Problem 1 - 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. \(x^{2}+y^{2}+z^{2}=17 ;(2,-2,3)\)

Step-by-Step Solution

Answer

Tangent plane: \(4x - 4y + 6z = 34\). Normal line: \(x = 2 + 4t\), \(y = -2 - 4t\), \(z = 3 + 6t\).

Step 1 - Understand the Surface Equation

The given surface is a sphere with the equation: \(x^2 + y^2 + z^2 = 17\).

Step 2 - Compute Partial Derivatives

To find the equation of the tangent plane and the normal line, compute the gradient of the surface at the given point (2, -2, 3). The gradient is: \(abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right)\). For \(F(x, y, z) = x^2 + y^2 + z^2 - 17\): \[\frac{\partial F}{\partial x} = 2x, \frac{\partial F}{\partial y} = 2y, \frac{\partial F}{\partial z} = 2z\].

Step 3 - Evaluate the Gradient at the Point

Substitute the point (2, -2, 3) into the gradient to find: \(abla F(2, -2, 3) = (2 \cdot 2, 2 \cdot (-2), 2 \cdot 3) = (4, -4, 6)\).

Step 4 - Equation of the Tangent Plane

The equation of the tangent plane to the surface at the point is given by: \(F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0\). Substituting the gradient and the point: \(4(x - 2) - 4(y + 2) + 6(z - 3) = 0\). Simplify this to: \(4x - 4y + 6z = 34\).

Step 5 - Equations of the Normal Line

The equations of the normal line can be derived from the point and the gradient. The parametric equations are: \(x = x_0 + t \cdot F_x, y = y_0 + t \cdot F_y, z = z_0 + t \cdot F_z\). With \((x_0, y_0, z_0) = (2, -2, 3)\) and \(abla F(2, -2, 3) = (4, -4, 6)\), the parametric form is: \(x = 2 + 4t, y = -2 - 4t, z = 3 + 6t\).

Key Concepts

gradient of a surface

The gradient of a surface is a vector that describes the direction and rate of fastest increase of a function. It points perpendicularly to the level surfaces of a function. The gradient is denoted as \(abla F\) and for a function \ F(x, y, z) \ can be computed using partial derivatives. For our example, where the surface is a sphere defined by \ F(x, y, z) = x^2 + y^2 + z^2 - 17 \,

• The gradient \( abla F \) involves computing the partial derivatives of \( F \) with respect to \( x, y, \) and \( z \).
• These partial derivatives represent the rates of change of \( F \) in the respective coordinate directions.

In our problem, the gradient is found to be \( abla F = (2x, 2y, 2z) \).

partial derivatives

Partial derivatives measure how a function changes as only one of its variables changes while the others remain fixed. For a function \( F(x, y, z) \) in the given problem, we found the following:

• \( \frac{abla F}{abla x} = 2x \)
• \( \frac{abla F}{abla y} = 2y \)
• \( \frac{abla F}{abla z} = 2z \)

These derivatives help us understand the behavior of the surface in each direction. At the specific point (2, -2, 3), substituting into each partial derivative gives us the components of the gradient vector.

normal line equation

The normal line to a surface at a given point is a line that is perpendicular to the tangent plane at that point. This line can be described parametrically using the gradient vector. For instance:

• The equation of the normal line uses the point (2, -2, 3) and direction vector given by the gradient \( (4, -4, 6) \).
• The parametric form of the normal line equations are:
  \( x = 2 + 4t \),
  \( y = -2 - 4t \),
  \( z = 3 + 6t \)

Here, \( t \) is a parameter that varies along the length of the normal line.

tangent plane equation

The tangent plane to a surface at a point is the plane that just 'touches' the surface at that point. It can be described using the gradient and point of tangency. The general formula for the tangent plane at \( (x_0, y_0, z_0) \) is:
\[ F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0 \]
For our surface \( x^2 + y^2 + z^2 = 17 \) at point \( (2, -2, 3) \), we substitute the gradient components and the point to get:
\[ 4(x - 2) - 4(y + 2) + 6(z - 3) = 0 \]
Simplifying this results in the tangent plane equation:

• \( 4x - 4y + 6z = 34 \)

This equation represents the plane that is tangent to the surface at the given point.