Problem 187 - 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 \(\left.x-x_{0}=a t, y-y_{0}=b t, z-z_{0}=c t .\right)\). \(z=x^{2}-2 x y+y^{2}\) at point \(P(1,2,1)\)

Step-by-Step Solution

Answer

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

Step 1: Find the normal vector

First, find the gradient of the surface function, which will be normal to the surface. The surface is given by \( z = f(x, y) = x^2 - 2xy + y^2 \). The gradient \( abla f \) is a vector of partial derivatives: \( \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1 \rangle \). Compute the partial derivatives: \( \frac{\partial f}{\partial x} = 2x - 2y \) and \( \frac{\partial f}{\partial y} = -2x + 2y \). At the point \( (1, 2) \), the gradients are \( \frac{\partial f}{\partial x} = 2(1) - 2(2) = -2 \) and \( \frac{\partial f}{\partial y} = -2(1) + 2(2) = 2 \). Thus, the normal vector is \( \mathbf{n} = \langle -2, 2, -1 \rangle \).

Step 2: Write parametric equations for the normal line

We have the point \( P_0(1, 2, 1) \) and a normal vector \( \mathbf{n} = \langle -2, 2, -1 \rangle \). The parametric equations for the line are given by \( x(t) = x_0 + at \), \( y(t) = y_0 + bt \), and \( z(t) = z_0 + ct \). Plug the given values into these equations: \( x(t) = 1 - 2t \), \( y(t) = 2 + 2t \), and \( z(t) = 1 - t \).

Key Concepts

Normal Line

A normal line to a surface is a line that is perpendicular, or orthogonal, to the surface at a given point. In the context of a function of two variables, like the surface defined by the equation \( z = x^2 - 2xy + y^2 \), the normal line at a point is particularly useful. It gives insight into the direction in which the surface is changing the fastest.
To describe any line in space, we need:

• A point on the line, such as \( P_0(1, 2, 1) \) in this exercise.
• A direction vector \( \mathbf{n} = \langle a, b, c \rangle \) that is parallel to the line.

The parametric equations of the line are derived using these, taking the form of \( x(t) = x_0 + at \), \( y(t) = y_0 + bt \), and \( z(t) = z_0 + ct \). The vector \( \mathbf{n} = \langle -2, 2, -1 \rangle \) found through partial derivatives, guides the line stemming from the surface through the chosen point.

Gradient

The gradient is a vector that points in the direction of the steepest ascent of a surface. It is composed of the partial derivatives of the function that defines the surface. For a function \( f(x, y) \), the gradient \( abla f \) is expressed as \( \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle \).
In our exercise, the surface's function is \( f(x, y) = x^2 - 2xy + y^2 \). The partial derivatives, \( \frac{\partial f}{\partial x} = 2x - 2y \) and \( \frac{\partial f}{\partial y} = -2x + 2y \), help in calculating the gradient. At point \( (1, 2) \), these become \( -2 \) and \( 2 \), respectively, forming the vector \( \langle -2, 2 \rangle \).
To complete the normal vector to the surface, the gradient in the \( z \)-direction is considered as \( -1 \). Thus, the full gradient vector \( abla f \) informs us about the surface's normal at that specific point.

Partial Derivatives

Partial derivatives tell us how a function changes as we vary one variable while keeping the others constant. They provide insight into the function's behavior and are essential in finding the gradient.
For the given function \( f(x, y) = x^2 - 2xy + y^2 \), we find:

• \( \frac{\partial f}{\partial x} = 2x - 2y \), which signifies how \( f \) changes with \( x \).
• \( \frac{\partial f}{\partial y} = -2x + 2y \), which signifies how \( f \) changes with \( y \).

At \( (1, 2) \), substituting the values gives \( \frac{\partial f}{\partial x} = -2 \) and \( \frac{\partial f}{\partial y} = 2 \). These derivatives are combined to form part of the normal vector \( \mathbf{n} = \langle -2, 2, -1 \rangle \), essential for defining the parametric equations for the normal line.

Surface Function

A surface function is essentially a mathematical representation of a surface in three-dimensional space. It is typically expressed in terms of \( x \) and \( y \) for a dependent variable \( z \). In this exercise, the surface function is given by \( z = f(x, y) = x^2 - 2xy + y^2 \).
Understanding surface functions is crucial since they describe the entire surface over which various mathematical problems are defined. These functions map out a surface in a 3D space using a formula that details how height \( z \) changes with respect to horizontal dimensions \( x \) and \( y \).
Working with surface functions involves differentiating them partially to understand slopes and changes, which was key in finding the normal line in this exercise. With knowledge about how the function behaves, particularly through the use of gradients and partial derivatives, one can determine any line's characteristics. This includes determining the normal line, which is vital in fields like geometry and physics.