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

Step-by-Step Solution

Answer

Tangent Plane: \(6x + 3y + 2z = 19\); Normal Line: \(x = 1 + 6t, y = 2 + 3t, z = 3 + 2t \).

Step 1: Define the Tangent Plane Equation

The general form for the equation of the tangent plane to a surface at a given point \((x_0, y_0, z_0)\) can be written as:\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \]where \((a, b, c)\) is the gradient of the function \(f(x, y, z)\) at the point \((x_0, y_0, z_0)\). Our given function is \(f(x, y, z) = xyz\).

Step 2: Calculate the Gradient Vector

The gradient vector \(abla f\) of the function \(f(x, y, z) = xyz\) is calculated by taking the partial derivatives:\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (yz, xz, xy) \]Evaluate the gradient at the point \((1, 2, 3)\):\[ abla f(1, 2, 3) = (2 \times 3, 1 \times 3, 1 \times 2) = (6, 3, 2) \]

Step 3: Write the Equation for the Tangent Plane

Using the gradient vector \((6, 3, 2)\) and the point \((1, 2, 3)\), the equation of the tangent plane is:\[ 6(x-1) + 3(y-2) + 2(z-3) = 0 \]Simplify this to:\[ 6x + 3y + 2z = 19 \]

Step 4: Define the Normal Line Equation

The equation for the normal line at a point \((x_0, y_0, z_0)\) with direction vector \((a, b, c)\) is given by:\[ x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct \]Substituting \((x_0, y_0, z_0) = (1, 2, 3)\) and \((a, b, c) = (6, 3, 2)\), we get:\[ x = 1 + 6t, \quad y = 2 + 3t, \quad z = 3 + 2t \]

Step 5: Write Down the Equations

The equations for the tangent plane and the normal line are:- **Tangent Plane:** \( 6x + 3y + 2z = 19 \)- **Normal Line:** \(\begin{align*}x &= 1 + 6t, \y &= 2 + 3t, \z &= 3 + 2t\end{align*} \)

Key Concepts

Normal Line

The normal line to a surface at a given point is an important geometric concept. It represents the line that is perpendicular to the tangent plane at that point. To determine the equation of the normal line, we need two main components:

• The point through which the line passes, in this case, our point \(1, 2, 3\).
• The direction vector, which is derived from the gradient vector at the point, here being \(6, 3, 2\).

A normal line equation takes the form:\[ x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct \]where \(x_0, y_0, z_0\) is the point on the surface and \(a, b, c\) is the direction vector. For our example, substituting the given values results in:\[ x = 1 + 6t, \quad y = 2 + 3t, \quad z = 3 + 2t \]

Gradient Vector

The gradient vector is crucial for understanding how a function changes at a point. It is essentially a collection of partial derivatives for each variable in the function. The gradient vector helps in finding the direction where the function increases most rapidly. For a function \(f(x, y, z) = xyz\), the gradient vector \(abla f\) is:\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (yz, xz, xy) \]Calculating this at our specific point \(1, 2, 3\) gives:\[ abla f(1, 2, 3) = (6, 3, 2) \]This gradient vector is perpendicular to the surface, making it not only crucial for calculating the tangent plane but also the normal line.

Partial Derivative

Partial derivatives form the basis for calculating the gradient vector and serve as the building blocks for understanding changes in multivariable functions. For the function \(f(x, y, z) = xyz\), each partial derivative is calculated by holding two variables constant and differentiating with respect to the third:

• \( \frac{\partial f}{\partial x} = yz \)
• \( \frac{\partial f}{\partial y} = xz \)
• \( \frac{\partial f}{\partial z} = xy \)

These partial derivatives show how the function changes as one variable changes, keeping the others fixed. Evaluating them at the point \(1, 2, 3\) gives us the components of the gradient vector, providing directions of maximum rate of change.

Surface Equation

A surface equation like \(f(x, y, z) = xyz = 6\) describes a three-dimensional shape in space. Finding the tangent plane to this surface at a given point helps us approximate the surface locally. In our exercise, the key step is finding the tangent plane using the gradient vector \(6, 3, 2\) calculated earlier. The equation of the tangent plane provides a flat, two-dimensional approximation of the surface at a specific point, making complex surfaces easier to understand and analyze. The resulting tangent plane equation at \(1, 2, 3\) is:\[ 6x + 3y + 2z = 19 \]This provides a linear approximation of the surface around the point \(1, 2, 3\), offering clear insights into its geometric properties around that area.