Problem 187 - Calculus 3

Question

For the following exercises, 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 \(x-x_{0}=a t, y-y_{0}=b t, z-z_{0}=c t . )\) $$z=x^{2}-2 x y+y^{2} \text { at point } P(1,2,1)$$

Step-by-Step Solution

Answer

The parametric equations for the normal line are \( x = 1 - 2t, y = 2 + 2t, z = 1 - t. \)

Step 1: Identify Point and Surface

We are given the surface equation \( z = x^2 - 2xy + y^2 \) and the point \( P(1, 2, 1) \). This means that \( (x_0, y_0, z_0) = (1, 2, 1) \). We need to find the parametric equations for the normal line to this surface at this particular point.

Step 2: Find Gradient Vector

The gradient vector \( abla f \) is normal to the surface. Calculate the partial derivatives: \[ f_x = \frac{\partial z}{\partial x} = 2x - 2y \] and \[ f_y = \frac{\partial z}{\partial y} = -2x + 2y. \] Evaluate at the point \( (1, 2) \): \( f_x(1, 2) = 2 \times 1 - 2 \times 2 = -2 \) and \( f_y(1, 2) = -2 \times 1 + 2 \times 2 = 2 \). Thus, the gradient vector is \( abla f = \langle -2, 2, -1 \rangle \).

Step 3: Write Parametric Equations

Using the point \( P(1, 2, 1) \) and the gradient vector \( \mathbf{n} = \langle a, b, c \rangle = \langle -2, 2, -1 \rangle \), we can write the parametric equations for the normal line as: \[ x = 1 - 2t, \] \[ y = 2 + 2t, \] \[ z = 1 - t. \]

Key Concepts

Gradient Vector

The gradient vector plays a crucial role in understanding how a multivariable function changes around a point. It is a vector that points in the direction of the greatest rate of increase of the function. For a function of two variables, like our exercise's equation for the surface, \( z = f(x, y) \), the gradient vector, \( abla f \), is comprised of the partial derivatives of \( f \) with respect to each variable.

• The component \( \frac{\partial f}{\partial x} \) indicates how the function changes as \( x \) changes, while \( y \) stays constant.
• Similarly, \( \frac{\partial f}{\partial y} \) shows how \( f \) changes with respect to \( y \), holding \( x \) constant.

In our problem, \( abla f \) is \( \langle -2, 2 \rangle \), evaluated at point \( (1, 2) \). This vector is perpendicular to the level curve at that point, which means it also represents the direction of a straight line normal to the surface at \( P(1, 2, 1) \). By computing the gradient, we find the normal line's direction vector, essential for setting up parametric equations.

Parametric Equations

Parametric equations are used to express lines in terms of parameters, which can describe motion along the line or geometric locations over time. For a line in 3D space, parametric equations are derived from a point on the line and a vector parallel to the line. This vector indicates the line's direction.

• In our exercise, the point is \( P(1, 2, 1) \), and the gradient vector \( \langle -2, 2, -1 \rangle \) serves as the direction vector.
• The parametric equations, derived as \( x = 1 - 2t \), \( y = 2 + 2t \), and \( z = 1 - t \), express coordinates \((x, y, z)\) in terms of a parameter \( t \).

These equations convey how the coordinates change as the parameter \( t \) varies. Such representation helps visualize or calculate points on a normal line effortlessly.

Partial Derivatives

Partial derivatives are fundamental in multivariable calculus to understand how functions change with respect to one variable while keeping others constant. For a function \( z = f(x, y) \), partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) are useful for analyzing changes in \( z \) with varying \( x \) or \( y \).

• In this exercise, we calculate the partial derivatives of the surface equation \( z = x^2 - 2xy + y^2 \).
• We get \( f_x = 2x - 2y \) and \( f_y = -2x + 2y \).

By substituting \( x = 1 \) and \( y = 2 \) into these derivatives, we find \( f_x(1, 2) = -2 \) and \( f_y(1, 2) = 2 \). These values form the gradient vector \( \langle -2, 2 \rangle \), which helps in determining the line normal to the surface. Understanding partial derivatives is key for gradient calculation and provides valuable insight into a function's behavior locally.