Problem 35 - 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. $$ z=4 x^{2}+9 y^{2}+1 ;\left(\frac{1}{2}, \frac{1}{3}, 3\right) $$

Step-by-Step Solution

Answer

The symmetric equations for the normal line are \(\frac{x - \frac{1}{2}}{4} = \frac{y - \frac{1}{3}}{6} = 3 - z\).

Step 1: Find the Gradient Vector at the Given Point

The gradient vector is the vector of partial derivatives, which for the function \( z = 4x^2 + 9y^2 + 1 \) is \( abla z = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, \frac{\partial z}{\partial z} \right) \). Compute these at the point \( \left( \frac{1}{2}, \frac{1}{3}, 3 \right) \).- \( \frac{\partial z}{\partial x} = 8x \), so at \( x = \frac{1}{2} \), it is \( 8 \times \frac{1}{2} = 4 \).- \( \frac{\partial z}{\partial y} = 18y \), so at \( y = \frac{1}{3} \), it is \( 18 \times \frac{1}{3} = 6 \).- Since \( z \) is only in terms of \( x \) and \( y \), \( \frac{\partial z}{\partial z} = -1 \) because we consider it implicitly from rising to \( z \) itself in spatial terms.Thus, the gradient vector is \( abla z = (4, 6, -1) \).

Step 2: Write Parametric Equations for the Normal Line

The normal line passes through the point \( \left( \frac{1}{2}, \frac{1}{3}, 3 \right) \) and has direction given by the gradient. The parametric equations are:\[x(t) = x_0 + at = \frac{1}{2} + 4t\]\[y(t) = y_0 + bt = \frac{1}{3} + 6t\]\[z(t) = z_0 + ct = 3 - 1t = 3 - t\]

Step 3: Convert Parametric Equations to Symmetric Equations

To convert to symmetric equations, eliminate the parameter \( t \) from the parametric equations.Equations: - \( t = x - \frac{1}{2}/4 \)- \( t = y - \frac{1}{3}/6 \)- \( t = 3 - z \)Therefore, the symmetric equations are:\[\frac{x - \frac{1}{2}}{4} = \frac{y - \frac{1}{3}}{6} = 3 - z\]

Key Concepts

Normal Line

A normal line is a concept that refers to a line that is perpendicular to a surface at a given point. When dealing with three-dimensional surfaces, normal lines are very useful in determining the orientation of the surface at any given point. In particular, it has the direction of the gradient vector at that point. The importance of the normal line lies in its perpendicular nature, offering a handy way to assess the slope of the surface or any tangent plane touching the surface.
To find the parametric equations of a normal line, the process involves these steps:

• Identify the surface equation and the point of interest on this surface.
• Compute the gradient vector, as it indicates the direction of the normal line.
• Use the point and the gradient to establish the parametric equations, which illustrate how the line extends away from the surface.

These parametric equations allow you to model how the line behaves in space, showing the trajectory of the normal line as it moves through the three-dimensional plane.

Gradient Vector

The gradient vector is a multi-dimensional concept that indicates the direction of the steepest ascent for a surface. It is derived from the partial derivatives of a function, indicating how the function changes with respect to each variable.

• Symbolically denoted as \( abla f \), where \( f \) is the function defining the surface.
• For a function \( z = f(x, y) \), the gradient is a vector comprised of \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).
• This gradient vector serves as the perpendicular (normal) direction to the surface at a given point.

Calculating the gradient vector involves evaluating these partial derivatives at the specific point of interest. In our exercise, for the surface defined by \( z = 4x^2 + 9y^2 + 1 \), the gradient at \( \left( \frac{1}{2}, \frac{1}{3}, 3 \right) \) is \( (4, 6, -1) \), which helps us find the direction of the normal line.

Symmetric Equations

Symmetric equations give a straightforward way of comparing different parts of a line by eliminating the parameter from its parametric forms. For those familiar with parametric equations, this conversion enhances understanding by aligning multiple linear expressions to a common form.

• Each part of the line relates dimensionally, providing a neat, collective expression.
• Symmetric equations are often found by setting each parametric equation equal to the parameter term \( t \) and then solving sequentially.

In the provided exercise, we transform parametric equations:\[x(t) = \frac{1}{2} + 4t, \quad y(t) = \frac{1}{3} + 6t, \quad z(t) = 3 - t\]into symmetric equations:\[\frac{x - \frac{1}{2}}{4} = \frac{y - \frac{1}{3}}{6} = 3 - z\]These allow for a holistic geometric interpretation of the normal line by correlating the changes in \( x \), \( y \), and \( z \), making it easier to analyze and visualize.