Problem 34 - Advanced Engineering Mathematics

Question

Find parametric equations for the normal line at the indicated point. In Problems 35 and 36, find symmetric equations for the normal line. $$ z=2 x^{2}-4 y^{2} ;(3,-2,2) $$

Step-by-Step Solution

Answer

Parametric: \(x = 3 - 12t, y = -2 - 16t, z = 2 + t\). Symmetric: \(\frac{x-3}{-12} = \frac{y+2}{-16} = \frac{z-2}{1}\).

Step 1: Determine the Normal Vector

To find the parametric equations for the normal line, we first need the normal vector of the surface at the given point. Since the surface is given by the equation \( F(x, y, z) = z - 2x^2 + 4y^2 \), the gradient of \( F(x, y, z) \) will give us the normal vector \( abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) = (-4x, 8y, 1) \). At the point \((3, -2, 2)\), the normal vector is \(abla F = (-12, -16, 1)\).

Step 2: Write Parametric Equations for the Normal Line

The parametric equations of the normal line can be written using the point \((3, -2, 2)\) and the normal vector \((-12, -16, 1)\). The parametric equations are given by: \[ x = 3 - 12t \]\[ y = -2 - 16t \]\[ z = 2 + t \]These equations describe a line starting at \((3, -2, 2)\) with the direction vector \((-12, -16, 1)\).

Step 3: Find Symmetric Equations

For the symmetric equations of the normal line, we equate the parameters \( t \) from the parametric equations: \[ \frac{x-3}{-12} = \frac{y+2}{-16} = \frac{z-2}{1} \]These symmetric equations show the relationship between \(x, y,\) and \(z\) along the normal line.

Key Concepts

Parametric Equations

Parametric equations are a way to describe geometric objects such as lines, curves, or surfaces in terms of parameters. They express each coordinate as a separate function of one or more parameters, offering us significant flexibility and insights into the dynamics of the shape.
For instance, in the context of a normal line on a surface, we often use a parameter, usually denoted by "t," to express each point on the line. Each coordinate (x, y, and z) is given as a function of "t," representing how the line progresses through these coordinates as "t" changes.
The parametric equations for the normal line described above are:

• \( x = 3 - 12t \)
• \( y = -2 - 16t \)
• \( z = 2 + t \)

These equations give us a vector form of a line, meaning that as "t" varies, we trace out a series of points on the line. Here, the coeffecients of "t" are derived from the normal vector, which indicates the line's direction, starting at the specific point (3, -2, 2). This highlights the beauty and simplicity of parametric equations— they provide a clear pathway to understand motion and trajectory.

Symmetric Equations

Symmetric equations present another approach to describe a line in three-dimensional space, and they are derived from parametric equations by eliminating the parameter. These equations relate all coordinates directly, showcasing the intrinsic relationship without referencing a parameter.
In a symmetric equation, all variables are expressed in terms of a common divider represented by the direction vector of the line. This results in equations of the form:
\[\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\]
Where \((x_0, y_0, z_0)\) is a specific point on the line, and \((a, b, c)\) is the direction vector.
For the normal line described, the symmetric equations are:

• \(\frac{x-3}{-12} = \frac{y+2}{-16} = \frac{z-2}{1}\)

These expressions are beneficial as they succinctly capture the essence of the line's orientation in space, offering clarity for analysis and calculation. Symmetric equations are thus an elegant way to understand connections in multidimensional settings.

Gradient Vector

The gradient vector is a crucial concept in multivariable calculus. When we have a function, the gradient vector provides the direction of steepest ascent at any given point. It is composed of all partial derivatives of the function with respect to each variable, making it a vector that gives us both magnitude and direction.
In the context of a surface described by the function \(z = 2x^2 - 4y^2\), we first reformulate it as \(F(x, y, z) = z - 2x^2 + 4y^2\) to use the gradient. Calculating the gradient involves determining:\(abla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)\).
In our example, this results in a gradient vector \((-4x, 8y, 1)\). Evaluating at the point (3, -2, 2) gives us the specific normal vector \((-12, -16, 1)\), which is used to define the direction of the normal line.
This vector is pivotal in expressing both parametric and symmetric equations of the normal line. It encapsulates vital geometric information, making the gradient vector a powerful tool for analyzing curves and surfaces.