Problem 183 - Calculus Volume 3

Question

Find parametric equations for the normal line to the surface at the indicated point. (Recall that to find the equation of a line in space, you need a point on the line, \(P_{0}\left(x_{0}, y_{0}, z_{0}\right),\) and a vector \(\mathbf{n}=\langle a, b, c\rangle\) that is parallel to the line. Then the equation of \(\quad\) the \(\quad\) line \(\quad\) is \(\left.x-x_{0}=a t, y-y_{0}=b t, z-z_{0}=c t .\right)\). \(z=5 x^{2}-2 y^{2}, P(2,1,18)\)

Step-by-Step Solution

Answer

The parametric equations are \( x = 2 + 20t \), \( y = 1 - 4t \), and \( z = 18 - t \).

Step 1: Find the Gradient of the Surface

The gradient vector \( abla z = \langle \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, -1 \rangle \) is normal to the surface at any point. For the function \( z = 5x^2 - 2y^2 \), compute the partial derivatives: - \( \frac{\partial z}{\partial x} = 10x \)- \( \frac{\partial z}{\partial y} = -4y \)Then, the gradient vector is \( abla z = \langle 10x, -4y, -1 \rangle \).

Step 2: Evaluate the Gradient at Given Point

Evaluate the gradient vector at the point \( P(2, 1, 18) \): - Calculate \( 10x = 10(2) = 20 \)- Calculate \( -4y = -4(1) = -4 \)Thus, the gradient vector \( \mathbf{n} = \langle 20, -4, -1 \rangle \) at point \( P(2, 1, 18) \).

Step 3: Write Parametric Equations for the Normal Line

The parametric equations for the normal line using the point \( P(2, 1, 18) \) and the direction vector \( \mathbf{n} = \langle 20, -4, -1 \rangle \) are:- \( x = 2 + 20t \)- \( y = 1 - 4t \)- \( z = 18 - t \)These equations describe the line normal to the surface at the given point.

Key Concepts

Normal Line

In the context of three-dimensional surfaces, a normal line is a line that is perpendicular to the tangent plane of the surface at a given point. Imagine placing a pencil on the surface of an egg; the direction the pencil stands upright, without slanting, is akin to the normal line at that point.

To define a normal line in space mathematically, we need two components:

• A point on the line, denoted as \( P_0(x_0, y_0, z_0) \).
• A vector that is parallel to the line, commonly the gradient vector, \( \mathbf{n} = \langle a, b, c \rangle \).

The parametric equations of a normal line are then given by:

• \( x - x_0 = at \)
• \( y - y_0 = bt \)
• \( z - z_0 = ct \)

These equations depict how the coordinates change as you move along the line with respect to a parameter \( t \). The coordinate changes are in the same direction as the vector \( \mathbf{n} \). This concept is crucial for analyzing surfaces in fields like physics and engineering.

Gradient Vector

The gradient vector is a fundamental tool in multivariable calculus that indicates the direction and rate of the steepest ascent of a function. For a function \( f(x, y) \), its gradient is denoted as \( abla f \) and is defined as:

• \( abla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle \)

When dealing with a surface described by \( z = f(x, y) \), the gradient vector input is crucial because it also aligns perpendicularly with the tangent plane of the surface. Hence, the vector \( abla z = \left\langle \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, -1 \right\rangle \) acts perpendicularly to the surface and can serve as a direction vector for the normal line.

The negative unit in \( \left\langle \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, -1 \right\rangle \) reflects the gradient's role in three-dimensional space, taking into account the change in the \( z \)-direction as well.

Partial Derivatives

Partial derivatives are derivatives of functions with multiple variables, taken with respect to one variable at a time while treating other variables as constants. This concept is crucial for understanding how a change in one variable affects the function's outcome.

• The partial derivative of \( z \) with respect to \( x \) is \( \frac{\partial z}{\partial x} \), measuring how \( z \) changes as \( x \) changes while holding \( y \) constant.
• Similarly, \( \frac{\partial z}{\partial y} \) measures the change in \( z \) with respect to \( y \), keeping \( x \) constant.

When the surface is described by \( z = 5x^2 - 2y^2 \), these derivatives help us find the gradient vector, essential for determining the normal line. By computing these derivatives, we interpret the influence each variable has at a specific point on the surface.

In our example, they guide us to determine how the surface curves around the point \( P(2, 1, 18) \), providing the directional components of the normal vector.

Surface Equations

Surface equations are mathematical representations of surfaces in three-dimensional space. They typically take the form \( z = f(x, y) \) or, in some cases, implicit equations like \( g(x, y, z) = 0 \). These equations describe how a surface stretches over a plane, providing a map of points in space.

For the given problem, the surface is represented by the equation \( z = 5x^2 - 2y^2 \). This specific type of equation is called an explicit function of \( x \) and \( y \), meaning the value of \( z \) is directly determined by \( x \) and \( y \).

Surface equations are needed to understand how to find normals at any given point. They enable us to apply gradients and partial derivatives effectively and connect the puzzle pieces in understanding surfaces' geometry. Through this, we can visualize and analyze complex shapes, crucial for applications in computer graphics, design, and scientific modeling.