By the end of this chapter, you will be able to:
Mastering these skills will help you tackle real-world engineering problems with confidence and precision.
Biomedical engineering professionals in Kenya rely heavily on mathematical tools to design, analyze, and troubleshoot biomedical devices and systems. Algebra forms the foundation for solving equations related to physiological measurements, device parameters, and system behaviors. Mastery of algebraic concepts such as indices and logarithms is essential for interpreting sensor data, modeling biological processes, and calibrating instruments accurately.
Algebra is the language of engineering calculations, enabling biomedical engineers to formulate relationships between variables such as voltage, current, resistance, and biological signals. It allows systematic manipulation of formulas to isolate unknowns, optimize device performance, and solve for critical parameters in complex systems. In Kenya’s healthcare infrastructure, accurate algebraic problem-solving supports the development and maintenance of equipment ranging from ECG machines to dialysis units. Algebraic skills also underpin data analysis, error correction, and algorithm development in biomedical instrumentation and diagnostics.
Algebra helps translate real-world biomedical scenarios into mathematical expressions that can be solved systematically.
Example 1: A pulse oximeter outputs a voltage \(V\) related to the oxygen saturation \(S\) by the equation \(V = 0.8S + 0.2\). Find \(S\) when \(V = 0.92\) volts.
Given: \(V = 0.92\) V, \(V = 0.8S + 0.2\)
$$0.92 = 0.8S + 0.2$$
$$0.8S = 0.92 - 0.2$$
$$0.8S = 0.72$$
$$S = \frac{0.72}{0.8}$$
$$S = 0.9$$
Answer: Oxygen saturation \(S = 0.9\) (or 90%)
Example 2: The resistance \(R\) of a biomedical sensor varies with temperature \(T\) as \(R = 100 + 0.5T\) ohms. Find \(T\) if the resistance measured is 125 ohms.
Given: \(R = 125\) Ω, \(R = 100 + 0.5T\)
$$125 = 100 + 0.5T$$
$$0.5T = 125 - 100$$
$$0.5T = 25$$
$$T = \frac{25}{0.5}$$
$$T = 50$$
Answer: Temperature \(T = 50^\circ C\)
Example 3: A glucose sensor’s output current \(I\) is related to glucose concentration \(C\) by \(I = 2C - 0.1\) mA. If the output current is 3.9 mA, find the glucose concentration.
Given: \(I = 3.9\) mA, \(I = 2C - 0.1\)
$$3.9 = 2C - 0.1$$
$$2C = 3.9 + 0.1$$
$$2C = 4.0$$
$$C = \frac{4.0}{2}$$
$$C = 2.0$$
Answer: Glucose concentration \(C = 2.0\) mmol/L
Biomedical engineering often involves systems with multiple variables requiring simultaneous equations to find unknowns.
Example 1: In a circuit, two equations are given: \(3V + 2I = 12\) and \(V - I = 2\). Find \(V\) and \(I\).
Given: \(3V + 2I = 12\), \(V - I = 2\)
From second equation:
$$V = I + 2$$
Substitute into first:
$$3(I + 2) + 2I = 12$$
$$3I + 6 + 2I = 12$$
$$5I + 6 = 12$$
$$5I = 6$$
$$I = \frac{6}{5} = 1.2$$
Then:
$$V = 1.2 + 2 = 3.2$$
Answer: \(V = 3.2\) volts, \(I = 1.2\) amperes
Example 2: Blood flow rate \(Q\) and pressure \(P\) satisfy \(2Q + 3P = 18\) and \(4Q - P = 10\). Find \(Q\) and \(P\).
Given: \(2Q + 3P = 18\), \(4Q - P = 10\)
From second equation:
$$P = 4Q - 10$$
Substitute into first:
$$2Q + 3(4Q - 10) = 18$$
$$2Q + 12Q - 30 = 18$$
$$14Q = 48$$
$$Q = \frac{48}{14} = \frac{24}{7} \approx 3.43$$
Then:
$$P = 4 \times 3.43 - 10 = 13.72 - 10 = 3.72$$
Answer: \(Q \approx 3.43\) L/min, \(P \approx 3.72\) kPa
Example 3: For a prosthetic joint, forces \(F_x\) and \(F_y\) satisfy \(5F_x - 2F_y = 7\) and \(3F_x + 4F_y = 1\). Find \(F_x\) and \(F_y\).
Given: \(5F_x - 2F_y = 7\), \(3F_x + 4F_y = 1\)
Multiply second by 2:
$$6F_x + 8F_y = 2$$
Multiply first by 4:
$$20F_x - 8F_y = 28$$
Add:
$$(20F_x - 8F_y) + (6F_x + 8F_y) = 28 + 2$$
$$26F_x = 30$$
$$F_x = \frac{30}{26} = \frac{15}{13} \approx 1.15$$
Substitute into first:
$$5(1.15) - 2F_y = 7$$
$$5.75 - 2F_y = 7$$
$$-2F_y = 7 - 5.75 = 1.25$$
$$F_y = -\frac{1.25}{2} = -0.625$$
Answer: \(F_x \approx 1.15\) N, \(F_y \approx -0.625\) N
Calibration of biomedical sensors requires algebraic rearrangement of formulas to determine calibration constants and unknown parameters.
Example 1: A temperature sensor output \(V\) relates to temperature \(T\) by \(V = aT + b\). Given \(V=2.5\) V at \(T=40^\circ C\) and \(V=1.5\) V at \(T=20^\circ C\), find \(a\) and \(b\).
Given: \(2.5 = 40a + b\), \(1.5 = 20a + b\)
Subtract second from first:
$$(2.5 - 1.5) = (40a - 20a) + (b, b)$$
$$1.0 = 20a$$
$$a = \frac{1.0}{20} = 0.05$$
Substitute \(a\) into second:
$$1.5 = 20 \times 0.05 + b$$
$$1.5 = 1.0 + b$$
$$b = 0.5$$
Answer: \(a = 0.05\) V/°C, \(b = 0.5\) V
Example 2: A sensor’s output voltage \(V\) follows \(V = kC^2 + m\), where \(C\) is concentration. Given \(V=10\) V at \(C=2\) and \(V=22.5\) V at \(C=3\), find \(k\) and \(m\).
Given: \(10 = 4k + m\), \(22.5 = 9k + m\)
Subtract first from second:
$$22.5 - 10 = 9k - 4k + m, m$$
$$12.5 = 5k$$
$$k = \frac{12.5}{5} = 2.5$$
Substitute into first:
$$10 = 4 \times 2.5 + m$$
$$10 = 10 + m$$
$$m = 0$$
Answer: \(k = 2.5\) V/(unit concentration)^2, \(m = 0\) V
Example 3: A pressure sensor output \(P\) satisfies \(P = a \sqrt{F} + b\), where \(F\) is force. Given \(P=5\) at \(F=16\) and \(P=7\) at \(F=36\), find \(a\) and \(b\).
Given: \(5 = a \times 4 + b\), \(7 = a \times 6 + b\)
Subtract first from second:
$$7 - 5 = 6a - 4a + b, b$$
$$2 = 2a$$
$$a = 1$$
Substitute into first:
$$5 = 1 \times 4 + b$$
$$5 = 4 + b$$
$$b = 1$$
Answer: \(a = 1\), \(b = 1\)
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