/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 The tangent plane at a point \(P... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(P_{0} .\) 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. 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 \((r, \theta)=(2, \pi / 4)\)

Short Answer

Expert verified
The tangent plane equation is \(-2x - 2y + 2\sqrt{2}z = 4\sqrt{2}\), and the surface equation is \(z^2 = x^2 + y^2\), forming a cone.

Step by step solution

01

Calculate Partial Derivatives

First, compute the partial derivatives of the given parametric representation of the cone. Let \( \mathbf{r}(r, \theta) = (r \cos \theta) \mathbf{i} + (r \sin \theta) \mathbf{j} + r \mathbf{k} \). The partial derivatives at any point \( (r, \theta) \) are:\[ \mathbf{r}_r = (\cos \theta) \mathbf{i} + (\sin \theta) \mathbf{j} + \mathbf{k} \]\[ \mathbf{r}_\theta = (-r \sin \theta) \mathbf{i} + (r \cos \theta) \mathbf{j} \]
02

Calculate the Cross Product

Compute the cross product of the partial derivatives at \( P_0 \) with \( r=2 \) and \( \theta=\pi/4 \):\[ \mathbf{r}_r(2, \pi/4) = \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 1 \right) \]\[ \mathbf{r}_\theta(2, \pi/4) = \left( -\sqrt{2}, \sqrt{2}, 0 \right) \]The cross product is:\[ \mathbf{n} = \mathbf{r}_r \times \mathbf{r}_\theta = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 1 \ -\sqrt{2} & \sqrt{2} & 0 \end{array} \right| \]Calculating the determinant, we get:\[ \mathbf{n} = -2 \mathbf{i} - 2 \mathbf{j} + 2 \sqrt{2} \mathbf{k} \]
03

Equation of the Tangent Plane

The equation of the plane tangent to the surface at \( P_0(\sqrt{2}, \sqrt{2}, 2) \) is given by:\[ -2(x - \sqrt{2}) - 2(y - \sqrt{2}) + 2\sqrt{2}(z - 2) = 0 \]Simplifying, we find:\[ -2x - 2y + 2\sqrt{2}z = 4\sqrt{2} \]
04

Cartesian Equation of the Surface

Convert the parametric surface to its Cartesian form. We have:\[ x = r \cos \theta \]\[ y = r \sin \theta \]\[ z = r \]The Cartesian equation of the surface is thus:\[ z^2 = x^2 + y^2 \]
05

Sketch the Surface and Tangent Plane

The surface described by the equation \( z^2 = x^2 + y^2 \) represents a cone. The tangent plane equation is \( -2x - 2y + 2\sqrt{2}z = 4\sqrt{2} \). To sketch both together, note the cone opens upwards with symmetry around the z-axis, and the tangent plane touches the cone at the point \( P_0(\sqrt{2}, \sqrt{2}, 2) \).

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Key Concepts

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

Parametrized Surface
A parametrized surface is a way of describing a surface in space where two parameters, often labeled as \(u\) and \(v\), each take on values that define coordinates in three-dimensional space. You can think of these parameters as coordinates in a new system that uniquely determines points on a surface. For instance, in the case of the cone given in the problem, the parameters \(r\) and \(\theta\) define every possible point on the surface of the cone through the use of trigonometric functions.
The parametrized vector function \(\mathbf{r}(u, v) = f(u, v) \mathbf{i} + g(u, v) \mathbf{j} + h(u, v) \mathbf{k}\) helps in expressing surfaces as sets of equations.
This allows us to describe complex surfaces in a simpler way, where \(f\), \(g\), and \(h\) dictate the shape and structure of the surface, and \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are the standard unit vectors in 3D Cartesian space. It is often used in conjunction with calculus to analyze the properties of surfaces, such as curvature and tangent planes.
Cross Product
The cross product is a vector operation fundamental in finding a vector perpendicular to two given vectors in three-dimensional space. Given two vectors \( \mathbf{a} \) and \( \mathbf{b} \), the cross product \( \mathbf{a} \times \mathbf{b} \) results in a third vector that is orthogonal to both \(\mathbf{a}\) and \(\mathbf{b}\). This property is extremely useful in finding normal vectors to surfaces.
For the exercise at hand, the tangent plane to a surface at a given point is specified using the normal vector derived from the cross product of the partial derivative vectors \( \mathbf{r}_u \) and \( \mathbf{r}_v \). In forming the determinant, the i, j, and k unit vectors help structure the cross product, resulting in a mathematically concise vector orthogonal to the surface at the point of interest.
This step is crucial for establishing the orientation of the tangent plane, as the plane's equation uses the normal vector to describe all points lying on it that are also tangent to the surface.
Partial Derivatives
Partial derivatives extend the concept of a derivative to functions with multiple variables. Instead of considering the rate of change with respect to one variable, partial derivatives look at how a function changes as just one of its multiple inputs varies.
In this context, dealing with the parametrized surface \(\mathbf{r}(r, \theta)\), we calculate partial derivatives \(\mathbf{r}_r\) and \(\mathbf{r}_\theta\), which represent the rate of change of the vector function in the directions of \(r\) and \(\theta\), respectively.
  • \(\mathbf{r}_r\) gives the change as \(r\) changes
  • \(\mathbf{r}_\theta\) gives the change as \(\theta\) changes
Derivative calculation is fundamental in assessing the directional growth of the surface and in this exercise, it serves as a preliminary step toward establishing the tangent plane by determining the vectors required for the cross product.
Cartesian Equation
A Cartesian equation is a mathematical representation of surfaces or curves in space using x, y, and z coordinates. For surfaces derived from parametric equations, it's often useful to express the relationship between coordinates directly.
In this exercise, the parametric cone can ultimately be described in Cartesian form with the equation \(z^2 = x^2 + y^2\). The transformation from parametric to Cartesian form provides a significant simplification in understanding the surface shape.
Such an equation is powerful for sketching surfaces and solving geometrical problems, as it directly relates the spaces in x, y, and z without additional parameters. Converting to a Cartesian form allows one to consider the symmetries and specific geometric features of the surface, making it more intuitive for spatial reasoning and further mathematical manipulation.

