Problem 10 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 12 , find an equation of the tangent plane and equations of the normal line to the given surface at the indicated point. \(z x^{2}-x y^{2}-y z^{2}=18 ;(0,-2,3)\)

Step-by-Step Solution

Answer

Tangent plane: \(-4x - 9y + 12z - 54 = 0\)Normal line: \(x = -4t\), \(y = -2 - 9t\), \(z = 3 + 12t\)

Step 1: Find the partial derivatives

Compute the partial derivatives of the function with respect to x, y, and z. Let the function be: \[ f(x, y, z) = zx^{2} - xy^{2} - yz^{2} - 18 \] We need: \[ f_{x}(x, y, z) = \frac{\partial f}{\partial x} = 2zx - y^{2} \] \[ f_{y}(x, y, z) = \frac{\partial f}{\partial y} = -2xy - z^{2} \] \[ f_{z}(x, y, z) = \frac{\partial f}{\partial z} = x^{2} - 2yz \]

Step 2: Evaluate the partial derivatives at the given point

Substitute the point (0, -2, 3) into the partial derivatives: \[ f_{x}(0, -2, 3) = 2(3)(0) - (-2)^{2} = -4 \] \[ f_{y}(0, -2, 3) = -2(0)(-2) - 3^{2} = -9 \] \[ f_{z}(0, -2, 3) = (0)^{2} - 2(-2)(3) = 12 \]

Step 3: Write the equation of the tangent plane

The general formula for the tangent plane at \((x_0, y_0, z_0)\) is: \[ f_{x}(x_0, y_0, z_0)(x - x_0) + f_{y}(x_0, y_0, z_0)(y - y_0) + f_{z}(x_0, y_0, z_0)(z - z_0) = 0 \] Substitute \( f_{x}(0, -2, 3) = -4 \), \( f_{y}(0, -2, 3) = -9 \), \( f_{z}(0, -2, 3) = 12 \), \( x_0 = 0 \), \( y_0 = -2 \), \( z_0 = 3 \): \[ -4(x - 0) - 9(y + 2) + 12(z - 3) = 0 \] Simplify: \[ -4x - 9y - 18 + 12z - 36 = 0 \] \[ -4x - 9y + 12z - 54 = 0 \]

Step 4: Write the equations of the normal line

The normal line can be expressed in parametric form: \[ x = x_0 + f_{x}(x_0, y_0, z_0)t \] \[ y = y_0 + f_{y}(x_0, y_0, z_0)t \] \[ z = z_0 + f_{z}(x_0, y_0, z_0)t \] Substitute \( x_0 = 0 \), \( y_0 = -2 \), \( z_0 = 3 \), \( f_{x}(0, -2, 3) = -4 \), \( f_{y}(0, -2, 3) = -9 \), \( f_{z}(0, -2, 3) = 12 \): \[ x = 0 - 4t \] \[ y = -2 - 9t \] \[ z = 3 + 12t \]

Key Concepts

Partial Derivatives

Partial derivatives are essential in multivariable calculus. They measure how a function changes as one of its input variables changes, while keeping the other variables constant.
For instance, consider a function of three variables:

• \( f(x, y, z) \)

The partial derivative of this function with respect to \( x \) is represented as \( \frac{\partial f}{\partial x} \). This measures the rate of change of \( f \) as \( x \) changes, with \( y \) and \( z \) held constant.
In our exercise, we found the partial derivatives of the function \( f(x, y, z) = zx^{2} - xy^{2} - yz^{2} - 18 \). Specifically, we computed:

• \( f_{x}(x, y, z) = 2zx - y^{2} \)
• \( f_{y}(x, y, z) = -2xy - z^{2} \)
• \( f_{z}(x, y, z) = x^{2} - 2yz \)

These partial derivatives give us a sense of how the function's value changes as each variable changes.

Normal Line

The normal line to a surface at a given point is a line that is perpendicular to the tangent plane at that point.
To find the equations of the normal line, we use the gradients obtained from the partial derivatives. These gradients act as the direction vector for the normal line.
The parametric equations for the normal line are as follows:

• \( x = x_0 + f_{x}(x_0, y_0, z_0)t \)
• \( y = y_0 + f_{y}(x_0, y_0, z_0)t \)
• \( z = z_0 + f_{z}(x_0, y_0, z_0)t \)

In our example, substituting the computed partial derivatives and the point \( (0, -2, 3) \), we find:

• \( x = 0 - 4t \)
• \( y = -2 - 9t \)
• \( z = 3 + 12t \)

These equations define the normal line to the surface at the given point.

Multivariable Calculus

Multivariable calculus deals with functions that have more than one variable. It includes concepts like partial derivatives, multiple integrals, and vector calculus.
Understanding how to work with these functions is crucial in fields like physics, engineering, and economics.
For example, tangents and normals are used to understand how surfaces behave in three-dimensional space.
In our exercise, we applied multivariable calculus concepts to find the tangent plane to the surface at the given point.
The general form for the equation of the tangent plane is:

• \( f_{x}(x_0, y_0, z_0)(x - x_0) + f_{y}(x_0, y_0, z_0)(y - y_0) + f_{z}(x_0, y_0, z_0)(z - z_0) = 0 \)

Using the values of the partial derivatives and the point \( (0, -2, 3) \), we obtain:

• \( -4(x-0) - 9(y+2) + 12(z-3) = 0 \)

Multiplying out and simplifying, we get:

• \( -4x - 9y + 12z - 54 = 0 \)

This is the equation of the tangent plane at the point \( (0, -2, 3) \).