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**