The “Mathematical Principles for Engineers” course is designed to provide engineers with a strong foundation in essential mathematical techniques. These techniques are critical for solving a wide range of engineering problems. The course is structured around three main areas:

1. Calculus
2. Algebra
3. Differential Equations

Each of these areas is tailored to meet the specific needs of engineering applications.

1. Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. For engineers, calculus is crucial for modeling and analyzing physical systems.

Key Topics and Examples

• Limits and Continuity: Understanding how functions behave near certain points and ensuring that they do not have abrupt changes.

Formulate the problem using calculus: Surface area constraint: \[ 2\pi rh + 2\pi r^2 = 300 \] Volume to maximize: \[ V = \pi r^2 h \] Express \( h \) in terms of \( r \) using the surface area constraint: \[ h = \frac{300 – 2\pi r^2}{2\pi r} \] Substitute \( h \) into the volume formula: \[ V = \pi r^2 \left( \frac{300 – 2\pi r^2}{2\pi r} \right) = \frac{300r – 2\pi r^3}{2} \] Differentiate the volume function with respect to \( r \) and set to zero to find critical points: \[ \frac{dV}{dr} = \frac{300 – 6\pi r^2}{2} = 0 \] \[ 300 – 6\pi r^2 = 0 \implies r^2 = \frac{300}{6\pi} \implies r = \frac{\sqrt{300}}{\sqrt{6\pi}} \] Determine \( h \) using the value of \( r \): \[ h = \frac{300 – 2\pi \left( \frac{\sqrt{300}}{\sqrt{6\pi}} \right)^2}{2\pi \left( \frac{\sqrt{300}}{\sqrt{6\pi}} \right)} = \frac{\sqrt{300}}{\sqrt{6\pi}} \] Result: The dimensions that maximize the volume are \( r = \frac{\sqrt{300}}{\sqrt{6\pi}} \) cm and \( h = \frac{\sqrt{300}}{\sqrt{6\pi}} \) cm.

**Problem**: Designing a container with maximum volume for a given surface area.

**Scenario**: An engineer needs to design a cylindrical can that has a surface area of 300 cm². The goal is to find the dimensions (radius \( r \) and height \( h \)) that maximize the volume.

**Solution**:

1. **Formulate the problem using calculus**: – Surface area constraint: \[2\pi r h + 2\pi r^2 = 300\] – Volume to maximize: \[V = \pi r^2 h\]

2. **Express \( h \) in terms of \( r \) using the surface area constraint**: \[h = \frac{300 – 2\pi r^2}{2\pi r}\]

3. **Substitute \( h \) into the volume formula**: \[V = \pi r^2 \left( \frac{300 – 2\pi r^2}{2\pi r} \right) = \frac{300r – 2\pi r^3}{2}\]

4. **Differentiate the volume function with respect to \( r \) and set to zero to find critical points**: \[\frac{dV}{dr} = \frac{300 – 6\pi r^2}{2} = 0\] \[300 – 6\pi r^2 = 0 \implies r^2 = \frac{300}{6\pi} \implies r = \sqrt{\frac{50}{\pi}}\]

5. **Determine \( h \) using the value of \( r \)**: \[h = \frac{300 – 2\pi \left( \frac{50}{\pi} \right)}{2\pi \sqrt{\frac{50}{\pi}}} = \sqrt{\frac{50}{\pi}}\]

**Result**: The dimensions that maximize the volume are \( r = \sqrt{\frac{50}{\pi}} \) cm and \( h = \sqrt{\frac{50}{\pi}} \) cm.

**Problem**: Solving a system of linear equations to find the currents in an electrical network.

**Scenario**: An electrical engineer needs to find the currents \( I_1 \) and \( I_2 \) in a circuit with two meshes.

– Mesh 1: \[10I_1 + 5(I_1 – I_2) = 15\]

– Mesh 2: \[5(I_2 – I_1) + 20I_2 = 0\]

**Solution**:

1. **Write the equations in standard form**: \[ \begin{cases} 15I_1 – 5I_2 = 15 \\ -5I_1 + 25I_2 = 0 \end{cases} \]

2. **Express as a matrix equation \( AX = B \)**: \[ A = \begin{pmatrix} 15 & -5 \\ -5 & 25 \end{pmatrix}, \quad X = \begin{pmatrix} I_1 \\ I_2 \end{pmatrix}, \quad B = \begin{pmatrix} 15 \\ 0 \end{pmatrix} \]

3. **Solve using matrix inversion**: \[X = A^{-1} B\]

Calculate \( A^{-1} \): \[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \] \[ \det(A) = 15 \cdot 25 – (-5) \cdot (-5) = 350 \] \[ \text{adj}(A) = \begin{pmatrix} 25 & 5 \\ 5 & 15 \end{pmatrix} \] \[ A^{-1} = \frac{1}{350} \begin{pmatrix} 25 & 5 \\ 5 & 15 \end{pmatrix} = \begin{pmatrix} \frac{1}{14} & \frac{1}{70} \\ \frac{1}{70} & \frac{3}{70} \end{pmatrix} \] \[ X = \begin{pmatrix} \frac{1}{14} & \frac{1}{70} \\ \frac{1}{70} & \frac{3}{70} \end{pmatrix} \begin{pmatrix} 15 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{15}{14} \\ \frac{15}{70} \end{pmatrix} = \begin{pmatrix} 1.071 \\ 0.214 \end{pmatrix} \]

**Result**: The currents are \( I_1 = 1.071 \) A and \( I_2 = 0.214 \) A.

**Problem**: Modeling temperature distribution in a metal rod over time.

**Scenario**: An engineer needs to determine the temperature distribution \( u(x, t) \) in a metal rod of length 1 m with boundary conditions held at 0°C and initial temperature distribution \( u(x, 0) = \sin(\pi x) \).

**Solution**:

1. **Formulate the problem using the heat equation**: \[\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}\] with boundary conditions \( u(0, t) = 0 \) and \( u(1, t) = 0 \), and initial condition \( u(x, 0) = \sin(\pi x) \).

2. **Solve using separation of variables**: \[u(x, t) = X(x)T(t)\] Substituting into the heat equation: \[X(x)T'(t) = \alpha X”(x)T(t)\] Dividing both sides by \( X(x)T(t) \): \[\frac{T'(t)}{\alpha T(t)} = \frac{X”(x)}{X(x)} = -\lambda\]

3. **Solve the spatial part \( X”(x) + \lambda X(x) = 0 \)** with boundary conditions: \[X(x) = \sin(n\pi x) \quad \text{for} \quad n = 1, 2, 3, \ldots\] Since \( \lambda = (n\pi)^2 \).

4. **Solve the temporal part \( T'(t) + \alpha (n\pi)^2 T(t) = 0 \)**: \[T(t) = e^{-\alpha (n\pi)^2 t}\]

5. **Combine solutions**: \[u(x, t) = \sum_{n=1}^{\infty} B_n e^{-\alpha (n\pi)^2 t} \sin(n\pi x)\] Using the initial condition \( u(x, 0) = \sin(\pi x) \), we find \( B_1 = 1 \) and \( B_n = 0 \) for \( n \neq 1 \).

**Result**