Problem 21 - Calculus Early Transcendentals

Question

Find the linear approximation of the function \(f(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}}\) at \((3,2,6)\) and use it to approximate the number \(\sqrt{(3.02)^{2}+(1.97)^{2}+(5.99)^{2}}\)

Step-by-Step Solution

Answer

The approximate value is 6.991.

Step 1: Find Partial Derivatives

The first step is to compute the partial derivatives of the function \( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} \). We need the partial derivatives with respect to \( x \), \( y \), and \( z \). These are:\[ f_x = \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \quad f_y = \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \quad f_z = \frac{z}{\sqrt{x^2 + y^2 + z^2}}. \]

Step 2: Evaluate Partial Derivatives at Given Point

Now we evaluate these partial derivatives at the point \((3, 2, 6)\):\[ f_x(3, 2, 6) = \frac{3}{\sqrt{3^2 + 2^2 + 6^2}} = \frac{3}{7}, \]\[ f_y(3, 2, 6) = \frac{2}{\sqrt{3^2 + 2^2 + 6^2}} = \frac{2}{7}, \]\[ f_z(3, 2, 6) = \frac{6}{\sqrt{3^2 + 2^2 + 6^2}} = \frac{6}{7}. \]

Step 3: Write the Linear Approximation Formula

The linear approximation (or tangent plane) of the function at the point \((3, 2, 6)\) is given by the formula:\[ L(x, y, z) = f(3, 2, 6) + f_x(3, 2, 6)(x-3) + f_y(3, 2, 6)(y-2) + f_z(3, 2, 6)(z-6). \]

Step 4: Evaluate the Function at the Point

Calculate \( f(3, 2, 6) \):\[ f(3, 2, 6) = \sqrt{3^2 + 2^2 + 6^2} = 7. \]

Step 5: Simplify Linear Approximation Expression

Substitute the values from the previous steps into the linear approximation formula:\[ L(x, y, z) = 7 + \frac{3}{7}(x-3) + \frac{2}{7}(y-2) + \frac{6}{7}(z-6). \]Simplifying this gives:\[ L(x, y, z) = 7 + \frac{3x - 9}{7} + \frac{2y - 4}{7} + \frac{6z - 36}{7}. \]

Step 6: Substitute Values to Approximate

Substitute \((3.02, 1.97, 5.99)\) into the linear approximation formula:\[ L(3.02, 1.97, 5.99) = 7 + \frac{3 \times 3.02 - 9}{7} + \frac{2 \times 1.97 - 4}{7} + \frac{6 \times 5.99 - 36}{7}. \]Calculate each part:\[ \frac{3 \times 3.02 - 9}{7} = \frac{0.06}{7}, \]\[ \frac{2 \times 1.97 - 4}{7} = \frac{-0.06}{7}, \]\[ \frac{6 \times 5.99 - 36}{7} = \frac{-0.06}{7}. \]Add these to the constant 7.

Step 7: Calculate Final Approximation

Finally, calculate \( L(3.02, 1.97, 5.99) \):\[ L(3.02, 1.97, 5.99) = 7 + \frac{0.06 - 0.06 - 0.06}{7} = 7 - \frac{0.06}{7}. \]This results in a linear approximation of approximately 6.991 for the given point.

Key Concepts

Partial Derivatives

In multivariable calculus, partial derivatives are crucial tools for analyzing functions with more than one variable. When we have a function like \( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} \), which depends on three variables, finding the rate of change with respect to each variable separately is fundamental. This is what partial derivatives do. They tell us how the function changes as we vary one variable while keeping the others constant.
In the exercise, the partial derivatives with respect to \( x \), \( y \), and \( z \) were calculated as follows:
\[ f_x = \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \quad f_y = \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \quad f_z = \frac{z}{\sqrt{x^2 + y^2 + z^2}}. \]
This set of equations defines the slope or the rate of change of the function in each direction. Understanding how to compute these derivatives helps us set the foundation for more advanced topics like finding the tangent plane.

Multivariable Calculus

Multivariable calculus extends the principles of calculus that are familiar from single-variable calculus to higher dimensions. Instead of dealing with one independent variable as in traditional calculus, multivariable calculus focuses on functions involving several variables. It allows us to explore concepts such as rates of change and optimization in more complex systems.
A central aspect of multivariable calculus is the concept of the gradient, a vector comprising all the partial derivatives of a function. This vector points in the direction of the steepest increase of a function. The gradient is integral in the construction of the linear approximation of a function which takes shape as the tangent plane.
Multivariable calculus is powerful for many applications in the real world, including physics, engineering, and economics, where multiple factors influence outcomes. Mastering these topics provides a deeper understanding of how multi-component systems evolve and interact.

Tangent Plane

The tangent plane is a flat surface that best approximates the surface of a multivariable function at a given point. When considering the function \( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} \), the tangent plane gives us a linear model which is useful for estimating function values near a particular point. This linear approximation is akin to the tangent line in single-variable calculus but extends it to higher dimensions.
In the original exercise, after calculating the partial derivatives, those values were used to determine the equation of the tangent plane, also known as the linear approximation formula:
\[ L(x, y, z) = f(3, 2, 6) + f_x(3, 2, 6)(x-3) + f_y(3, 2, 6)(y-2) + f_z(3, 2, 6)(z-6). \]
This formula allows for quick estimation of the function's output by substituting small deviations from the point \((3, 2, 6)\), making it a powerful tool for approximating values without having to calculate the function's output directly.