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Chapter 14: Problem 4

In Exercises \(1-4,\) find the specific function values. $$ f(x, y, z)=\sqrt{49-x^{2}-y^{2}-z^{2}} $$ $$ \begin{array}{ll}{\text { a. } f(0,0,0)} & {\text { b. } f(2,-3,6)} \\ {\text { c. } f(-1,2,3)} & {\text { d. } f\left(\frac{4}{\sqrt{2}}, \frac{5}{\sqrt{2}}, \frac{6}{\sqrt{2}}\right)}\end{array} $$

Short Answer

Expert verified
f(0,0,0)=7; f(2,-3,6)=0; f(-1,2,3)=\sqrt{35}; f\left(\frac{4}{\sqrt{2}}, \frac{5}{\sqrt{2}}, \frac{6}{\sqrt{2}}\right)=\sqrt{\frac{21}{2}}.

Step by step solution

01

Calculate f(0,0,0)

Plug in the values \(x = 0\), \(y = 0\), and \(z = 0\) into the function. The expression becomes:\[f(0,0,0) = \sqrt{49 - 0^2 - 0^2 - 0^2} = \sqrt{49} = 7.\]
02

Calculate f(2,-3,6)

Substitute \(x = 2\), \(y = -3\), and \(z = 6\) into the function:\[f(2,-3,6) = \sqrt{49 - 2^2 - (-3)^2 - 6^2} = \sqrt{49 - 4 - 9 - 36} = \sqrt{49 - 49} = \sqrt{0} = 0.\]
03

Calculate f(-1,2,3)

Insert \(x = -1\), \(y = 2\), and \(z = 3\) into the formula:\[f(-1,2,3) = \sqrt{49 - (-1)^2 - 2^2 - 3^2} = \sqrt{49 - 1 - 4 - 9} = \sqrt{35}.\]Simplify: The exact form is \(\sqrt{35}\).
04

Calculate \(f\left(\frac{4}{\sqrt{2}}, \frac{5}{\sqrt{2}}, \frac{6}{\sqrt{2}}\right)\)

Substitute \(x = \frac{4}{\sqrt{2}}\), \(y = \frac{5}{\sqrt{2}}\), and \(z = \frac{6}{\sqrt{2}}\) into the function:\[f\left(\frac{4}{\sqrt{2}}, \frac{5}{\sqrt{2}}, \frac{6}{\sqrt{2}}\right) = \sqrt{49 - \left(\frac{4}{\sqrt{2}}\right)^2 - \left(\frac{5}{\sqrt{2}}\right)^2 - \left(\frac{6}{\sqrt{2}}\right)^2}.\]Calculate each:\[\left(\frac{4}{\sqrt{2}}\right)^2 = \frac{16}{2} = 8,\quad \left(\frac{5}{\sqrt{2}}\right)^2 = \frac{25}{2},\quad \left(\frac{6}{\sqrt{2}}\right)^2 = \frac{36}{2} = 18.\]Combine and simplify:\[49 - 8 - \frac{25}{2} - 18 = \frac{98}{2} - \frac{16}{2} - \frac{25}{2} - \frac{36}{2} = \frac{21}{2}.\]Finally calculate the square root:\[f\left(\frac{4}{\sqrt{2}}, \frac{5}{\sqrt{2}}, \frac{6}{\sqrt{2}}\right) = \sqrt{\frac{21}{2}}.\]The exact form remains \(\sqrt{\frac{21}{2}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Functions of Several Variables
Functions of several variables are an extension of single-variable functions, allowing us to evaluate more complex relationships. Instead of having just one input, these functions operate on multiple variables.
  • Inputs: In our example, the function is defined with three variables: \( x \), \( y \), and \( z \).
  • Function Notation: The function is written as \( f(x, y, z) \), where the result depends on the values of each parameter.
The ability to handle multiple inputs is useful in fields like physics, engineering, and data science. These functions help model the world around us, by taking into consideration different factors simultaneously.
Understanding how these variables interact within the function is key to successfully applying multivariable calculus.When you're evaluating a function of several variables, substitute values for each variable, then calculate the result. This is similar to single-variable functions but in a three-dimensional context. It adds a depth to the problem, simulating real-world scenarios where a single factor isn't enough to define an outcome.
Square Root Function
The square root function appears often in calculus and describes the operation of taking the square root of a number. This concept is crucial when dealing with square relationships in equations, especially when working with functions of several variables.
  • Definition: The square root function is denoted by \( \sqrt{} \), and it returns the number that, when multiplied by itself, gives the original number. For instance, \( \sqrt{49} = 7 \).
  • Properties: Square root functions provide only the principal (non-negative) root. Hence, \( \sqrt{x^2} = |x| \).
When evaluating a multivariable function, such as \( f(x,y,z) = \sqrt{49-x^2-y^2-z^2} \), you are essentially finding the square root of the remaining portion of 49 after accounting for the sum of the squares of the variables.
This concept is further extended in calculus to understand gradients and directions of greatest increase of multivariable functions, emphasizing its importance in higher learning and applications.
Evaluating Functions
Evaluating functions involves finding the value of a function for specific input parameters. With multivariable functions, this process consists of three main steps:
  • Substitution: Begin by inserting the given values of the variables into the function. For example, if calculating \( f(0,0,0) \), substitute \( x = 0 \), \( y = 0 \), and \( z = 0 \).
  • Simplification: Once substituted, use basic algebra to simplify the expression. It's about combining like terms and reducing the equation without losing any information.
  • Evaluation: Finally, carry out the final computations, such as calculating the square root if necessary, to determine the function's final output.
When you evaluate a function of several variables, you're seeking to understand how different variables influence the result. This is crucial for interpreting mathematical models and solving practical problems. Through practice and understanding, you can efficiently handle even complex functions and compute their values with accuracy.

