Problem 305 - Calculus 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. $$ f(x, y, z)=x e^{y} \cos z-z=1 \text { at point }(1,0,0) $$

Step-by-Step Solution

Answer

The tangent plane equation is \( x + y - z = 1 \) and the normal line is \( x = 1 + t, y = t, z = -t \).

Step 1: Calculate the Gradient of the Surface

We need the partial derivatives of the function with respect to each variable. The function is given by: \[ f(x, y, z) = x e^y \cos z - z \]- **Partial derivative with respect to** \(x\): \[ \frac{\partial f}{\partial x} = e^y \cos z \]- **Partial derivative with respect to** \(y\): \[ \frac{\partial f}{\partial y} = x e^y \cos z \]- **Partial derivative with respect to** \(z\): \[ \frac{\partial f}{\partial z} = -x e^y \sin z - 1 \]The gradient at any point \((x, y, z)\) is:\[ abla f = \left( e^y \cos z, x e^y \cos z, -x e^y \sin z - 1 \right) \].

Step 2: Evaluate the Gradient at the Given Point

Substitute \((x, y, z) = (1, 0, 0)\) into the gradient:- **Evaluate** \(\frac{\partial f}{\partial x}(1,0,0)\): \[ e^0 \cos 0 = 1 \times 1 = 1 \]- **Evaluate** \(\frac{\partial f}{\partial y}(1,0,0)\): \[ 1 \times e^0 \cos 0 = 1 \times 1 = 1 \]- **Evaluate** \(\frac{\partial f}{\partial z}(1,0,0)\): \[ -(1) \times e^0 \sin 0 - 1 = 0 - 1 = -1 \]Thus the gradient at point \((1, 0, 0)\) is \((1, 1, -1)\).

Step 3: Form the Equation of the Tangent Plane

The equation of the tangent plane at point \((x_0, y_0, z_0)\) is given by:\[ abla f \cdot ((x-x_0), (y-y_0), (z-z_0)) = 0 \]Substitute \((x_0, y_0, z_0) = (1, 0, 0)\) and the gradient \((1, 1, -1)\):\[ 1(x-1) + 1(y-0) - 1(z-0) = 0 \]Simplify the equation:\[ x + y - z = 1 \]This is the equation of the tangent plane.

Step 4: Form the Equation of the Normal Line

The normal line has a direction vector given by the gradient at the point. For point \((1, 0, 0)\) with gradient \((1, 1, -1)\), the parametric equations for the line are:- **x-component**: \( x = 1 + t \)- **y-component**: \( y = 0 + t \)- **z-component**: \( z = 0 - t \)The equations of the normal line are:\[ x = 1 + t, \quad y = t, \quad z = -t \]

Key Concepts

Gradient Vector

The gradient vector is a powerful tool in calculus. It helps us find the direction of the steepest ascent on a surface. Think of it as a pointer that indicates where the function increases most rapidly. For a multivariable function like \( f(x, y, z) \), the gradient vector is composed of all the partial derivatives of the function:
\( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).
In our exercise, the partial derivatives were calculated as follows:

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

Evaluating these at the specific point \( (1, 0, 0) \) gave us the gradient vector \( (1, 1, -1) \).
This vector not only tells us the direction of steepest ascent at that point but also gives us the normal vector to the tangent plane.

Partial Derivatives

Partial derivatives are used to differentiate functions of multiple variables. In simpler terms, they measure how the function changes as one of the variables changes, while keeping others constant. For instance, in the function \( f(x, y, z) \), the partial derivative \( \frac{\partial f}{\partial x} \) tells us how \( f \) changes as \( x \) changes, with \( y \) and \( z \) being constant.

• Partial derivatives help us understand the local behavior of multivariable functions.
• They are essential for finding the gradient vector, an important aspect in optimization and geometry.

In our exercise, partial derivatives played a crucial role in determining the gradient vector.
Each partial derivative contributed a component to the gradient, indicating the influence of each variable on the behavior of the function.

Normal Line

In geometry, a normal line is a line that is perpendicular to a surface at a given point. For a surface described by a function \( f(x, y, z) = 1 \), finding the equation of the normal line involves using the gradient vector. This vector serves as the direction vector for the normal line.
Here’s how it works:

• The normal line at a point is directed along the gradient at that point.
• To form the equation of the normal line, we use the point itself and the direction given by the gradient vector.

In our example, the normal line at the point \( (1, 0, 0) \) had the direction vector \( (1, 1, -1) \), leading to the parametric equations:
\( x = 1 + t \), \( y = t \), \( z = -t \).
These equations describe a line extending in both positive and negative directions, perpendicular to the tangent plane at the specified point.