Problem 30 - Thomas Calculus

Question

The tangent plane at a point \(P_{0}\left(f\left(u_{0}, v_{0}\right), g\left(u_{0}, v_{0}\right), h\left(u_{0}, v_{0}\right)\right)\) on a parametrized surface \(\mathbf{r}(u, \boldsymbol{v})=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) is the plane through \(P_{0}\) normal to the vector \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right) \times \mathbf{r}_{v}\left(u_{0}, v_{0}\right),\) the cross product of the tangent vectors \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right)\) and \(\mathbf{r}_{v}\left(u_{0}, v_{0}\right)\) at \(P_{0}\) . Find an equation for the plane tangent to the surface at \(P_{0} .\) Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. Parabolic cylinder \(\quad\) The parabolic cylinder surface \(\mathbf{r}(x, y)=\) \(x \mathbf{i}+y \mathbf{j}-x^{2} \mathbf{k},-\infty

Step-by-Step Solution

Answer

The tangent plane equation is \(2x + z = 1\). The surface equation is \(z = -x^2\).

Step 1: Calculate the Tangent Vectors

Given the surface \( \mathbf{r}(x, y) = x \mathbf{i} + y \mathbf{j} - x^2 \mathbf{k} \), we first find the tangent vectors. The partial derivative with respect to \(x\) gives \( \mathbf{r}_x = \frac{\partial \mathbf{r}}{\partial x} = \mathbf{i} - 2x \mathbf{k} \), and with respect to \(y\), we have \( \mathbf{r}_y = \frac{\partial \mathbf{r}}{\partial y} = \mathbf{j} \).

Step 2: Calculate the Normal Vector using the Cross Product

The normal vector to the tangent plane at \( P_0 \) is obtained by the cross product \( \mathbf{n} = \mathbf{r}_x \times \mathbf{r}_y \). Compute this as follows: \[ \mathbf{n} = (\mathbf{i} - 2x \mathbf{k}) \times \mathbf{j} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & -2x \ 0 & 1 & 0 \end{vmatrix} = 2x \mathbf{i} + \mathbf{k}. \] At \((x, y) = (1, 2)\), \( \mathbf{n} = 2 \mathbf{i} + \mathbf{k} \).

Step 3: Form the Equation of the Tangent Plane

The equation of the tangent plane at the point \( P_0(1, 2, -1) \) can be formed using the point-normal form of a plane \( \mathbf{n} \cdot ((x, y, z) - (1, 2, -1)) = 0 \). Substituting the point and normal vector, we have: \[ (2 \mathbf{i} + \mathbf{k}) \cdot \begin{pmatrix} x - 1 \ y - 2 \ z + 1 \end{pmatrix} = 0 \] which simplifies to \( 2(x - 1) + (z + 1) = 0 \), or \( 2x + z = 1 \).

Step 4: Write the Cartesian Equation of the Surface

The given parametric equations are already in Cartesian coordinates where \( z = -x^2 \). Thus, the Cartesian equation of the surface is \( z = -x^2 \).

Step 5: Visualize the Surface and Tangent Plane

To sketch them together, plot the parabolic cylinder defined by \( z = -x^2 \). At the point \((1, 2, -1)\), draw the tangent plane defined by \( 2x + z = 1 \). These will intersect at this tangent point on the surface, visually showing how the plane just 'touches' the surface.

Key Concepts

Parametric Surface

A parametric surface is a mathematical representation of a surface defined using parameters. Instead of describing a surface with fixed equations in Cartesian coordinates, parametric surfaces use functions of one or more variables. In our exercise, the surface is given by the equation \( \mathbf{r}(x, y) = x \mathbf{i} + y \mathbf{j} - x^2 \mathbf{k} \). Here, \( x \) and \( y \) are parameters that vary across the surface, and the output is a vector in a three-dimensional space.
Parametric surfaces provide flexibility in representing complex shapes. For example, you can model surfaces that have intricate features or curves, which are challenging to describe with simple algebraic equations. By using parameters, you can focus on specific sections of a surface or express motions and transformations easily.
When dealing with parametric surfaces, one benefit is the ability to easily calculate derivatives. These derivatives are essential for tasks like finding tangent vectors as it involves computing partial derivatives with respect to the parameters. In this exercise, we use the parametric form of the surface to find the tangent and normal vectors.

Cross Product

In vector calculus, the cross product is a vital operation used to find a vector that is perpendicular to two other vectors. In the context of our exercise, the cross product helps us in determining the normal vector to the tangent plane.
Here's how it works. For any two vectors \( \mathbf{a} \) and \( \mathbf{b} \), the cross product \( \mathbf{a} \times \mathbf{b} \) results in a new vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). This property is very useful in physics and engineering, especially when dealing with three-dimensional spaces.
In this exercise, we took the tangent vectors \( \mathbf{r}_x \) and \( \mathbf{r}_y \) derived from the parametric surface, and performed the cross product to obtain \( \mathbf{n} = 2x \mathbf{i} + \mathbf{k} \), which gives the normal to the tangent plane at that point \( P_0 \). This normal vector is then crucial for writing the equation of the tangent plane.

Parabolic Cylinder

A parabolic cylinder is a simple yet intriguing surface in three-dimensional space. Unlike a circular cylinder, which is closed and curved around, a parabolic cylinder is open and shaped like an elongated cup.
The standard equation for a parabolic cylinder is \( z = ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants. In our exercise, the equation is \( z = -x^2 \), which represents a downward-facing parabola opening in the \( z \)-direction, extended infinitely along the \( y \)-axis.
Understanding the nature of a parabolic cylinder can help in visualizing problems related to surfaces and planes. Because it extends infinitely along the \( y \)-axis, it is considered to have constant cross-sections in this direction. This makes operations like finding the tangent plane easier as we can consider "slices" along constant \( y \) values.
In terms of application, knowing the geometry of a parabolic cylinder allows us to predict how it will interact with other mathematical objects, like tangent planes in this case. The ability to sketch both these elements together enhances understanding of the points of contact and the plane's orientation relative to the surface.