Math Problem Statement

Find the linearizations L(x, y, z) of the function f(x, y, z) = sqrt(x^2 + y^2 + z^2) at the given points: (1, 0, 0), (5, 1, 0), (1, 2, 2).

Solution

We are tasked with finding the linearization L(x,y,z)L(x, y, z) of the function

f(x,y,z)=x2+y2+z2f(x, y, z) = \sqrt{x^2 + y^2 + z^2}

at the given points: (1, 0, 0), (5, 1, 0), and (1, 2, 2).

Step 1: General formula for linearization

The linearization of a function f(x,y,z)f(x, y, z) around a point (x0,y0,z0)(x_0, y_0, z_0) is given by:

L(x,y,z)=f(x0,y0,z0)+fx(x0,y0,z0)(xx0)+fy(x0,y0,z0)(yy0)+fz(x0,y0,z0)(zz0)L(x, y, z) = f(x_0, y_0, z_0) + 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)

Where:

  • fxf_x, fyf_y, and fzf_z are the partial derivatives of ff with respect to xx, yy, and zz, respectively.

Step 2: Compute partial derivatives of f(x,y,z)f(x, y, z)

Given f(x,y,z)=x2+y2+z2f(x, y, z) = \sqrt{x^2 + y^2 + z^2}, we calculate the partial derivatives.

  1. Partial derivative with respect to xx:

fx=xx2+y2+z2=xx2+y2+z2f_x = \frac{\partial}{\partial x} \sqrt{x^2 + y^2 + z^2} = \frac{x}{\sqrt{x^2 + y^2 + z^2}}

  1. Partial derivative with respect to yy:

fy=yx2+y2+z2f_y = \frac{y}{\sqrt{x^2 + y^2 + z^2}}

  1. Partial derivative with respect to zz:

fz=zx2+y2+z2f_z = \frac{z}{\sqrt{x^2 + y^2 + z^2}}

Step 3: Compute linearizations at the given points

a. At (1,0,0)(1, 0, 0):

  • f(1,0,0)=12+02+02=1f(1, 0, 0) = \sqrt{1^2 + 0^2 + 0^2} = 1
  • fx(1,0,0)=11=1f_x(1, 0, 0) = \frac{1}{1} = 1
  • fy(1,0,0)=01=0f_y(1, 0, 0) = \frac{0}{1} = 0
  • fz(1,0,0)=01=0f_z(1, 0, 0) = \frac{0}{1} = 0

The linearization is:

L(x,y,z)=1+1(x1)+0(y0)+0(z0)L(x, y, z) = 1 + 1(x - 1) + 0(y - 0) + 0(z - 0) L(x,y,z)=1+(x1)=xL(x, y, z) = 1 + (x - 1) = x

b. At (5,1,0)(5, 1, 0):

  • f(5,1,0)=52+12+02=26f(5, 1, 0) = \sqrt{5^2 + 1^2 + 0^2} = \sqrt{26}
  • fx(5,1,0)=526f_x(5, 1, 0) = \frac{5}{\sqrt{26}}
  • fy(5,1,0)=126f_y(5, 1, 0) = \frac{1}{\sqrt{26}}
  • fz(5,1,0)=0f_z(5, 1, 0) = 0

The linearization is:

L(x,y,z)=26+526(x5)+126(y1)+0(z0)L(x, y, z) = \sqrt{26} + \frac{5}{\sqrt{26}}(x - 5) + \frac{1}{\sqrt{26}}(y - 1) + 0(z - 0) L(x,y,z)=26+526(x5)+126(y1)L(x, y, z) = \sqrt{26} + \frac{5}{\sqrt{26}}(x - 5) + \frac{1}{\sqrt{26}}(y - 1)

c. At (1,2,2)(1, 2, 2):

  • f(1,2,2)=12+22+22=3f(1, 2, 2) = \sqrt{1^2 + 2^2 + 2^2} = 3
  • fx(1,2,2)=13f_x(1, 2, 2) = \frac{1}{3}
  • fy(1,2,2)=23f_y(1, 2, 2) = \frac{2}{3}
  • fz(1,2,2)=23f_z(1, 2, 2) = \frac{2}{3}

The linearization is:

L(x,y,z)=3+13(x1)+23(y2)+23(z2)L(x, y, z) = 3 + \frac{1}{3}(x - 1) + \frac{2}{3}(y - 2) + \frac{2}{3}(z - 2) L(x,y,z)=3+13(x1)+23(y2)+23(z2)L(x, y, z) = 3 + \frac{1}{3}(x - 1) + \frac{2}{3}(y - 2) + \frac{2}{3}(z - 2)


Do you want more details on any part of this?

Here are five related questions:

  1. How do we compute the gradient of a multivariable function?
  2. What is the geometric meaning of a function's linearization?
  3. How does the linearization differ from a tangent plane?
  4. Can we generalize this approach to higher dimensions?
  5. What role do partial derivatives play in optimization problems?

Tip: When computing partial derivatives, always remember that the other variables are treated as constants while differentiating with respect to one variable.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Linearization
Partial Derivatives

Formulas

L(x, y, z) = f(x_0, y_0, z_0) + 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)
Partial derivatives: f_x = x/sqrt(x^2 + y^2 + z^2), f_y = y/sqrt(x^2 + y^2 + z^2), f_z = z/sqrt(x^2 + y^2 + z^2)

Theorems

Multivariable Function Linearization Theorem

Suitable Grade Level

Undergraduate Calculus