Problem 43 - 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. $$z+1=x e^{y} \cos z, \quad(1,0,0)$$

Step-by-Step Solution

Answer

Tangent plane: \(x - y - z = 1\); Normal line: \(x = 1 - t, y = -t, z = t\).

Step 1: Understand the Given Surface Equation

The given surface equation is \(z + 1 = x e^{y} \cos z\). We need to find the tangent plane and normal line for this surface at the point \((1, 0, 0)\).

Step 2: Find the Gradient of the Surface Function

First, express the surface as a level surface \( F(x, y, z) = z + 1 - x e^{y} \cos z = 0 \). The gradient \( abla F \) is \( \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \).

Step 3: Evaluate Partial Derivatives

Compute each partial derivative. - \( \frac{\partial F}{\partial x} = -e^y \cos z \) - \( \frac{\partial F}{\partial y} = -x e^y \cos z \) - \( \frac{\partial F}{\partial z} = 1 + x e^y \sin z \).

Step 4: Evaluate Gradients at the Point

Evaluate the partials at \((1, 0, 0)\):- \( \frac{\partial F}{\partial x} = -e^0 \cos 0 = -1 \)- \( \frac{\partial F}{\partial y} = -1 \cdot e^0 \cdot \cos 0 = -1 \)- \( \frac{\partial F}{\partial z} = 1 + 1 \cdot e^0 \cdot \sin 0 = 1 \).This gives \( abla F(1, 0, 0) = (-1, -1, 1) \).

Step 5: Equation of the Tangent Plane

The equation of the tangent plane at \((x_0, y_0, z_0) = (1, 0, 0)\) is given by:\[-1(x - 1) - 1(y - 0) + 1(z - 0) = 0\]Simplifying gives \(-x + y + z = -1\) or \(x - y - z = 1\).

Step 6: Equation of the Normal Line

The normal line at \((1, 0, 0)\) can be given in parametric form using the gradient as direction vectors:\[ x = 1 - t, \quad y = 0 - t, \quad z = 0 + t \]Or explicitly: \( x = 1 - t, y = -t, z = t \).

Key Concepts

Normal Line

A normal line to a surface at a given point is a line that is perpendicular to the tangent plane at that point. In the context of a three-dimensional surface, the normal line extends outward in both directions. It's an important concept because it helps us understand how a surface behaves at a specific location.
To find the normal line, we can use the gradient of the surface, which offers the direction vector for the normal line. Suppose we have evaluated the gradient as \( abla F = (-1, -1, 1) \) at the point \((1, 0, 0)\).

• The direction vector for the normal line is then \( (-1, -1, 1) \).
• Using this vector, we can establish the parametric equations for the normal line:
• \( x = 1 - t \)
• \( y = -t \)
• \( z = t \)

Here, \( t \) represents a parameter which can vary, defining the position along the line.
This representation gives us a vivid picture of how the normal line is oriented and helps in visualizing the spatial structure of the surface at the specific point.

Gradient of a Surface

The gradient of a surface is a vector that points in the direction of greatest increase of a function of two or more variables. For a surface represented by a function \( F(x, y, z) \), the gradient is computed as \( abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \).
The gradient not only provides the direction of maximum slope but also its magnitude.

• In the exercise, the function is transformed into a level surface: \( F(x, y, z) = z + 1 - x e^{y} \cos z = 0 \).
• Its gradient is calculated using partial derivatives for each of \( x \), \( y \), and \( z \):

After computing, at the point \((1, 0, 0)\), the gradient is \( (-1, -1, 1) \).
This vector is perpendicular to the tangent plane of the surface at that point, making it central to both tangent plane and normal line calculations.

Partial Derivatives

Partial derivatives are foundational in multivariable calculus, involving functions with more than one variable. They represent how a function changes as one particular variable changes, holding others constant.
In this exercise, partial derivatives are pivotal in finding the gradient of the surface:

• _{\( \frac{\partial F}{\partial x} \)} measures the change in \( F \) as \( x \) changes, resulting in \( -e^y \cos z \) for our equation.
• _{\( \frac{\partial F}{\partial y} \)} measures the change in \( F \) with varying \( y \), here \( -x e^y \cos z \).
• _{\( \frac{\partial F}{\partial z} \)} focuses on changes in \( z \), giving \( 1 + x e^y \sin z \).

Evaluated at the point \((1, 0, 0)\), these derivatives yield values that make up the gradient vector. Thus, partial derivatives are essential calculations that help us understand how the function behaves locally at each point on the surface.