Problem 41 - Calculus Early Transcendentals

Question

\(39-44\) Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point. $$x^{2}-2 y^{2}+z^{2}+y z=2, \quad(2,1,-1)$$

Step-by-Step Solution

Answer

The tangent plane is \( 4x - 5y - z = 10 \). The normal line is \( x = 2 + 4t, y = 1 - 5t, z = -1 - t \).

Step 1: Identify the function and point

The given surface is determined by the function \( f(x, y, z) = x^2 - 2y^2 + z^2 + yz \). We need to evaluate this at the point \((2, 1, -1)\) to find the tangent plane and normal line.

Step 2: Calculate the partial derivatives

To find the tangent plane, compute the partial derivatives of \( f \). - \( f_x = \frac{\partial}{\partial x}(x^2 - 2y^2 + z^2 + yz) = 2x \)- \( f_y = \frac{\partial}{\partial y}(x^2 - 2y^2 + z^2 + yz) = -4y + z \)- \( f_z = \frac{\partial}{\partial z}(x^2 - 2y^2 + z^2 + yz) = 2z + y \)

Step 3: Evaluate the derivatives at the point

At the point \((2, 1, -1)\):- \( f_x(2, 1, -1) = 2(2) = 4 \)- \( f_y(2, 1, -1) = -4(1) + (-1) = -5 \)- \( f_z(2, 1, -1) = 2(-1) + (1) = -1 \)

Step 4: Formulate the equation of the tangent plane

Using the point \((2, 1, -1)\) and the gradient \( abla f = (4, -5, -1) \), the equation of the tangent plane is:\[ 4(x - 2) - 5(y - 1) - 1(z + 1) = 0 \]which simplifies to:\[ 4x - 5y - z = 10 \]

Step 5: Calculate the equation of the normal line

The normal line at the point \((2, 1, -1)\) can be expressed using the gradient as its direction vector:\[ x = 2 + 4t \]\[ y = 1 - 5t \]\[ z = -1 - t \]

Key Concepts

Partial Derivatives

Partial derivatives come into play when we want to examine how a function changes as one specific variable transforms, while keeping all other variables constant. For a multivariable function like \( f(x, y, z) = x^2 - 2y^2 + z^2 + yz \), partial derivatives help us understand how the function responds to small changes in one of its three inputs: \( x \), \( y \), or \( z \).

In our exercise, to find the tangent plane at a specified point \((2, 1, -1)\), we first calculate the partial derivatives \( f_x \), \( f_y \), and \( f_z \). These represent the rates of change of the function with respect to each variable:

• \( f_x = \frac{\partial}{\partial x}(x^2 - 2y^2 + z^2 + yz) = 2x \)
• \( f_y = \frac{\partial}{\partial y}(x^2 - 2y^2 + z^2 + yz) = -4y + z \)
• \( f_z = \frac{\partial}{\partial z}(x^2 - 2y^2 + z^2 + yz) = 2z + y \)

These derivatives are evaluated at the point \((2, 1, -1)\), allowing us to apply them directly to compute the tangent plane and normal line equations, making partial derivatives a crucial first step in analyzing the geometry of surfaces.

Normal Line

A normal line to a surface at a given point is a line which is perpendicular to the tangent plane at that point. It provides insight into the orientation of a surface in three-dimensional space.

To formulate the equation of a normal line, we need the gradient vector at the specified point, as the vector gives the direction in which the surface rises most steeply. This is essentially the direction of our normal line.

In our exercise, the gradient vector at the point \((2, 1, -1)\) is obtained from the partial derivatives evaluated at that point: \( abla f = (4, -5, -1) \). This vector serves as the direction vector for the normal line. With this, the set of parametric equations for our normal line becomes:

• \( x = 2 + 4t \)
• \( y = 1 - 5t \)
• \( z = -1 - t \)

These equations tell us how the coordinates change linearly with parameter \( t \), along the normal line. This approach not only provides the equation, but also visualizes the line's trajectory in three-dimensional space.

Gradient Vector

The gradient vector, denoted as \( abla f \), is a powerful tool in multivariable calculus. It consists of the partial derivatives of a function, thus providing a concise way to express the directions and rates at which a function increases.

In the context of our surface \( f(x, y, z) = x^2 - 2y^2 + z^2 + yz \), the gradient at any point is the vector composed of its partial derivatives: \( abla f = (f_x, f_y, f_z) \). For our specified point \((2, 1, -1)\), we've calculated:

• \( f_x(2, 1, -1) = 4 \)
• \( f_y(2, 1, -1) = -5 \)
• \( f_z(2, 1, -1) = -1 \)

Thus, the gradient vector at the point is \( abla f = (4, -5, -1) \). This vector is orthogonal to the tangent plane, meaning it points in the direction of maximum increase of the function, forming the basis for finding both the equation of the tangent plane and the normal line.

The role of the gradient vector is pivotal since it interlinks various geometric and analytical aspects, such as optimizing functions and understanding their critical points.