Problem 5 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 12 , find an equation of the tangent plane and equations of the normal line to the given surface at the indicated point. \(y=e^{x} \cos z ;(1, e, 0)\)

Step-by-Step Solution

Answer

The tangent plane is \( y = ex\) and the normal line equations are \( x = 1 - et \), \( y = e + t \), \( z = 0 \).

Step 1: Find the partial derivatives

Calculate the partial derivatives of the function with respect to x, y, and z.\[ f(x, y, z) = y - e^x \, \text{cos} \, z \]\[ f_x = -e^x \, \text{cos} \, z \]\[ f_y = 1 \]\[ f_z = e^x \, \text{sin} \, z \]

Step 2: Evaluate the partial derivatives at the given point

Substitute the given point (1, e, 0) into the partial derivatives calculated in Step 1.\[ f_x(1, e, 0) = -e \, \text{cos} \, (0) = -e \]\[ f_y(1, e, 0) = 1 \]\[ f_z(1, e, 0) = e \, \text{sin} \, (0) = 0 \]

Step 3: Form the equation of the tangent plane

Use the point (1, e, 0) and the evaluated partial derivatives to form the equation of the tangent plane.\[ -e(x - 1) + 1(y - e) + 0(z - 0) = 0 \]\[ -ex + e + y - e = 0 \]Simplify the equation: \[ -ex + y = 0 \]Thus, \[ y = ex \]

Step 4: Write the parametric equations of the normal line

The parametric equations of the normal line can be found using the given point and the gradient vector \( (f_x, f_y, f_z) \).\[ x = 1 - et \]\[ y = e + t \]\[ z = 0 \]

Key Concepts

partial derivatives

Before we can find the tangent plane or normal line to a surface, we need to understand partial derivatives. Partial derivatives measure how a function changes as one of its variables changes, while the other variables are held constant.
For a function of three variables, like our given function, the partial derivatives with respect to each variable can be found by differentiating the function in terms of one variable at a time. This results in three different partial derivatives: \( \frac{\text{d}f}{\text{d}x} \, \frac{\text{d}f}{\text{d}y}, \text{ and } \frac{\text{d}f}{\text{d}z} \).
In our exercise, we started with the function \( f(x, y, z) = y - e^x \text{cos} z \) and calculated its partial derivatives:

• \( f_x = -e^x \text{cos} z \)
• \( f_y = 1 \)
• \( f_z = e^x \text{sin} z \)

These derivatives tell us how the function changes in the \( x \), \( y \), and \( z \) directions, respectively.

normal line

Once we've found the partial derivatives, we can determine the normal line to the surface. The normal line is perpendicular to the tangent plane at a given point.
We use the gradient vector, which is made up of the partial derivatives, to find the direction of the normal line. For our function \( f(x, y, z) \) at the point \( (1, e, 0) \), the gradient vector is \( (-e, 1, 0) \).
The equations of the normal line are then found by using the point and the gradient vector:

• \( x = 1 - et \)
• \( y = e + t \)
• \( z = 0 \)

Here, \( t \) is a parameter that moves along the normal line.

tangent plane equation

The next step is to find the equation of the tangent plane to the surface at the given point. The tangent plane approximates the surface near that point. The general formula for the equation of the tangent plane to \( f(x, y, z) = 0 \) at \( (x_0, y_0, z_0) \) is: \[ f_x(x_0, y_0, z_0) (x - x_0) + f_y(x_0, y_0, z_0) (y - y_0) + f_z(x_0, y_0, z_0) (z - z_0) = 0 \]
For our function, using \( f_x = -e \), \( f_y = 1 \), and \( f_z = 0 \) at point \( (1, e, 0) \), we form the tangent plane equation:
\( -e(x - 1) + 1(y - e) + 0(z - 0) = 0 \)
Simplifying, we get:
\( -ex + y = 0 \), which is the equation of the tangent plane.

gradient vector

The gradient vector is a crucial concept for understanding both the tangent plane and the normal line. The gradient vector of a function \( f(x, y, z) \) is given by:
\( abla f = (f_x, f_y, f_z) \)
This vector points in the direction of the steepest ascent of the function and is perpendicular to the tangent plane.
In our problem, we calculated the gradient vector at point \( (1, e, 0) \) as:
\( abla f = (-e, 1, 0) \)
This vector was then used to determine both the direction of the normal line and to form the equation of the tangent plane. Understanding the gradient vector helps us connect the behavior of the function in different directions and form a precise geometric interpretation of the surface.