Problem 9 - 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. \(x^{1 / 2}+y^{1 / 2}+z^{1 / 2}=4 ;(4,1,1)\)

Step-by-Step Solution

Answer

The equation of the tangent plane is: \( x + 2y + 2z = 12 \). The equations of the normal line are: \( \frac{x - 4}{1/4} = \frac{y - 1}{1/2} = \frac{z - 1}{1/2} \).

Step 1: Find the partial derivatives

Determine the partial derivatives of the given function with respect to x, y, and z. The function is: a(x, y, z) = x^{1/2} + y^{1/2} + z^{1/2}. So, calculate \(f_x = \frac{\partial f}{\partial x} = \frac{1}{2}x^{-1/2}\), \(f_y = \frac{\partial f}{\partial y} = \frac{1}{2}y^{-1/2} \), and \(f_z = \frac{\partial f}{\partial z} = \frac{1}{2}z^{-1/2} \).

Step 2: Evaluate the partial derivatives at the given point

Evaluate the partial derivatives at the point (4,1,1). For \( f_x \), substitute x = 4: \( f_x (4, 1, 1) = \frac{1}{2} × (4)^{-1/2} = \frac{1}{2} × \frac{1}{2} = \frac{1}{4} \).For \(f_y\), substitute y = 1: \( f_y (4, 1, 1) = \frac{1}{2} × (1)^{-1/2} = \frac{1}{2} \).For \(f_z\), substitute z = 1: \( f_z (4, 1, 1) = \frac{1}{2} × (1)^{-1/2} = \frac{1}{2} \).

Step 3: Write the equation of the tangent plane

The equation of the tangent plane at a given point (x_0, y_0, z_0) can be written as: \( f_x (x_0, y_0, z_0) (x-x_0) + f_y (x_0, y_0, z_0) (y-y_0) + f_z (x_0, y_0, z_0) (z-z_0) = 0 \).Using the values from previous steps: \( \frac{1}{4}(x-4) + \frac{1}{2}(y-1) + \frac{1}{2}(z-1) = 0 \), simplify to get \( \frac{1}{4}x + \frac{1}{2}y + \frac{1}{2}z = 3 \)or equivalently, \( x + 2y + 2z = 12 \).

Step 4: Write the equations of the normal line

The equations of the normal line to the surface at a given point can be written as: \( \frac{x-x_0}{f_x (x_0, y_0, z_0)} = \frac{y-y_0}{f_y (x_0, y_0, z_0)} = \frac{z-z_0}{f_z (x_0, y_0, z_0)}\). Using \( f_x (4, 1, 1) = \frac{1}{4} \), \( f_y (4, 1, 1) = \frac{1}{2} \), and \( f_z (4, 1, 1) = \frac{1}{2} \), we get: \( \frac{x - 4}{1/4} = \frac{y - 1}{1/2} = \frac{z - 1}{1/2} \).

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a function changes with respect to one variable while keeping the others constant. Imagine you have a surface defined by a function, and you want to see how changes in each direction (x, y, and z) affect the surface.
In the given exercise, the surface is described by the function: \( f(x, y, z) = x^{1/2} + y^{1/2} + z^{1/2} \). The partial derivative of this function with respect to each variable is:

• With respect to x: \(\frac{\frac{\text{d}}{\text{d}x}[x^{1/2}]}{\text{d}x} = \frac{1}{2}x^{-1/2}\)
• With respect to y: \(\frac{\frac{\text{d}}{\text{d}y}[y^{1/2}]}{\text{d}y} = \frac{1}{2}y^{-1/2}\)
• With respect to z: \(\frac{\frac{\text{d}}{\text{d}z}[z^{1/2}]}{\text{d}z} = \frac{1}{2}z^{-1/2}\)

These derivatives tell us the rate of change of the function with respect to x, y, and z. To evaluate these at a specific point, say \( (4, 1, 1) \), we just plug these values into the equations:

• For x = 4, \( f_x = \frac{1}{4} \)
• For y = 1, \( f_y = \frac{1}{2} \)
• For z = 1, \( f_z = \frac{1}{2} \)

Knowing these helps us understand how steep the surface is at that exact point.

Tangent Plane

A tangent plane is a plane that touches a surface at exactly one point. Think of it like laying a flat piece of paper on a hill at a single spot. The plane matches the hill's slope exactly at that point.
The general formula for the tangent plane to a surface at a specific point \( (x_0, y_0, z_0) \rightarrow (4, 1, 1) \) is:

• \( f_x(x_0, y_0, z_0)(x - x_0) + f_y(x_0, y_0, z_0)(y - y_0) + f_z(x_0, y_0, z_0)(z - z_0) = 0 \)

Substituting our partial derivatives from the previous steps, we get:

• \( \frac{1}{4}(x - 4) + \frac{1}{2}(y - 1) + \frac{1}{2}(z - 1) = 0 \)

Simplify the expression to get the equation of the tangent plane:

• \( x + 2y + 2z = 12 \)

This represents a flat surface that just grazes our original surface at the point (4, 1, 1). It is useful for approximations and understanding the behavior of the surface around that point.

Normal Line

The normal line is a line perpendicular to the tangent plane at a given point. If the tangent plane is your flat piece of paper, the normal line is like a pencil sticking straight up through the paper.
The general formula for the normal line at a specific point \( (x_0, y_0, z_0) = (4, 1, 1) \) can be written as:

• \( \frac{x - x_0}{f_x(x_0, y_0, z_0)} = \frac{y - y_0}{f_y(x_0, y_0, z_0)} = \frac{z - z_0}{f_z(x_0, y_0, z_0)} \)

Using our partial derivatives:

• \( \frac{x - 4}{1/4} = \frac{y - 1}{1/2} = \frac{z - 1}{1/2} \)

Simplifying each part, we get the parametric equations for the line:

• x = 4 + 4t
• y = 1 + 2t
• z = 1 + 2t

These equations describe how x, y, and z change together as you move along the normal line starting from point (4, 1, 1). The variable t represents how far you go in the direction of the normal line.