91Ó°ÊÓ

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. Hemisphere The hemisphere surface \(\mathbf{r}(\phi, \theta)=(4 \sin \phi \cos \theta) \mathbf{i}\) \(+(4 \sin \phi \sin \theta) \mathbf{j}+(4 \cos \phi) \mathbf{k}, 0 \leq \phi \leq \pi / 2,0 \leq \theta \leq 2 \pi\) at the point \(P_{0}(\sqrt{2}, \sqrt{2}, 2 \sqrt{3})\) corresponding to \((\phi, \theta)=\) \((\pi / 6, \pi / 4)\)

Step by step solution

01

Identify Components and Parameters

The surface is given by \(\mathbf{r}(\phi, \theta) = (4 \sin \phi \cos \theta) \mathbf{i} + (4 \sin \phi \sin \theta) \mathbf{j} + (4 \cos \phi) \mathbf{k}\). With parameters \(\phi\) and \(\theta\), we identify the point \(P_0(\sqrt{2}, \sqrt{2}, 2\sqrt{3})\) corresponds to \((\phi, \theta) = (\pi/6, \pi/4)\).

02

Calculate Partial Derivatives

Compute \(\mathbf{r}_{\phi}\) and \(\mathbf{r}_{\theta}\):\[\mathbf{r}_{\phi} = \left(4 \cos \phi \cos \theta\right) \mathbf{i} + \left(4 \cos \phi \sin \theta\right) \mathbf{j} - 4 \sin \phi \mathbf{k}\] \[\mathbf{r}_{\theta} = \left(-4 \sin \phi \sin \theta\right) \mathbf{i} + \left(4 \sin \phi \cos \theta\right) \mathbf{j} + 0 \mathbf{k}\].

03

Find the Normal Vector

Calculate the cross product \(\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}\):\[\mathbf{r}_{\phi}(\pi/6, \pi/4) = \left(\sqrt{2} \right) \mathbf{i} + \left(\sqrt{2} \right) \mathbf{j} - 2 \sqrt{3} \mathbf{k}\] \[\mathbf{r}_{\theta}(\pi/6, \pi/4) = -2 \mathbf{i} + 2 \mathbf{j}\]The cross product is \[\mathbf{n} = \mathbf{r}_{\phi} \times \mathbf{r}_{\theta} = (4 \sqrt{3}) \mathbf{i} + (4 \sqrt{3}) \mathbf{j} + (4) \mathbf{k}\].

04

Write the Tangent Plane Equation

Using \(\mathbf{n} = (4 \sqrt{3}, 4 \sqrt{3}, 4)\) and point \(P_0 (\sqrt{2}, \sqrt{2}, 2\sqrt{3})\), the tangent plane equation is \[4\sqrt{3}(x - \sqrt{2}) + 4\sqrt{3}(y - \sqrt{2}) + 4(z - 2\sqrt{3}) = 0\].

05

Simplify Tangent Plane Equation

Simplifying the equation:\[4\sqrt{3}x + 4\sqrt{3}y + 4z - 16 = 0\].

06

Convert Parametric Surface to Cartesian Equation

Converting hemispherical parametric equations \(x = 4 \sin \phi \cos \theta\), \(y = 4 \sin \phi \sin \theta\), \(z = 4 \cos \phi\) by squaring and adding:\[x^2 + y^2 + z^2 = 16\].

07

Sketch the Surface and Tangent Plane

The hemisphere \(x^2 + y^2 + z^2 = 16\) lives above the xy-plane (\(z \geq 0\)). The tangent plane "touches" the hemisphere at \((\sqrt{2}, \sqrt{2}, 2\sqrt{3})\). Sketch the hemisphere as the upper half of a sphere, and the plane intersecting the hemisphere at its tangent point.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametrized Surface

A parametrized surface uses parameters, typically denoted as \( u \) and \( v \), to describe a surface in three-dimensional space. In this context, these parameters help express complex surfaces with simple, mathematical functions. The surface might be presented as a vector function \( \mathbf{r}(u, v) = f(u, v) \mathbf{i} + g(u, v) \mathbf{j} + h(u, v) \mathbf{k} \), which implies that the surface is traced by varying \( u \) and \( v \).

For our hemisphere example, the parameters \( \phi \) and \( \theta \) are specifically chosen to exploit spherical coordinates, making it easier to express a hemisphere. These types of surfaces are powerful in that they allow us to manipulate and explore geometrically complex shapes through simple mathematical tools.

With parametrized surfaces, you can:

• Describe curved surfaces with simple equations.
• Calculate tangent vectors and then normal vectors.
• Use parameters to dynamically model surface behaviors.

Cross Product

The cross product is a vector operation that takes two vectors in three-dimensional space and produces another vector that is perpendicular to both of the original vectors. This new vector is often called the normal vector, and its length corresponds to the area of the parallelogram formed by the two vectors.

In the context of finding a tangent plane, we start with tangent vectors \( \mathbf{r}_{u} \) and \( \mathbf{r}_{v} \) derived from the partial derivatives of the surface function. The cross product \( \mathbf{r}_{u} \times \mathbf{r}_{v} \) gives us the normal vector - a crucial component used to define the plane that is tangent to our surface at a specific point.

Key properties of the cross product include:

• Your resulting vector is orthogonal to both original vectors.
• Allows calculation of planes and determining orientations.
• Vector has a magnitude that helps interpret spatial interactions.

Cartesian Equation

A Cartesian equation is an equation that relates \( x \), \( y \), and \( z \) coordinates directly, typically using algebra without involving any parameters. Once a parametrized surface is defined, converting it to a Cartesian form helps visualize the surface in regular coordinate operations.

In the exercise, we converted the parameterized hemisphere equations \( x = 4 \sin \phi \cos \theta \), \( y = 4 \sin \phi \sin \theta \), and \( z = 4 \cos \phi \) into the Cartesian equation \( x^2 + y^2 + z^2 = 16 \). This is achieved by using trigonometric identities to eliminate parameters, showing the hemisphere as part of a sphere in a clearer algebraic form.

Benefits of Cartesian equations include:

• Easier visualization and graph plotting.
• Simplified algebraic manipulation for calculations.
• Direct interpretation without needing parameter adjustments.

Partial Derivatives

Partial derivatives are essential in multivariable calculus, allowing us to understand changes along specific dimensions while treating other variables as constants. They are especially prevalent in calculating the slope of surfaces.

For a parametrized surface \( \mathbf{r}(u, v) = f(u, v) \mathbf{i} + g(u, v) \mathbf{j} + h(u, v) \mathbf{k} \), partial derivatives \( \mathbf{r}_{u} \) and \( \mathbf{r}_{v} \) give us the tangent vectors at any point \( (u, v) \) on the surface. These tangent vectors demonstrate how the surface changes along each of the parameter directions.

In practice, understanding partial derivatives assists with:

• Finding and analyzing tangent planes.
• Determining surface behavior and shape.
• Solving optimization problems in higher dimensions.