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Chapter 13: Problem 14

Find an equation for the line normal to \(x^{2}+y^{2}+9 z^{2}=56\) at (4,2,-2) .

Short Answer

Expert verified
The normal line's equation is \((x, y, z) = (4 + 8t, 2 + 4t, -2 - 36t)\).

Step by step solution

01

Understand the Surface

The surface described by the equation \(x^2 + y^2 + 9z^2 = 56\) represents an ellipsoid. Understanding the geometry is vital because we need to find a normal to this surface at a given point.
02

Compute the Gradient

To find the normal line, we calculate the gradient of the equation \(x^2 + y^2 + 9z^2 - 56 = 0\). The gradient \(abla f(x, y, z)\) is \((2x, 2y, 18z)\).
03

Evaluate Gradient at the Point

Substitute the coordinates (4, 2, -2) into the gradient formula. This gives: \(abla f(4, 2, -2) = (8, 4, -36)\). This vector is normal to the surface at the given point.
04

Form the Equation of the Normal Line

A parametric equation for the line through the point (4, 2, -2) in the direction of \((8, 4, -36)\) can be given by: \((x, y, z) = (4 + 8t, 2 + 4t, -2 - 36t)\), where \(t\) is a parameter.

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Key Concepts

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

Ellipsoid
An ellipsoid is a three-dimensional surface that looks somewhat like a stretched or compressed sphere. It's defined by a quadratic equation, often of the form \( ax^2 + by^2 + cz^2 = d \). In our case, the equation \( x^2 + y^2 + 9z^2 = 56 \) describes an ellipsoid. Here, the coefficients of \( x^2 \) and \( y^2 \) are 1, while \( z^2 \) is multiplied by 9.
This means the ellipsoid is stretched more along the z-axis than along the x- or y-axes. The term 56 on the right side is a constant that generally represents how the ellipsoid is scaled.
Understanding this scaling and stretching is crucial, as it affects where and how the normal line intersects the surface at a given point.
Gradient Vector
The gradient vector is a crucial concept in calculus, especially when dealing with surfaces and curves. It is a vector that points in the direction of the steepest ascent on a surface.
For a function \( f(x, y, z) \), the gradient, denoted as \( abla f \), is a vector composed of the partial derivatives of the function. Specifically, if \( f(x, y, z) = x^2 + y^2 + 9z^2 - 56 \),
then the gradient is \( (2x, 2y, 18z) \).
  • The first component (partial derivative with respect to \( x \)) is \( 2x \).
  • The second component (partial with respect to \( y \)) is \( 2y \).
  • And the third component (partial with respect to \( z \)) is \( 18z \).
At the point (4, 2, -2), substituting the values yields the gradient \( abla f(4, 2, -2) = (8, 4, -36) \).
This vector is perpendicular to the ellipsoid at that point, making it the ideal direction for forming the normal line.
Parametric Equation of a Line
The parametric equation of a line provides a smooth way to describe the direction and position of a line in space using a parameter, usually denoted by \( t \).
For a line passing through a point \( (x_0, y_0, z_0) \) in the direction of a vector \( (a, b, c) \), the parametric equations are:
  • \( x = x_0 + at \)
  • \( y = y_0 + bt \)
  • \( z = z_0 + ct \)
In our exercise, the point is \( (4, 2, -2) \) and the direction is given by the gradient vector \( (8, 4, -36) \).
Plugging these into the parametric form results in:
  • \( x = 4 + 8t \)
  • \( y = 2 + 4t \)
  • \( z = -2 - 36t \)
This set of equations fully characterizes the trajectory of the line normal to the ellipsoid at the specified point.

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