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 \(\mathbf{r}(u, v)=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) 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 \(P_{0} .\) In Exercises \(27-30,\) 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. $$ \begin{array}{l}{\text { Cone The cone } \mathbf{r}(r, \theta)=(r \cos \theta) \mathbf{i}+(r \sin \theta) \mathbf{j}+r \mathbf{k}, r \geq 0} \\ {0 \leq \theta \leq 2 \pi \text { at the point } P_{0}(\sqrt{2}, \sqrt{2}, 2) \text { corresponding to }} \\ {(r, \theta)=(2, \pi / 4)}\end{array} $$

Step-by-Step Solution

Answer

Tangent plane: \(-\sqrt{2}x - \sqrt{2}y + 2z = 2\). Surface: \(z = \sqrt{x^2 + y^2}\).

Step 1: Calculate Partial Derivatives

First, let's find the partial derivatives of \( \mathbf{r} \) with respect to \( r \) and \( \theta \).\[ \mathbf{r}_r = \left( \cos \theta \right) \mathbf{i} + \left( \sin \theta \right) \mathbf{j} + \mathbf{k} \] \[ \mathbf{r}_\theta = \left( -r \sin \theta \right) \mathbf{i} + \left( r \cos \theta \right) \mathbf{j} \]

Step 2: Evaluate at Given Parameters

Evaluate \( \mathbf{r}_r \) and \( \mathbf{r}_\theta \) at \( (r, \theta) = (2, \pi/4) \).\[ \mathbf{r}_r(2, \pi/4) = \left( \cos(\pi/4) \right) \mathbf{i} + \left( \sin(\pi/4) \right) \mathbf{j} + \mathbf{k} = \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} + \mathbf{k} \] \[ \mathbf{r}_\theta(2, \pi/4) = \left( -2 \sin(\pi/4) \right) \mathbf{i} + \left( 2 \cos(\pi/4) \right) \mathbf{j} = -\sqrt{2} \mathbf{i} + \sqrt{2} \mathbf{j} \]

Step 3: Calculate Cross Product

Find the cross product of \( \mathbf{r}_r(2, \pi/4) \) and \( \mathbf{r}_\theta(2, \pi/4) \).\[ \mathbf{r}_r \times \mathbf{r}_\theta = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 1 \-\sqrt{2} & \sqrt{2} & 0 \end{vmatrix} = -\sqrt{2} \mathbf{i} - \sqrt{2} \mathbf{j} + 2 \mathbf{k} \]

Step 4: Equation of Tangent Plane

Use the normal vector to find the equation of the plane: \( -\sqrt{2} (x - \sqrt{2}) - \sqrt{2} (y - \sqrt{2}) + 2 (z - 2) = 0 \).Expanding, we have \( -\sqrt{2}x - \sqrt{2}y + 2z = 2 \).

Step 5: Cartesian Equation of the Surface

Convert the parametric form to Cartesian form. Note that the surface is defined as \( z = \sqrt{x^2 + y^2} \).

Step 6: Sketch Surface and Plane

Sketch a cone opening upwards along the z-axis. The tangent plane will touch the cone at the point \((\sqrt{2}, \sqrt{2}, 2)\). The plane is inclined, intersecting the cone in a manner that locally approximates the cone shape. The plane's slope and position derive from the normal vector and point of tangency.

Key Concepts

Parametrized Surface

A parametrized surface is a way of representing a surface using parameters. Unlike traditional 3D coordinates, parametrized surfaces use two parameters, often denoted as \(u\) and \(v\), to describe any point on the surface.
Think of these parameters as sliders: adjust one, and you move along one direction of the surface. Adjust both, and you move freely across the surface. For example, in our exercise, the cone is described using parameters \(r\) and \(\theta\). These help define every point on the cone by moving a distance \(r\) from the origin in the plane and rotating by the angle \(\theta\).
This method allows complex surfaces that might be difficult to describe using traditional Cartesian coordinates to be simplified and analyzed. Choosing the right parameters can make defining and working with the surface easier and more intuitive.

Partial Derivatives

Partial derivatives are crucial in understanding changes on parametrized surfaces. They measure how a function changes as one of its inputs changes, keeping others constant. In the context of surfaces, partial derivatives help us understand the surface's slope or tilt at any given point.
For instance, calculating the partial derivative of our surface \(\mathbf{r}\) with respect to parameter \(r\) or \(\theta\) tells us how the surface changes in the direction of increasing \(r\) or \(\theta\). This gives rise to tangent vectors along those directions.

• The partial derivative with respect to \(r\), \(\mathbf{r}_r\), provides the direction and rate of change when moving along the radius of the cone.
• The partial derivative with respect to \(\theta\), \(\mathbf{r}_\theta\), helps us understand the changes when rotating around the cone's axis.

Understanding these changes is essential for determining the tangent plane at any given point on the surface.

Cross Product

The cross product is a mathematical operation used to find a vector perpendicular to two given vectors. In the context of surfaces, it's a powerful tool to find the normal vector to the surface at a certain point. This vector is crucial for determining the tangent plane.
In our exercise, we have two tangent vectors, \(\mathbf{r}_r\) and \(\mathbf{r}_\theta\), which lie on the surface. Their cross product \(\mathbf{r}_r \times \mathbf{r}_\theta\) gives us a vector that is orthogonal, or perpendicular, to the surface.

• This normal vector is used to derive the equation of the tangent plane.
• The cross product offers insight into the surface's orientation in space.

Calculating the cross product helps visualize how the surface interacts with the surrounding space, which is key in many engineering and physics applications.

Cartesian Equation

The Cartesian equation is a way of expressing surfaces or curves using traditional x, y, and z coordinates. It’s the familiar algebraic format that helps visualize and understand the geometric properties of an object.
In our task, converting the parametrized surface of the cone into a Cartesian equation was necessary to better analyze its features and interactions. \(z = \sqrt{x^2 + y^2}\) clearly describes how the cone's height relates to the x and y positions, and thus transforms the parametric definition into something more familiar.
This form is also useful if you want to plot or calculate points on the surface using standard graphing techniques or software. Additionally, by comparing both parametric and Cartesian forms, deeper insights into the shape's structure and characteristics are revealed. Through these different perspectives, you can solve more complex geometrical problems.