Problem 28 - 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. $$ \begin{array}{l}{\text { Hemisphere } \quad \text { The hemisphere surface } \mathbf{r}(\phi, \theta)=(4 \sin \phi \cos \theta) \mathbf{i}} \\ {+(4 \sin \phi \sin \theta) \mathbf{j}+(4 \cos \phi) \mathbf{k}, 0 \leq \phi \leq \pi / 2,0 \leq \theta \leq 2 \pi} \\ {\text { at the point } P_{0}(\sqrt{2}, \sqrt{2}, 2 \sqrt{3}) \text { corresponding to }(\phi, \theta)=} \\ {(\pi / 6, \pi / 4)}\end{array} $$

Step-by-Step Solution

Answer

The tangent plane equation is \( 4\sqrt{3}x + 4\sqrt{3}y + 8z = 16 \).

Step 1: Find the Tangent Vectors

First, find the tangent vectors \( \mathbf{r}_\phi \) and \( \mathbf{r}_\theta \) of the surface by differentiating \( \mathbf{r}(\phi, \theta) \) with respect to \( \phi \) and \( \theta \), respectively. Compute:\[ \mathbf{r}_\phi (\phi, \theta) = \frac{\partial \mathbf{r}}{\partial \phi} = (4 \cos \phi \cos \theta) \mathbf{i} + (4 \cos \phi \sin \theta) \mathbf{j} - (4 \sin \phi) \mathbf{k} \]\[ \mathbf{r}_\theta (\phi, \theta) = \frac{\partial \mathbf{r}}{\partial \theta} = (-4 \sin \phi \sin \theta) \mathbf{i} + (4 \sin \phi \cos \theta) \mathbf{j} \]

Step 2: Calculate the Normal Vector

Find the normal vector to the plane by taking the cross product of the tangent vectors \( \mathbf{r}_\phi \) and \( \mathbf{r}_\theta \) at the point \((\phi, \theta) = (\pi/6, \pi/4)\). Substitute and compute:\[ \mathbf{r}_\phi (\pi/6, \pi/4) = (2 \sqrt{2}) \mathbf{i} + (2 \sqrt{2}) \mathbf{j} - 2 \mathbf{k} \]\[ \mathbf{r}_\theta (\pi/6, \pi/4) = (-2 \sqrt{2}) \mathbf{i} + 2 \sqrt{2} \mathbf{j} \]Cross product yields:\[ \mathbf{n} = \mathbf{r}_\phi \times \mathbf{r}_\theta = (4 \sqrt{3}) \mathbf{i} + (4 \sqrt{3}) \mathbf{j} + (8) \mathbf{k} \]

Step 3: Equation of the Tangent Plane

With the normal vector \( \mathbf{n} = (4 \sqrt{3}, 4 \sqrt{3}, 8) \), the equation of the tangent plane at point \( P_0(\sqrt{2}, \sqrt{2}, 2\sqrt{3}) \) is given by:\[ 4 \sqrt{3} (x - \sqrt{2}) + 4 \sqrt{3} (y - \sqrt{2}) + 8 (z - 2\sqrt{3}) = 0 \]Simplify this equation:\[ 4\sqrt{3}x + 4\sqrt{3}y + 8z = 16 \]

Step 4: Find Cartesian Equation of the Hemisphere

To find the Cartesian equation, recognize that the surface is the hemisphere of radius 4. For a full sphere, \( x^2 + y^2 + z^2 = 16 \). For a hemisphere, considering \( z \geq 0 \), the equation becomes:\[ x^2 + y^2 + z^2 = 16 \] with \( z \geq 0 \).

Step 5: Sketch the Surface and Tangent Plane

Draw the hemisphere with the center at the origin and radius 4. The tangent plane is a flat plane touching the hemisphere at \( (\sqrt{2}, \sqrt{2}, 2\sqrt{3}) \). This point lies on the hemisphere and satisfies the tangent plane equation. The plane forms a flat surface that is tangent (just touches) the outer surface of the hemisphere.

Key Concepts

Parametrized Surface

A parametrized surface is a mathematical representation of a surface using two parameters, often denoted as \(u\) and \(v\). It is expressed as a vector function \( \mathbf{r}(u, v) \) that describes every point on the surface in three-dimensional space.

• The function \( \mathbf{r}(u, v) = f(u, v) \mathbf{i} + g(u, v) \mathbf{j} + h(u, v) \mathbf{k} \) involves three coordinate functions: \( f(u, v) \), \( g(u, v) \), and \( h(u, v) \).
• These functions produce the \(x\), \(y\), and \(z\) coordinates, allowing for a smooth transition across the surface.

To deal with surfaces more intuitively, think of these parameters as analogous to latitude and longitude on Earth. They provide a grid-like structure that accurately pins down every point on the surface, offering a clear picture of its topology.

Hemisphere Surface

A hemisphere is half of a sphere, specifically taking its cut along the great circle. In this context, the surface is parameterized in terms of spherical coordinates.

• The spherical coordinates \( \phi \) and \( \theta \) act as our parameters, where \( \phi \) ranges from 0 to \( \pi/2 \) to cover the top half, and \( \theta \) circles the equator from 0 to \( 2\pi \).
• The parameterization is \( \mathbf{r}(\phi, \theta) = (4 \sin \phi \cos \theta) \mathbf{i} + (4 \sin \phi \sin \theta) \mathbf{j} + (4 \cos \phi) \mathbf{k} \).

This parameterization is useful because it is akin to peeling an orange, laying its peel flat; each point on the hemisphere maps to a unique (\( \phi \), \( \theta \)) pair, allowing easy description and manipulation of the surface.

Cross Product

The cross product is a crucial operation when dealing with vectors in 3D space. It takes two vectors and produces another vector, orthogonal to both.

• To find the cross product, you need to arrange the components of the vectors in a determinant involving unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \).
• For vectors \( \mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{B} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \), the cross product \( \mathbf{A} \times \mathbf{B} \) is calculated as follows:\[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3\end{vmatrix}\]
• It results in a new vector which helps describe planes and surfaces.

The property of orthogonality proves beneficial when determining vectors normal to surfaces, making cross products indispensable in 3D geometry and physics.

Normal Vector

A normal vector is a vector that is perpendicular to a surface at a given point. This concept is pivotal during analysis of tangent planes.

• For a surface described by \( \mathbf{r}(u, v) \), the normal vector at a point can be found using the cross product of the partial derivatives: \( \mathbf{r}_u \) and \( \mathbf{r}_v \).
• Considering our hemisphere, these derivatives \( \mathbf{r}_\phi \) and \( \mathbf{r}_\theta \) are computed first, and then \( \mathbf{r}_\phi \times \mathbf{r}_\theta \) gives the normal vector at the point.

The normal vector plays a critical role in equations of tangent planes because it defines the plane's orientation in space. At the hemispheric application point \( P_0(\sqrt{2}, \sqrt{2}, 2\sqrt{3}) \), it ensures that the tangent plane just skims the surface without cutting through it.