Let's break this problem into its parts and solve it:

Part (i): Briefly Explain Why the Vectors $\mathbf{T}_u$ and $\mathbf{T}_v$ Look Like This

Given the parametrization: $x = u, \quad y = v, \quad z = g(u, v),$ the surface can be expressed as $\sigma(u, v) = \langle u, v, g(u, v) \rangle$.

Now, calculate the partial derivatives:

1. Partial with respect to $u$: $\mathbf{T}_u = \frac{\partial \sigma}{\partial u} = \left\langle \frac{\partial u}{\partial u}, \frac{\partial v}{\partial u}, \frac{\partial g}{\partial u} \right\rangle = \langle 1, 0, g_u \rangle,$ where $g_u = \frac{\partial g}{\partial u}$.

2. Partial with respect to $v$: $\mathbf{T}_v = \frac{\partial \sigma}{\partial v} = \left\langle \frac{\partial u}{\partial v}, \frac{\partial v}{\partial v}, \frac{\partial g}{\partial v} \right\rangle = \langle 0, 1, g_v \rangle,$ where $g_v = \frac{\partial g}{\partial v}$.

Thus, the given forms of $\mathbf{T}_u = \langle 1, 0, g_u \rangle$ and $\mathbf{T}_v = \langle 0, 1, g_v \rangle$ directly result from these calculations.

Part (ii): Use (i) to Show the Surface Area Element

To find the surface area element $dS$, compute the magnitude of the cross product $\mathbf{T}_u \times \mathbf{T}_v$:

1. Cross Product $\mathbf{T}_u \times \mathbf{T}_v$: Using the determinant method:

   \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & g_u \\ 0 & 1 & g_v \end{vmatrix},$$ expand this determinant: $$\mathbf{T}_u \times \mathbf{T}_v = \mathbf{i} \begin{vmatrix} 0 & g_u \\ 1 & g_v \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & g_u \\ 0 & g_v \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}.$$ Compute each minor: - For $$\mathbf{i}$$: $$\begin{vmatrix} 0 & g_u \\ 1 & g_v \end{vmatrix} = (0)(g_v) - (1)(g_u) = -g_u$$, - For $$\mathbf{j}$$: $$\begin{vmatrix} 1 & g_u \\ 0 & g_v \end{vmatrix} = (1)(g_v) - (0)(g_u) = g_v$$, - For $$\mathbf{k}$$: $$\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1$$. Thus: $$\mathbf{T}_u \times \mathbf{T}_v = \langle -g_u, -g_v, 1 \rangle.$$

2. Magnitude of the Cross Product: $\|\mathbf{T}_u \times \mathbf{T}_v\| = \sqrt{(-g_u)^2 + (-g_v)^2 + 1^2} = \sqrt{g_u^2 + g_v^2 + 1}.$

3. Surface Area Element: Since $dS = \|\mathbf{T}_u \times \mathbf{T}_v\| du dv$, substitute: $dS = \sqrt{1 + \left(\frac{\partial g}{\partial u}\right)^2 + \left(\frac{\partial g}{\partial v}\right)^2} \, du dv.$

When replacing $u$ and $v$ with $x$ and $y$, the formula becomes: $dS = \sqrt{1 + \left(\frac{\partial g}{\partial x}\right)^2 + \left(\frac{\partial g}{\partial y}\right)^2} \, dx dy.$

Let me know if you need further clarification!

Follow-Up Questions:

1. How do partial derivatives relate to the tangent vectors of a surface?
2. Why is the magnitude of the cross product used to find the surface area element?
3. Can this process be extended to surfaces parametrized differently than $\sigma(u, v)$?
4. How does the formula for $dS$ simplify when the surface is flat ($g(x, y) = 0$)?
5. What happens if $g(u, v)$ has discontinuities or undefined derivatives?

Tip:

Visualizing the tangent vectors $\mathbf{T}_u$ and $\mathbf{T}_v$ can help better understand how the cross product captures the local orientation and scaling of the surface!