Problem 39 - 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. $$2(x-2)^{2}+(y-1)^{2}+(z-3)^{2}=10, \quad(3,3,5)$$

Step-by-Step Solution

Answer

Tangent plane: \(x + y + z = 11\); Normal line: \(x = 3 + 4t, y = 3 + 4t, z = 5 + 4t\).

Step 1: Identify Surface Equation

The given surface is a level surface of the function \( F(x, y, z) = 2(x-2)^2 + (y-1)^2 + (z-3)^2 \). It is represented by the equation \( F(x, y, z) = 10 \).

Step 2: Compute Gradient of F

The gradient of \( F \), denoted \( abla F \), gives the normal vector to the tangent plane. Compute the partial derivatives:\[ \frac{\partial F}{\partial x} = 4(x-2), \quad \frac{\partial F}{\partial y} = 2(y-1), \quad \frac{\partial F}{\partial z} = 2(z-3) \]

Step 3: Evaluate Gradient at Given Point

Substitute \((x, y, z) = (3, 3, 5)\) into the gradient to find the normal vector at this point:\[ abla F(3, 3, 5) = (4(3-2), 2(3-1), 2(5-3)) = (4, 4, 4) \].

Step 4: Equation of Tangent Plane

The equation of the tangent plane at a point \((x_0, y_0, z_0)\) is given by:\[4(x-x_0) + 4(y-y_0) + 4(z-z_0) = 0 \] Substitute \( (x_0, y_0, z_0) = (3, 3, 5) \):\[4(x-3) + 4(y-3) + 4(z-5) = 0 \]Simplifying gives:\[x + y + z = 11 \]

Step 5: Equation of Normal Line

The normal line can be parameterized using the point \((3, 3, 5)\) and direction \((4, 4, 4)\):\[ x = 3 + 4t, \quad y = 3 + 4t, \quad z = 5 + 4t \]

Key Concepts

Level Surface

A level surface is a set of points that make a function equal to a constant. Think of it like a 3D version of contour lines. In this particular exercise, the surface is described by the equation \( 2(x-2)^2 + (y-1)^2 + (z-3)^2 = 10 \). Here, the function \( F(x, y, z) \) is constant, equal to 10. This means that all points \((x, y, z)\) satisfying the equation are on the same 'level.' This concept helps you visualize the 3D shape in space, like a sphere that results from distance constraints relative to a point. Understanding level surfaces is crucial for problems involving curvature and shapes, as it aids in identifying tangent planes and more.

Gradient

The gradient of a multivariable function points in the direction of the greatest rate of increase of the function. It is a vector composed of all partial derivatives of the function. If you have a function \( F(x, y, z) \), its gradient, \( abla F \), is a vector that consists of the partial derivatives:

• \( \frac{\partial F}{\partial x} \)
• \( \frac{\partial F}{\partial y} \)
• \( \frac{\partial F}{\partial z} \)

For the given exercise, the gradient at the specified point \((3, 3, 5)\) is calculated as \( (4, 4, 4) \). This gradient is crucial because it serves as the normal vector to the tangent plane at this point. Calculating the gradient helps pinpoint directions and rates of change in multivariable calculus.

Normal Line

The normal line is a line that is perpendicular to a surface at a given point. It provides insight into how the surface is oriented in space. To find the normal line, use the gradient vector at the specified point as the direction vector. In this exercise, the normal line can be written parametrically:

• \( x = 3 + 4t \)
• \( y = 3 + 4t \)
• \( z = 5 + 4t \)

Here, \(t\) is a parameter that describes different points along the line. This normal line is essentially a path showing you how the surface 'sticks out' at that point. It's like seeing which way a stick pokes when stuck perpendicularly into the ground at a point on the surface.

Partial Derivatives

Partial derivatives represent the rate of change of a multivariable function with respect to one variable while keeping others constant. They are crucial in obtaining the gradient. For the function \( F(x, y, z) \), the partial derivatives are:

• \( \frac{\partial F}{\partial x} = 4(x-2) \)
• \( \frac{\partial F}{\partial y} = 2(y-1) \)
• \( \frac{\partial F}{\partial z} = 2(z-3) \)

By evaluating these expressions at the point \((x, y, z) = (3, 3, 5)\), we obtain the components of the gradient vector. Partial derivatives help in understanding how changes in one particular dimension affect the surface, enabling the calculation of the tangent plane through the gradient.