Problem 304 - 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. $$ f(x, y, z)=x y z=6 \text { at point }(1,2,3) $$

Step-by-Step Solution

Answer

Tangent plane equation: \(6x + 3y + 2z = 20\); Normal line equations: \(x = 1 + 6t\), \(y = 2 + 3t\), \(z = 3 + 2t\).

Step 1: Calculate the Partial Derivatives

To find the tangent plane and normal line, we begin by calculating the partial derivatives of function \( f(x, y, z) = xyz \) at the given point \((1, 2, 3)\). \( f(x, y, z) = x y z \), so the partial derivatives are: 1. \( f_x = yz \)2. \( f_y = xz \)3. \( f_z = xy \)

Step 2: Evaluate Partial Derivatives at the Point

Substitute \((1, 2, 3)\) into the partial derivatives:1. \( f_x(1, 2, 3) = 2 \cdot 3 = 6 \)2. \( f_y(1, 2, 3) = 1 \cdot 3 = 3 \)3. \( f_z(1, 2, 3) = 1 \cdot 2 = 2 \)This gives us the gradient at the point \( (1, 2, 3) \): \( abla f(1, 2, 3) = (6, 3, 2) \).

Step 3: Write the Equation of the Tangent Plane

The equation of the tangent plane at a point \((x_0, y_0, z_0)\) on the surface is given by:\[ f_x(x_0, y_0, z_0)(x - x_0) + f_y(x_0, y_0, z_0)(y - y_0) + f_z(x_0, y_0, z_0)(z - z_0) = 0 \]Using \( f_x = 6 \), \( f_y = 3 \), and \( f_z = 2 \) at \((1, 2, 3)\), substitute into the equation:\[ 6(x - 1) + 3(y - 2) + 2(z - 3) = 0 \]Simplify to get the equation of the tangent plane:\[ 6x + 3y + 2z = 20 \]

Step 4: Write the Equation of the Normal Line

The normal line to the surface at a point \((x_0, y_0, z_0)\) can be found using the direction from the gradient \( (f_x, f_y, f_z) \). The parameter \( t \) determines the position along the normal line:\[ (x, y, z) = (x_0, y_0, z_0) + t(f_x, f_y, f_z) \]Substitute \((x_0, y_0, z_0) = (1, 2, 3)\) and \( (f_x, f_y, f_z) = (6, 3, 2) \):\[ (x, y, z) = (1 + 6t, 2 + 3t, 3 + 2t) \]This represents the parametric equations of the normal line.

Key Concepts

Partial Derivatives

In calculus, partial derivatives play a key role when dealing with functions of multiple variables. These derivatives measure how a function changes as each variable is varied separately, keeping all other variables constant. For a function like \( f(x,y,z) = xyz \), the partial derivatives with respect to \(x\), \(y\), and \(z\) are:

• \( f_x = yz \)
• \( f_y = xz \)
• \( f_z = xy \)

Partial derivatives are crucial in surface analyses, as they help us understand the rate of change of the surface at a specific point. For example, evaluating these partial derivatives at the point \((1, 2, 3)\) gives us specific values \(6, 3,\) and \(2\) respectively. These values are vital for calculating the gradient vector.

Gradient Vector

The gradient vector of a function is a vector composed of all its partial derivatives. It points in the direction of the greatest rate of increase of the function. For \( f(x, y, z) = xyz \), at the point \((1, 2, 3)\), the gradient vector is:

• \( abla f(1, 2, 3) = (f_x, f_y, f_z) = (6, 3, 2) \)

This vector not only highlights how steeply the surface rises but also operates as a normal vector to the surface at that point.
The importance of the gradient vector in this context is twofold:

• It aids in the determination of the tangent plane equation.
• It provides the direction for the normal line equation.

Normal Line

The normal line to a surface at a given point is an intuitive concept. It is the line that is perpendicular to the surface at that point. Utilizing the gradient vector, we can derive the equation of the normal line.
Given the gradient \( (6, 3, 2) \) at \((1, 2, 3)\), the parametric equations for the normal line can be expressed as:

• \( x = 1 + 6t \)
• \( y = 2 + 3t \)
• \( z = 3 + 2t \)

Here, \(t\) is a parameter. As \(t\) varies, it describes each point along the normal line. It's a straightforward but powerful method to comprehend how a surface behaves at a specific point.

Equation of the Surface

The equation of a surface in space is an equation that every point on the surface satisfies. For the surface defined by \( f(x, y, z) = x y z = 6 \), the task was to find its tangent plane at \((1, 2, 3)\).
The equation of the tangent plane at any point \((x_0, y_0, z_0)\) can be written using partial derivatives and the point itself:

• \( f_x(x_0, y_0, z_0)(x - x_0) + f_y(x_0, y_0, z_0)(y - y_0) + f_z(x_0, y_0, z_0)(z - z_0) = 0 \)

Plugging in our computed partial derivatives \( (6, 3, 2) \) and point \((1, 2, 3)\), the equation simplifies to:

• \( 6(x - 1) + 3(y - 2) + 2(z - 3) = 0 \)

This form highlights how the surface locally "touches" space at that point, and is essential for a deeper understanding of the surface's spatial orientation.