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Most popular questions from this chapter

Two dependent variables Find \(\partial x / \partial u\) and \(\partial y / \partial u\) if the equations \(u=x^{2}-y^{2}\) and \(v=x^{2}-y\) define \(x\) and \(y\) as functions of the independent variables \(u\) and \(v,\) and the partial derivatives exist. (See the hint in Exercise \(69 . )\) Then let \(s=x^{2}+y^{2}\) and find \(\partial s / \partial u .\)

Maximizing a utility function: an example from economics In economics, the usefulness or utility of amounts \(x\) and \(y\) of two capital goods \(G_{1}\) and \(G_{2}\) is sometimes measured by a function \(U(x, y) .\) For example, \(G_{1}\) and \(G_{2}\) might be two chemicals a pharmaceutical company needs to have on hand and \(U(x, y)\) the gain from manufacturing a product whose synthesis requires different amounts of the chemicals depending on the process used. If \(G_{1}\) costs \(a\) dollars per kilogram, \(G_{2}\) costs \(b\) dollars per kilogram, and the total amount allocated for the purchase of \(G_{1}\) and \(G_{2}\) together is \(c\) dollars, then the company's managers want to maximize \(U(x, y)\) given that \(a x+b y=c .\) Thus, they need to solve a typical Lagrange multiplier problem. Suppose that $$U(x, y)=x y+2 x$$ and that the equation \(a x+b y=c\) simplifies to $$2 x+y=30.$$ Find the maximum value of \(U\) and the corresponding values of \(x\) and \(y\) subject to this latter constraint.

Find the linearization \(L(x, y, z)\) of the function \(f(x, y, z)\) at \(P_{0} .\) Then find an upper bound for the magnitude of the error \(E\) in the approximation \(f(x, y, z) \approx L(x, y, z)\) over the region \(R\) $$ \begin{array}{l}{f(x, y, z)=x y+2 y z-3 x z \text { at } P_{0}(1,1,0)} \\ {R :|x-1| \leq 0.01, \quad|y-1| \leq 0.01, \quad|z| \leq 0.01}\end{array} $$

Extrema on a circle of intersection Find the extreme values of the function \(f(x, y, z)=x y+z^{2}\) on the circle in which the plane \(y-x=0\) intersects the sphere \(x^{2}+y^{2}+z^{2}=4.\)

Minimum distance to the origin Find the point closest to the origin on the curve of intersection of the plane \(2 y+4 z=5\) and the cone \(z^{2}=4 x^{2}+4 y^{2}.\)
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