Problem 42 - Calculus Early Transcendentals

Question

\(39-44\) Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point. $$x-z=4 \arctan (y z), \quad(1+\pi, 1,1)$$

Step-by-Step Solution

Answer

(a)\( x - 2y - 3z = -3-\pi \), (b) \( x=1+\pi+t, y=1-2t, z=1-3t \)."

Step 1: Understand the Surface Equation

The given surface equation is \( x - z = 4 \arctan(yz) \). This is a level surface equation which we will use to find the tangent plane and normal line at the specified point \((1+\pi, 1, 1)\).

Step 2: Find Partial Derivatives

To find the equation of the tangent plane and the normal line, we need the gradient of the implicit function \( F(x, y, z) = x - z - 4 \arctan(yz) \), which represents the surface. Calculate the partial derivatives: \( F_x = 1 \), \( F_y = -\frac{4z}{1+(yz)^2} \), and \( F_z = -1 - \frac{4y}{1+(yz)^2} \).

Step 3: Evaluate Gradient at the Point

Evaluate the gradient, \( abla F = (F_x, F_y, F_z) \), at the point \((1+\pi, 1, 1)\). Substitute these coordinates into the partial derivatives to find: \( F_x = 1 \), \( F_y = -2 \), and \( F_z = -3 \). Thus, \( abla F(1+\pi, 1, 1) = (1, -2, -3) \).

Step 4: Write the Tangent Plane Equation

The equation of the tangent plane at a point \((x_0, y_0, z_0)\) on a surface is \( F_x(x - x_0) + F_y(y - y_0) + F_z(z - z_0) = 0 \). For the point \((1+\pi, 1, 1)\), this becomes: \( 1(x - (1+\pi)) - 2(y - 1) - 3(z - 1) = 0 \). Simplify to get the tangent plane equation: \( x - 2y - 3z = -3 - \pi \).

Step 5: Write the Normal Line Equation

The normal line can be expressed parametrically using the point and direction given by the gradient. If \((x_0, y_0, z_0)\) is the point and \((a, b, c)\) is the gradient, then the parametric equations are \( x = x_0 + at \), \( y = y_0 + bt \), \( z = z_0 + ct \). Thus, we have: \( x = 1+\pi+t \), \( y = 1-2t \), \( z = 1-3t \).

Key Concepts

Partial Derivatives

When dealing with equations of surfaces, partial derivatives come into play as they help us understand how a multivariable function changes with respect to each of its variables. Let's break it down:

• If you have a function like \( F(x, y, z) \), the partial derivative \( F_x \) represents the rate of change of \( F \) with respect to \( x \), while keeping the other variables \( y \) and \( z \) constant.
• Similarly, \( F_y \) and \( F_z \) represent the rate of change with respect to \( y \) and \( z \), respectively, while holding the others constant.

In the context of our problem, calculating partial derivatives of \( F(x, y, z) = x - z - 4 \arctan(yz) \) gives us insights into the slopes of the surface at different directions. This is crucial for subsequent calculations like determining the tangent plane and normal line.

Normal Line

The concept of a normal line is tied closely to understanding how a surface behaves in three-dimensional space. At any given point on a surface, the normal line is perpendicular to the tangent plane.

• Think of it as the way a toothpick might point directly out from a piece of paper if the paper were bent.
• The direction of the normal line is given by the gradient of the surface's equation at that point.

For a surface represented by \( F(x, y, z) = 0 \), once you have the gradient vector—consisting of the partial derivatives \( F_x, F_y, \) and \( F_z \)—this vector not only shows the steepest direction of the surface at that point but also serves as the direction vector for the normal line. Using the point of interest and this gradient vector, we write the parametric equations of the normal line. For instance, in our exercise, they are \( x = 1+\pi+t \), \( y = 1-2t \), and \( z = 1-3t \).

Gradient

The gradient is a vector that encapsulates all the partial derivatives of a function, and it plays a pivotal role when analyzing surfaces. Here's why it's important:

• The gradient vector \( abla F = (F_x, F_y, F_z) \) points in the direction of the steepest ascent of the function \( F \).
• Its magnitude tells us how steep the slope is in that direction.
• For surfaces, the gradient at a point not only indicates how the surface is inclined but also serves as the normal to the tangent plane at that point.

In our problem, after calculating the partial derivatives, the gradient at the point \((1+\pi, 1, 1)\) was determined to be \((1, -2, -3)\). This vector became crucial in writing the equations for both the tangent plane and the normal line as it dictates their orientation.

Implicit Function

An implicit function is one where the relationship between variables is given without solving for any single variable explicitly. Surfaces in three-dimensional space are often described this way.For example, in the equation \( x - z = 4 \arctan(yz) \), \( x, y, \) and \( z \) interact in a way that isn't straightforwardly solved for one of the variables. Instead, we treat the equation as defining a set of points fulfilling this relation.

• This form is particularly valuable as it directly describes the surface or curve without isolating a single variable.
• To find features like the tangent plane or normal line at a particular point, we consider \( F(x, y, z) = x - z - 4 \arctan(yz) \) as it represents all possible points (\( x, y, z \)) that form the surface.

This implicit approach requires us mostly to focus on derivatives, making partial derivatives and the gradient essential tools in our toolkit for exploring and understanding the behavior of the surface.