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Most popular questions from this chapter

Let \(S\) be the portion of the cylinder \(y=\ln x\) in the first octant whose projection parallel to the \(y\) -axis onto the \(x z\) -plane is the rectangle \(R_{k}: 1 \leq x \leq e, 0 \leq z \leq 1 .\) Let \(n\) be the unit vector normal to \(S\) that points away from the \(x z\) -plane. Find the flux of \(\mathbf{F}=2 y \mathbf{j}+z \mathbf{k}\) through \(S\) in the direction of \(\mathbf{n}\)

Find a potential function for \(\mathbf{F}\). $$\mathbf{F}=\left(e^{x} \ln y\right) \mathbf{i}+\left(\frac{e^{x}}{y}+\sin z\right) \mathbf{j}+(y \cos z) \mathbf{k}$$

Calculating Area with Green's Theorem If a simple closed curve C in the plane and the region \(R\) it encloses satisfy the hypotheses of Green's Theorem, the area of \(R\) is given by Green's Theorem Area Formula Area of \(R=\frac{1}{2} \oint_{c} x d y-y d x\) The reason is that, by Equation (4) run backward, Area of \(R=\iint_{k} d y d x=\iint_{k}\left(\frac{1}{2}+\frac{1}{2}\right) d y d x\) $$=\oint_{c} \frac{1}{2} x d y-\frac{1}{2} y d x$$ Use the Green's Theorem area formula given above to find the areas of the regions enclosed by the curves in Exercises \(31-34\) The astroid \(\mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{i}+\left(\sin ^{3} t\right) \mathbf{j}, \quad \mathbf{0} \leq t \leq 2 \pi\)

Conservation of mass \(\quad\) Let \(\mathbf{v}(t, x, y, z)\) be a continuously differentiable vector field over the region \(D\) in space, and let \(p(t, x,\) \(y, z)\) be a continuously differentiable scalar function. The variable t represents the time domain. The Law of Conservation of Mass asserts that $$\frac{d}{d t} \iiint_{D} p(t, x, y, z) d V=-\iint_{S} p \mathbf{v} \cdot \mathbf{n} d \sigma,$$ where \(S\) is the surface enclosing \(D\). a. Give a physical interpretation of the conservation of mass law if \(\mathbf{v}\) is a velocity flow field and \(p\) represents the density of the fluid at point \((x, y, z)\) at time \(t\) b. Use the Divergence Theorem and Leibniz's Rule, $$\frac{d}{d t} \iiint_{D} p(t, x, y, z) d V=\iiint_{D} \frac{\partial p}{\partial t} d V,$$ to show that the Law of Conservation of Mass is equivalent to the continuity equation, $$\nabla \cdot p \mathbf{v}+\frac{\partial p}{\partial t}=0.$$ (In the first term \(\nabla \cdot p \mathbf{v},\) the variable \(t\) is held fixed, and in the second term \(\partial p / \partial t,\) it is assumed that the point \((x, y, z)\) in \(D\) is held fixed.)

a. Show that the outward flux of the position vector field \(\mathbf{F}=\) \(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) through a smooth closed surface \(S\) is three times the volume of the region enclosed by the surface. b. Let \(n\) be the outward unit normal vector field on \(S\). Show that it is not possible for \(\mathbf{F}\) to be orthogonal to \(\mathbf{n}\) at every point of \(S .\)

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