/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 For the given function and value... [FREE SOLUTION] | 91Ó°ÊÓ

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For the given function and values, find: a. \(\Delta f \quad\) b. \(d f\) $$ \begin{array}{ll} f(x, y, z)=x^{2}+y^{2}+z^{2}, & x=3, \quad \Delta x=d x=0.1 \\ y=4, \quad \Delta y=d y=0.1, & z=5, \quad \Delta z=d z=0.1 \end{array} $$

Short Answer

Expert verified
\(\Delta f = 2.43\); \(df = 2.4\).

Step by step solution

01

Calculate the change in the function

To find \(\Delta f\), calculate the change in the function when the variables change by \(\Delta x\), \(\Delta y\), and \(\Delta z\).The formula for \(\Delta f\) is:\[\Delta f = f(x + \Delta x, y + \Delta y, z + \Delta z) - f(x, y, z)\]Substitute the values:\[f(x, y, z) = x^2 + y^2 + z^2 = 3^2 + 4^2 + 5^2 = 9 + 16 + 25 = 50\]\[f(x + \Delta x, y + \Delta y, z + \Delta z) = (3 + 0.1)^2 + (4 + 0.1)^2 + (5 + 0.1)^2\]\[= 3.1^2 + 4.1^2 + 5.1^2\]\[= 9.61 + 16.81 + 26.01 = 52.43\]\[\Delta f = 52.43 - 50 = 2.43\]
02

Calculate the differential of the function

The differential \(df\) can be calculated using the formula:\[df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz\]First, find the partial derivatives:\[\frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = 2y, \quad \frac{\partial f}{\partial z} = 2z\]Substitute the values:\[\frac{\partial f}{\partial x} = 2 \times 3 = 6, \quad \frac{\partial f}{\partial y} = 2 \times 4 = 8, \quad \frac{\partial f}{\partial z} = 2 \times 5 = 10\]Now calculate \(df\):\[df = 6 \times 0.1 + 8 \times 0.1 + 10 \times 0.1\]\[= 0.6 + 0.8 + 1.0\]\[= 2.4\]
03

Interpret Results

The changes in the function and its differential have been calculated. \(\Delta f = 2.43\), a direct computation of the change, while \(df = 2.4\) is an approximate linear estimate. The values are close, indicating the differential is a good linear approximation for small changes.

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Key Concepts

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

Differentials
Differentials provide an approximate way of measuring how a function changes as its input variables change. For a function of several variables, differentials are crucial to understanding how small changes in input affect the function's output. In the context of this problem, we explore the function \[f(x, y, z) = x^2 + y^2 + z^2\] with given changes in variables \(\Delta x = \Delta y = \Delta z = 0.1\). The differential \(df\) is the linear approximation of \(\Delta f\), the actual change in the function.
The formula for the differential of a function \(f(x, y, z)\) is:
\[df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz\]
  • The partial derivative \(\frac{\partial f}{\partial x}\) is \(2x\), giving the rate of change of \(f\) with respect to \(x\).
  • Similarly, \(\frac{\partial f}{\partial y} = 2y\) and \(\frac{\partial f}{\partial z} = 2z\).
For small changes, \(df\) is calculated as:\[df = 6 \times 0.1 + 8 \times 0.1 + 10 \times 0.1 = 2.4\]
This implies that a total change of 2.4 units is expected in the function according to the linear approximation provided by the differential.
Multivariable Functions
Multivariable functions depend on two or more variables and can describe various phenomena in engineering and physics. In this case, the function \(f(x, y, z) = x^2 + y^2 + z^2\) is a multivariable function that combines three variables into a single output value.
To understand how these functions change, it's important to compute partial derivatives for each variable. These derivatives tell us how changes in only one variable affect the output when all other variables are kept constant.
  • The partial derivative \(\frac{\partial f}{\partial x} = 2x\) shows the sensitivity of the function to changes in \(x\).
  • For \(y\), \(\frac{\partial f}{\partial y} = 2y\) tells us how sensitive \(f\) is to changes in \(y\).
  • And for \(z\), \(\frac{\partial f}{\partial z} = 2z\) reveals the sensitivity to changes in \(z\).
By substituting specific values of \(x=3\), \(y=4\), and \(z=5\), we observe that each variable contributes to the overall behavior of the function.
This multidimensional approach allows us to predict how simultaneous changes in all variables impact the function's value.
Calculus Problem Solving
Solving calculus problems, especially in multivariable settings, involves understanding changes and approximations.
The process of identifying changes in the function \(\Delta f\) helps us measure the effect of the entire input changes, while the differential \(df\) is useful for understanding local linear approximation.
Steps to solve such problems can be broken down as:
  • Calculate the original value of the function at specific points, \(f(x, y, z)\).
  • Evaluate new values after changing the input, \(f(x + \Delta x, y + \Delta y, z + \Delta z)\).
  • Compute the difference \(\Delta f\) as the real change in the function.
  • Find the differential \(df\) using partial derivatives, providing a linear estimate.
  • Compare \(\Delta f\) with \(df\) to see how close the approximation is.
Seeing that in our problem \(\Delta f = 2.43\) and \(df = 2.4\) shows that the differential is quite accurate for small changes. This reaffirms that differentials can efficiently approximate changes under minor perturbations.

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