Problem 27 - 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. Cone The cone \(\mathbf{r}(r, \theta)=(r \cos \theta) \mathbf{i}+(r \sin \theta) \mathbf{j}+r \mathbf{k}, r \geq 0\) \(0 \leq \theta \leq 2 \pi\) at the point \(P_{0}(\sqrt{2}, \sqrt{2}, 2)\) corresponding to \((r, \theta)=(2, \pi / 4)\)

Step-by-Step Solution

Answer

The tangent plane equation is \( -\sqrt{2}x - \sqrt{2}y + z = 2 \) for the cone \( z = \sqrt{x^2 + y^2} \).

Step 1: Calculate Partial Derivatives

First, calculate the partial derivatives of the parameterized surface \( \mathbf{r}(r, \theta) = (r \cos \theta) \mathbf{i} + (r \sin \theta) \mathbf{j} + r \mathbf{k} \) with respect to \( r \) and \( \theta \). They are:\[ \mathbf{r}_r = (\cos \theta) \mathbf{i} + (\sin \theta) \mathbf{j} + \mathbf{k} \]\[ \mathbf{r}_\theta = (-r \sin \theta) \mathbf{i} + (r \cos \theta) \mathbf{j} \]

Step 2: Evaluate Partial Derivatives at the Point

Evaluate the partial derivatives at the specific point \( (r, \theta) = (2, \pi/4) \):\[ \mathbf{r}_r(2, \pi/4) = \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} + \mathbf{k} \]\[ \mathbf{r}_\theta(2, \pi/4) = -\sqrt{2} \mathbf{i} + \sqrt{2} \mathbf{j} \]

Step 3: Calculate the Normal Vector

Find the cross product of the evaluated partial derivatives to obtain the normal vector:\[ \mathbf{N} = \mathbf{r}_r \times \mathbf{r}_\theta = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 1 \ -\sqrt{2} & \sqrt{2} & 0 \end{vmatrix} \]Evaluating the determinant, we find:\[ \mathbf{N} = (-2\sqrt{2}) \mathbf{i} - (2\sqrt{2}) \mathbf{j} + 2 \mathbf{k} \]

Step 4: Write the Equation of the Tangent Plane

Using the point \( P_0 = (\sqrt{2}, \sqrt{2}, 2) \) and the normal vector \( \mathbf{N} \), write the equation of the tangent plane:\[ -2\sqrt{2}(x - \sqrt{2}) - 2\sqrt{2}(y - \sqrt{2}) + 2(z - 2) = 0 \]Simplifying gives:\[ -\sqrt{2}x - \sqrt{2}y + z = 2 \]

Step 5: Cartesian Equation of the Surface

The parameterized surface \( \mathbf{r}(r, \theta) = (r \cos \theta, r \sin \theta, r) \) can be written in Cartesian coordinates by substituting \( x = r \cos \theta \) and \( y = r \sin \theta \). This gives:\[ z = \sqrt{x^2 + y^2} \], representing a cone.

Step 6: Sketch the Surface and Tangent Plane

Sketch the cone given by \( z = \sqrt{x^2 + y^2} \) and the tangent plane \( -\sqrt{2}x - \sqrt{2}y + z = 2 \). The surface is a cone extending upwards, centered around the z-axis, and the tangent plane is positioned based on the normal vector and equation obtained.

Key Concepts

Parametrized Surface

A parametrized surface is a way to describe surfaces in a three-dimensional space using two parameters. Think of it as spreading a surface onto a flat sheet, so you can navigate through it using coordinates like a map.
In this context, each point on the surface can be represented by a set of functions. For example, \[ \mathbf{r}(u, v) = f(u, v) \mathbf{i} + g(u, v) \mathbf{j} + h(u, v) \mathbf{k} \]where \( u \) and \( v \) are parameters. This means every point is placed using these two parameters through functions \( f \), \( g \), and \( h \).
For the problem at hand, the surface of the cone is parametrized by radial distance \( r \) and angle \( \theta \): \[ \mathbf{r}(r, \theta) = (r \cos \theta) \mathbf{i} + (r \sin \theta) \mathbf{j} + r \mathbf{k} \] which encompasses all points over the surface of the cone for certain ranges of \( r \) and \( \theta \).

• Understanding how to parameterize a surface allows us to explore its properties mathematically and simplifies the process of finding tangent planes and other geometric descriptors.

Partial Derivatives

Partial derivatives are essential tools in calculus that measure how a function changes as one of its input variables changes, while keeping others constant.
For a parametrized surface, like \( \mathbf{r}(r, \theta) \), you can compute partial derivatives with respect to each parameter to determine how the surface slopes or changes in different directions.
For example, when computing the partial derivatives of the cone surface with respect to \( r \) and \( \theta \):\[ \mathbf{r}_r = (\cos \theta) \mathbf{i} + (\sin \theta) \mathbf{j} + \mathbf{k} \] \[ \mathbf{r}_\theta = (-r \sin \theta) \mathbf{i} + (r \cos \theta) \mathbf{j} \] these vectors provide tangent directions at any given point, indicating how the surface is changing in the \( r \) and \( \theta \) directions respectively.

• The computed partial derivatives give us a framework to understand the local geometry of the surface, important for complex surfaces like cones where normals and tangential directions vary widely.

Cross Product

The cross product is a mathematical operation performed on two vectors that results in another vector which is perpendicular to both. It is mainly used to find the normal vector to a surface in this context.
In the case of a surface that is described by a parametrization, the vectors are typically the derivatives with respect to each parameter of interest.
For example, to find a vector normal to the surface at a point on the cone, take the partial derivative vectors \( \mathbf{r}_r \) and \( \mathbf{r}_\theta \) and compute their cross product:\[ \mathbf{N} = \mathbf{r}_r \times \mathbf{r}_\theta = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 1 \-\sqrt{2} & \sqrt{2} & 0 \end{vmatrix} \]Evaluating this determinant provides the vector \[-2\sqrt{2} \mathbf{i} - 2\sqrt{2} \mathbf{j} + 2 \mathbf{k}\], a normal vector to the surface at that point.

• This process helps to identify the orientation and inclination of the surface in space, which is crucial for constructing tangent planes and assessing geometric properties.

Normal Vector

A normal vector is an important concept in geometry and calculus, as it is perpendicular to a surface at a given point.
When working with surfaces such as the parametrized cone, normal vectors help define planes tangent to the surface at specific points.
In our example with a cone, after computing the cross product of the partial derivative vectors \( \mathbf{r}_r \) and \( \mathbf{r}_\theta \), the resulting vector tells us the orthogonal direction to the surface:\[ \mathbf{N} = (-2\sqrt{2}) \mathbf{i} - (2\sqrt{2}) \mathbf{j} + 2 \mathbf{k} \].
This normal vector can then be used to formulate the equation of a tangent plane.

• The tangent plane is essentially flat and "touches" the surface at precisely one point. In our specific example, it’s constructed such that it stretches out from \( P_0(\sqrt{2}, \sqrt{2}, 2) \) following the normal direction vector \( \mathbf{N} \).