Problem 305 - Calculus Volume 3

Question

For the following exercises, find equations of a. the tangent plane and b. the normal line to the given surface at the given point. \(f(x, y, z)=x e^{y} \cos z-z=1\) at point (1,0,0).

Step-by-Step Solution

Answer

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

Step 1: Identify the Given Function and Point

We have the surface described by the equation \( f(x, y, z) = x e^{y} \cos z - z = 1 \), and we need to find the tangent plane and normal line at the point \((1, 0, 0)\). The point \((1, 0, 0)\) is on the surface since substituting it into \( f \) gives: \( 1 \cdot e^0 \cdot \cos 0 - 0 = 1 \).

Step 2: Calculate the Gradient of f

The gradient of \( f(x, y, z) = x e^y \cos z - z \) is the vector of partial derivatives: \[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right). \]Calculate each partial derivative:- \( \frac{\partial f}{\partial x} = e^y \cos z \),- \( \frac{\partial f}{\partial y} = x e^y \cos z \),- \( \frac{\partial f}{\partial z} = -x e^y \sin z - 1 \).

Step 3: Evaluate the Gradient at the Given Point

Substitute \( (1, 0, 0) \) into each component of \( abla f \):- \( \frac{\partial f}{\partial x}(1, 0, 0) = e^0 \cdot \cos 0 = 1 \),- \( \frac{\partial f}{\partial y}(1, 0, 0) = 1 \cdot e^0 \cdot \cos 0 = 1 \),- \( \frac{\partial f}{\partial z}(1, 0, 0) = -1 \cdot e^0 \cdot \sin 0 - 1 = -1 \).Therefore, \( abla f(1, 0, 0) = (1, 1, -1) \).

Step 4: Find the Equation of the Tangent Plane

The equation of the tangent plane to the surface at the point \((x_0, y_0, z_0)\) is:\[ (x - x_0) \frac{\partial f}{\partial x} + (y - y_0) \frac{\partial f}{\partial y} + (z - z_0) \frac{\partial f}{\partial z} = 0. \]At \((1, 0, 0)\), this becomes:\[ (x - 1) \cdot 1 + (y - 0) \cdot 1 + (z - 0) \cdot (-1) = 0 \] Simplifying, we have:\[ x + y - z = 1. \]

Step 5: Find the Equation of the Normal Line

The normal line to the surface at \((x_0, y_0, z_0)\) uses the direction of \(abla f\). Parametrize the line using the point \((1, 0, 0)\) and direction \((1, 1, -1)\):\[ x = 1 + t, \quad y = t, \quad z = 0 - t. \]This gives us the parametric equations for the normal line.

Key Concepts

Normal Line

In geometry, a normal line is a line that is perpendicular to a tangent plane at a given point. For a surface, this line represents the direction in which the surface rises most steeply from the tangent plane.

To find the normal line to the given surface equation, we first determine the gradient vector at the specified point. Once we have this gradient vector, which is also normal to the tangent plane, we use it as the direction vector for the line.

For the surface defined by the equation:

• \( f(x, y, z) = x e^y \cos z - z = 1 \)

At the point

• (1, 0, 0)

We found the direction of the normal line to be \((1, 1, -1)\). We can express the normal line as a parametric equation:

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

Here, "\(t\)" is a parameter that scales the direction of the gradient vector and shifts the line away or towards the given point, constructing the normal line.

Gradient Vector

The gradient vector is a key mathematical tool that provides the direction of steepest ascent or descent for a function. It is comprised of partial derivatives with respect to each variable.

For the function defining our surface, the gradient vector \( abla f \) is:

• \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \)

Specifically:

• \( \frac{\partial f}{\partial x} = e^y \cos z \)
• \( \frac{\partial f}{\partial y} = x e^y \cos z \)
• \( \frac{\partial f}{\partial z} = -x e^y \sin z - 1 \)

Evaluating these at the point

• (1, 0, 0)

results in a gradient vector of:

• \( (1, 1, -1) \)

This vector not only points in the direction of greatest change of the function value but also serves as the normal vector for the tangent plane.

Partial Derivatives

Partial derivatives measure how a function changes as each variable changes, while keeping the other variables constant. They are the foundation of the gradient vector, giving us insight into the sensitivity of the function with respect to each variable.

For our surface equation:

• \( f(x, y, z) = x e^y \cos z - z \)

The partial derivatives are:

• \( \frac{\partial f}{\partial x} = e^y \cos z \)
• \( \frac{\partial f}{\partial y} = x e^y \cos z \)
• \( \frac{\partial f}{\partial z} = -x e^y \sin z - 1 \)

These derivatives give us the rates at which the function is changing in the \(x\), \(y\), and \(z\) directions specifically at any point on this surface. For a point like

• (1, 0, 0)

we substitute into these derivatives to evaluate their values and find the gradient vector.

Surface Equation

The surface equation is the mathematical representation of a three-dimensional shape defined by an equation in three variables. In this exercise, the surface is described by:

• \( f(x, y, z) = x e^y \cos z - z = 1 \)

This equation implies a specific spatial form or surface in 3D space. The task requires understanding the position and shape of this surface.

At a given point, such as

• (1, 0, 0)

we check if it satisfies the surface equation, which it does, confirming the point lies on the surface. These equations are essential for defining boundaries or constraints in physical problems, like finding tangent planes or normal lines.
In the course of solving geometrical problems like finding tangent planes or normal lines, the surface equation serves as the constraint from which important derivative information is derived. This information is crucial in visualizing properties like tangency and orthogonality, or in solving further related mathematical problems.