91Ó°ÊÓ

A parametric surface is a way of representing a surface in three dimensions using parameters. These parameters, often denoted by \( u \) and \( v \), define the coordinates of any point on the surface through functions incorporating these two variables. In the given exercise, the parametric surface is described by:

• \( \mathbf{r}(u, v) = uv\mathbf{i} + ue^v\mathbf{j} + ve^u\mathbf{k} \)

Here, \( u \) and \( v \) are parameters that help locate a particular point on this surface. At a given \( u \) and \( v \), we can plug these values into the equations to find a specific point. For our exercise, we need to determine the tangent plane at the point where \( u = \ln 2 \) and \( v = 0 \). This means substitute these values into the parametric equations to get the location on the surface.

The normal vector to a surface at a given point is crucial for determining the equation of the tangent plane. A normal vector is perpendicular to the surface at that point and can be found using a cross product of the partial derivative vectors of the parametric equations. For our exercise, we compute:

• The partial derivatives \( \frac{\partial \mathbf{r}}{\partial u} \) and \( \frac{\partial \mathbf{r}}{\partial v} \).
• Evaluate these derivatives at the point \((u=\ln 2, v=0)\).
• Perform a cross product to obtain the normal vector \( \mathbf{N} \).

After computing, the normal vector \( \mathbf{N} \) turns out to be:\[ \mathbf{N} = 2\mathbf{i} + 0.5\ln 2\mathbf{j} - \ln 2\mathbf{k} \]This vector provides the necessary orientation to describe the tangent plane at the given point.

Partial derivatives are pivotal to understanding how a surface changes with respect to one parameter while keeping the other constant. In a parametric surface context, taking a partial derivative with respect to \( u \) gives us a direction vector that shows how the surface changes as \( u \) changes, keeping \( v \) constant, and vice versa. Take a look at these two critical derivatives in the solution:

• \( \frac{\partial \mathbf{r}}{\partial u} = v\mathbf{i} + e^v\mathbf{j} + ve^u\mathbf{k} \)
• \( \frac{\partial \mathbf{r}}{\partial v} = u\mathbf{i} + ue^v\mathbf{j} + e^u\mathbf{k} \)

When evaluated at \( u = \ln 2 \) and \( v = 0 \), they help us understand the local structure of the surface:\[ \frac{\partial \mathbf{r}}{\partial u} \bigg|_{(u=\ln 2, v=0)} = 0\mathbf{i} + 1\mathbf{j} + 0.5\mathbf{k} \]\[ \frac{\partial \mathbf{r}}{\partial v} \bigg|_{(u=\ln 2, v=0)} = \ln 2 \mathbf{i} + 0\mathbf{j} + 2\mathbf{k} \]These derivatives help in setting up the required vectors for determining the normal vector and subsequently, the tangent plane's equation.