Chapter 20: Problem 9
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. \(x^{1 / 2}+y^{1 / 2}+z^{1 / 2}=4 ;(4,1,1)\)
Short Answer
Expert verified
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 by step solution
01
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 \(f_z = \frac{\partial f}{\partial z} = \frac{1}{2}z^{-1/2} \).
02
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: \( f_z (4, 1, 1) = \frac{1}{2} × (1)^{-1/2} = \frac{1}{2} \).
03
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, \( x + 2y + 2z = 12 \).
04
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: \( \frac{x - 4}{1/4} = \frac{y - 1}{1/2} = \frac{z - 1}{1/2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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: \(\frac{\frac{\text{d}}{\text{d}z}[z^{1/2}]}{\text{d}z} = \frac{1}{2}z^{-1/2}\)
- For x = 4, \( f_x = \frac{1}{4} \)
- For y = 1, \( f_y = \frac{1}{2} \)
- For z = 1, \( f_z = \frac{1}{2} \)
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:
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:
- \( f_x(x_0, y_0, z_0)(x - x_0) + f_y(x_0, y_0, z_0)(y - y_0) + f_z(x_0, y_0, z_0)(z - z_0) = 0 \)
- \( \frac{1}{4}(x - 4) + \frac{1}{2}(y - 1) + \frac{1}{2}(z - 1) = 0 \)
- \( x + 2y + 2z = 12 \)
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:
The general formula for the normal line at a specific point \( (x_0, y_0, z_0) = (4, 1, 1) \) can be written as:
- \( \frac{x - x_0}{f_x(x_0, y_0, z_0)} = \frac{y - y_0}{f_y(x_0, y_0, z_0)} = \frac{z - z_0}{f_z(x_0, y_0, z_0)} \)
- \( \frac{x - 4}{1/4} = \frac{y - 1}{1/2} = \frac{z - 1}{1/2} \)
- x = 4 + 4t
- y = 1 + 2t
- z = 1 + 2t