\). Find the rate at \(T = 3^\circ C\).
Given: \(r(T) = 3T^2 + 2T + 1\), \(T = 3^\circ C\)
$$r(3) = 3 \times 3^2 + 2 \times 3 + 1$$
$$= 3 \times 9 + 6 + 1$$
$$= 27 + 6 + 1$$
Answer: 34 units
Differentiation calculates the rate of change of a function with respect to its variable, which in laboratory settings models how quantities such as concentration or temperature vary over time.
The derivative of a function \(f(x)\) with respect to \(x\) is denoted as \(\frac{df}{dx}\) or \(f'(x)\).
$$\frac{d}{dx}[f(x)] = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
Common differentiation rules include:
Example 1: Differentiate \(f(x) = 4x^3\).
Given: \(f(x) = 4x^3\)
$$\frac{d}{dx}[4x^3] = 4 \times \frac{d}{dx}[x^3]$$
$$= 4 \times 3x^{2}$$
$$= 12x^{2}$$
Answer: 12x²
Example 2: Find \(\frac{d}{dt}[7t^2 + 5t + 3]\).
Given: \(f(t) = 7t^2 + 5t + 3\)
$$\frac{d}{dt}[7t^2 + 5t + 3] = 7 \times 2t + 5 \times 1 + 0$$
$$= 14t + 5$$
Answer: 14t + 5
Example 3: Differentiate the reaction concentration function \(C(V) = 50V^{-1}\).
Given: \(C(V) = 50V^{-1}\)
$$\frac{d}{dV}[50V^{-1}] = 50 \times (-1) V^{-2}$$
$$= -50 V^{-2}$$
Answer: \(-\frac{50}{V^2}\)
Example 4: Differentiate \(f(x) = 3x^4 - 2x^3 + x - 5\).
Given: \(f(x) = 3x^4 - 2x^3 + x - 5\)
$$\frac{d}{dx}[3x^4 - 2x^3 + x - 5] = 3 \times 4x^{3} - 2 \times 3x^{2} + 1 - 0$$
$$= 12x^{3} - 6x^{2} + 1$$
Answer: \(12x^3 - 6x^2 + 1\)
Example 5: Differentiate the pressure function \(P(t) = 2t^3 + 4t^2 + 6t + 8\).
Given: \(P(t) = 2t^3 + 4t^2 + 6t + 8\)
$$\frac{d}{dt}[2t^3 + 4t^2 + 6t + 8] = 2 \times 3t^{2} + 4 \times 2t + 6 + 0$$
$$= 6t^{2} + 8t + 6$$
Answer: \(6t^2 + 8t + 6\)
Differential equations relate a function to its derivatives and are used to model dynamic laboratory systems such as reaction kinetics, heat transfer, and population growth of microorganisms.
A first-order differential equation has the form:
$$\frac{dy}{dx} = f(x, y)$$
Solutions to differential equations can be explicit functions or implicit relations defining the behaviour of the system.
Example 1: Solve \(\frac{dy}{dx} = 3x^2\) with initial condition \(y(0) = 4\).
Given: \(\frac{dy}{dx} = 3x^2\), \(y(0) = 4\)
Integrate both sides:
$$y = \int 3x^2 dx = 3 \times \frac{x^3}{3} + C = x^3 + C$$
Use initial condition:
$$y(0) = 0^3 + C = 4 \Rightarrow C = 4$$
Answer: \(y = x^3 + 4\)
Example 2: Solve \(\frac{dC}{dt} = -0.5C\), where \(C\) is concentration and \(C(0) = 10\).
Given: \(\frac{dC}{dt} = -0.5C\), \(C(0) = 10\)
Separate variables:
$$\frac{dC}{C} = -0.5 dt$$
Integrate both sides:
$$\int \frac{1}{C} dC = \int -0.5 dt$$
$$\ln |C| = -0.5 t + K$$
Exponentiate:
$$C = e^{K} e^{-0.5 t} = C_0 e^{-0.5 t}$$
Use initial condition:
$$C(0) = C_0 = 10$$
Answer: \(C = 10 e^{-0.5 t}\)
Example 3: Solve \(\frac{dy}{dx} = 2y\) with \(y(1) = 3\).
