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Chapter 14: Problem 43

Find a parametric equation for the line that is perpendicular to the graph of the given equation at the given point. \begin{equation}x^{2}+y^{2}+z^{2}=14, \quad(3,-2,1)\end{equation}

Short Answer

Expert verified
The parametric equations are: \( x = 3 + 6t \), \( y = -2 - 4t \), \( z = 1 + 2t \).

Step by step solution

01

Understand the Surface Equation

The equation \( x^2 + y^2 + z^2 = 14 \) describes a sphere centered at the origin with radius \( \sqrt{14} \). A line perpendicular to the surface at a given point is aligned with the normal vector at that point.
02

Find the Gradient Vector

The gradient vector \( abla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \) represents the normal vector to the surface. For \( f(x, y, z) = x^2 + y^2 + z^2 \), we have \( abla f = (2x, 2y, 2z) \).
03

Calculate the Normal Vector at the Point

Substitute the point \( (3, -2, 1) \) into the gradient to find the normal vector: \( abla f(3, -2, 1) = (2 \cdot 3, 2 \cdot (-2), 2 \cdot 1) = (6, -4, 2) \). This vector \((6, -4, 2)\) is the direction vector of the line.
04

Write the Parametric Equations of the Line

The parametric equations of a line with point \((x_0, y_0, z_0)\) and direction vector \((a, b, c)\) are given by: \( x = x_0 + at \), \( y = y_0 + bt \), \( z = z_0 + ct \). Substituting the point \((3, -2, 1)\) and the direction vector \((6, -4, 2)\), we have:\[ x = 3 + 6t \]\[ y = -2 - 4t \]\[ z = 1 + 2t \]

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Key Concepts

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

Normal Vector
A normal vector is a fundamental concept in geometry and vector calculus. It is a vector that is perpendicular to a surface at a given point. The term "normal" indicates that it is orthogonal to every tangent vector in that area.

Understanding normal vectors is crucial because they help in determining the orientation of a surface, be it a simple plane or a more complex shape like a sphere. In the context of a sphere, any normal vector at a given point will always point radially outward from the center of the sphere.
  • The normal vector is unique in being perpendicular to the tangent plane at the point of interest.
  • It provides a direction that is locally orthogonal to the surface.
  • This property is vital in various applications, including physics and computer graphics.
Sphere
A sphere is a perfectly symmetrical three-dimensional shape where every point on its surface is equidistant from its center.

The standard equation of a sphere with a center at the origin (0, 0, 0) and radius \( r \) is given by:\[ x^2 + y^2 + z^2 = r^2 \]For this particular exercise, the sphere equation is \( x^2 + y^2 + z^2 = 14 \), which implies a sphere centered at the origin with a radius of \( \sqrt{14} \).
  • The distance from the center of the sphere to any point on its surface is constant.
  • Spheres are often encountered in physics, as they are a good representation of idealized celestial bodies like planets.
  • Using the geometry of spheres, one can easily define directions such as radial (like the normal vector discussed above).
Gradient Vector
The gradient vector is a key player in multivariable calculus, acting as a tool to navigate the steepest ascent of a function.

Given a function \( f(x, y, z) \), the gradient \( abla f \) is calculated by taking the partial derivatives with respect to each variable. The gradient vector essentially tells us how the function changes the fastest and in which direction.
  • For the function \( f(x, y, z) = x^2 + y^2 + z^2 \), the gradient is \( abla f = (2x, 2y, 2z) \).
  • The gradient vector at any point on the sphere gives the normal vector for that point because it is perpendicular to the sphere's surface.
  • This property is particularly useful for optimization problems and in finding tangent planes and normals.
Perpendicular Line
A perpendicular line is a line that intersects another line or surface forming a right angle (90 degrees).

In the context of the given exercise, the objective is to find a line that is perpendicular to the surface of the sphere at a given point. This is accomplished by using the normal vector, which provides the direction for this perpendicular line.
  • For the sphere \( x^2 + y^2 + z^2 = 14 \) at the point \( (3, -2, 1) \), the normal vector (calculated via the gradient) is \( (6, -4, 2) \).
  • Using this normal vector as the direction, a line can be described parametrically.
  • The parametric equations for the perpendicular line are:
    \( x = 3 + 6t \),
    \( y = -2 - 4t \),
    \( z = 1 + 2t \).

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Most popular questions from this chapter

Let \(f(x, y)=\left\\{\begin{array}{ll}{0,} & {x^{2} < y < 2 x^{2}} \\ {1,} & {\text { otherwise. }}\end{array}\right.\) Show that \(f_{x}(0,0)\) and \(f_{y}(0,0)\) exist, but \(f\) is not differentiable at \((0,0) .\)

Find the dimensions of the rectangular box of maximum volume that can be inscribed inside the sphere \(x^{2}+y^{2}+z^{2}=4\) .

Locating a radio telescope You are in charge of erecting a radio telescope on a newly discovered planet. To minimize interference, you want to place it where the magnetic field of the planet is weakest. The planet is spherical, with a radius of 6 units. Based on a coordinate system whose origin is at the center of the planet, the strength of the magnetic field is given by \(M(x, y, z)=6 x-\) \(y^{2}+x z+60 .\) Where should you locate the radio telescope?

In Exercises \(63-66,\) use the limit definition of partial derivative to compute the partial derivatives of the functions at the specified points. $$f(x, y)=\sqrt{2 x+3 y-1}, \quad \frac{\partial f}{\partial x} \quad \text { and } \quad \frac{\partial f}{\partial y} \quad \text { at }(-2,3)$$

Find two numbers \(a\) and \(b\) with \(a \leq b\) such that $$\int_{a}^{b}\left(6-x-x^{2}\right) d x$$ has its largest value.
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