Given: \(\frac{dy}{dx} = 2y\), \(y(1) = 3\)
Separate variables:
$$\frac{dy}{y} = 2 dx$$
Integrate:
$$\int \frac{1}{y} dy = \int 2 dx$$
$$\ln |y| = 2x + C$$
Exponentiate:
$$y = e^{C} e^{2x} = A e^{2x}$$
Use initial condition:
$$y(1) = A e^{2} = 3 \Rightarrow A = \frac{3}{e^{2}}$$
Answer: \(y = \frac{3}{e^{2}} e^{2x}\)
Example 4: Solve the differential equation \(\frac{dT}{dt} = 4 - 0.1T\) with \(T(0) = 20\).
Given: \(\frac{dT}{dt} = 4 - 0.1T\), \(T(0) = 20\)
Rewrite:
$$\frac{dT}{dt} + 0.1 T = 4$$
Integrating factor:
$$\mu(t) = e^{\int 0.1 dt} = e^{0.1 t}$$
Multiply both sides:
$$e^{0.1 t} \frac{dT}{dt} + 0.1 e^{0.1 t} T = 4 e^{0.1 t}$$
Left side is derivative:
$$\frac{d}{dt}[T e^{0.1 t}] = 4 e^{0.1 t}$$
Integrate both sides:
$$T e^{0.1 t} = \int 4 e^{0.1 t} dt + C$$
$$= 4 \times \frac{e^{0.1 t}}{0.1} + C = 40 e^{0.1 t} + C$$
Divide both sides by \(e^{0.1 t}\):
$$T = 40 + C e^{-0.1 t}$$
Use initial condition:
$$20 = 40 + C e^{0} \Rightarrow C = -20$$
Answer: \(T = 40 - 20 e^{-0.1 t}\)
Example 5: Solve \(\frac{dy}{dx} = \frac{x}{y}\) with \(y(0) = 1\).
Given: \(\frac{dy}{dx} = \frac{x}{y}\), \(y(0) = 1\)
Separate variables:
$$y dy = x dx$$
Integrate both sides:
$$\int y dy = \int x dx$$
$$\frac{y^2}{2} = \frac{x^2}{2} + C$$
Multiply both sides by 2:
$$y^2 = x^2 + 2C$$
Use initial condition:
$$1^2 = 0^2 + 2C \Rightarrow 2C = 1 \Rightarrow C = \frac{1}{2}$$
Answer: \(y^2 = x^2 + 1\)
For the function \(f(t) = 6t^2 + 3t + 2\), find the value of \(f(5)\). (2 marks)
Differentiate \(g(x) = 8x^5 - 4x^3 + x\). (4 marks)
Given \(\frac{dy}{dx} = 4x^3\) and \(y(0) = 2\), find \(y\) as a function of \(x\). (5 marks)
Solve the differential equation \(\frac{dC}{dt} = -0.3 C\), with \(C(0) = 15\). (5 marks)
For the temperature model \(\frac{dT}{dt} = 5 - 0.2 T\), find \(T\) if \(T(0) = 25\). (6 marks)
Create a free account to open more of this chapter.
Free: practical guides, quick cards, workplace scenarios and more.
Create a free accountThis chapter introduced differential calculus by first explaining functional notation and the concept of standard differentiation before exploring differential equations. Various methods of differentiation were covered, including differentiation by first principle, followed by the product rule, quotient rule, and chain rule to handle more complex functions. The chapter then extended to derivatives of higher order functions, highlighting how successive differentiation provides deeper insights into function behavior. Applications of differentiation were discussed in detail, focusing on finding normals and tangents to curves, identifying stationary points such as maxima, minima, and points of inflection, as well as calculating rates of change and analyzing small changes in variables. Finally, the differentiation of inverse trigonometric functions was examined, completing the comprehensive overview of differential calculus techniques and their practical uses in scientific and engineering contexts.
Differentiate the function \( f(x) = 3x^2 + 5x - 7 \). (2 marks)
Find \(\frac{dy}{dx}\) if \( y = \frac{4}{x} + 2x^3 \). (2 marks)