/*! 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 59 Suppose that certain measured qu... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that certain measured quantities \(x\) and \(y\) have errors of at most \(r \%\) and \(s \%\), respectively. For each of the following formulas in \(x\) and \(y,\) use differentials to approximate the maximum possible error in the calculated result. $$ \begin{array}{llll}{\text { (a) } x y} & {\text { (b) } x / y} & {\text { (c) } x^{2} y^{3}} & {\text { (d) } x^{3} \sqrt{y}}\end{array} $$

Short Answer

Expert verified
(a) \( r+s \)%, (b) \( r+s \)%, (c) \( 2r+3s \)%, and (d) \( 3r+\frac{s}{2} \)%.

Step by step solution

01

Understanding Differentials

Differentials are used to approximate the error in a function based on the small changes in variables. If the function is given as \( f(x, y) \), then the differential \( df \) is given by \( df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \).
02

Apply Differentials to Part (a)

For the product \( f = xy \), we have \( \frac{\partial f}{\partial x} = y \) and \( \frac{\partial f}{\partial y} = x \). The differential, \( df = y \cdot dx + x \cdot dy \). Converting this to relative error, \( \frac{df}{f} = \frac{dy}{y} + \frac{dx}{x} \), which is approximately \( r + s \) percent. Thus, the maximum error in \( xy \) is \( r + s \) percent.
03

Apply Differentials to Part (b)

For \( f = x/y \), \( \frac{\partial f}{\partial x} = \frac{1}{y} \) and \( \frac{\partial f}{\partial y} = -\frac{x}{y^2} \). The differential is \( df = \frac{1}{y}dx - \frac{x}{y^2}dy \). The relative error is \( \frac{df}{f} = \frac{dx}{x} + \frac{dy}{y} \), which is \( r + s \) percent. Thus, the maximum error in \( x/y \) is \( r + s \) percent.
04

Apply Differentials to Part (c)

Given \( f = x^2 y^3 \), we calculate \( \frac{\partial f}{\partial x} = 2x y^3 \) and \( \frac{\partial f}{\partial y} = 3x^2 y^2 \). The differential is \( df = 2x y^3 dx + 3x^2 y^2 dy \) and the relative error is \( \frac{df}{f} = 2 \frac{dx}{x} + 3 \frac{dy}{y} \). This becomes approximately \( 2r + 3s \), giving a maximum error of \( 2r + 3s \) percent.
05

Apply Differentials to Part (d)

For \( f = x^3 \sqrt{y} \), \( \frac{\partial f}{\partial x} = 3x^2 \sqrt{y} \) and \( \frac{\partial f}{\partial y} = \frac{x^3}{2\sqrt{y}} \). The differential is \( df = 3x^2 \sqrt{y} dx + \frac{x^3}{2\sqrt{y}} dy \), and the relative error \( \frac{df}{f} = 3 \frac{dx}{x} + \frac{1}{2} \frac{dy}{y} \). This results in a maximum relative error of \( 3r + \frac{s}{2} \) percent.

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Key Concepts

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

Error Approximation in Calculus
In calculus, error approximation plays a crucial role in understanding how uncertainties in measurements can affect calculations. It's all about estimating how much a small error in an input value can cause a change in the output of a function. This is particularly important when dealing with functions of multiple variables, like in our exercise with the functions of two measured quantities, \(x\) and \(y\), each with their own potential measurement error.

When we talk about error approximation in the context of calculus, we often refer to the differential approximation technique. By using the differentials, we can estimate the maximum error in a calculated result as a proportion of the errors in the input values. The differential, denoted as \(df\), provides an approximation of how a small change \((dx, dy)\) in \((x, y)\) affects the function \(f(x, y)\).
  • For a function \(f(x)\), the differential is \(df = f'(x) \cdot dx\).
  • For functions of two variables \(f(x, y)\), it becomes \(df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy\).

Using differentials is beneficial to obtain an approximate value of the error, which can then be translated into terms like percentage errors to interpret the impact on practical measurements.
Understanding Partial Derivatives
Partial derivatives are the building blocks of understanding how a multi-variable function changes with respect to its variables individually. They show us the rate of change of the function with respect to one variable, while keeping other variables constant.

In our exercise scenario, where we apply partial derivatives to calculate the differential, each partial derivative captures the effect on the function caused by changing one of the input variables at a time. For instance, if \(f(x, y) = xy\), then:
  • \(\frac{\partial f}{\partial x} = y\) describes how \(f\) changes as \(x\) changes with \(y\) being constant.
  • \(\frac{\partial f}{\partial y} = x\) describes how \(f\) changes as \(y\) changes with \(x\) being constant.

By using these partial derivatives, we build a differential formula \(df\) that accounts for changes in both \(x\) and \(y\) simultaneously. This is essential for understanding the compound effect of both variables' measurements errors in practical applications.
Relative Error and Its Importance
The concept of relative error is indispensable when you want to understand how impactful a measurement error is compared to the actual value itself. Unlike absolute error, which gives you just a numerical value of error, relative error is expressed as a percentage, making it easier to grasp how significant that error is in the context of the entire measurement.

When we deal with formulas of two variables, like in the exercise where functions of \(x\) and \(y\) were explored, calculating the relative error helps explain how the errors in \(x\) and \(y\) translate into an error in the result. The relative error of a function \(f\) is given by:\
\[\frac{df}{f} = \frac{\frac{df}{\text{(function)}}}{\text{(function)}}\]
This equation tells you that the total relative error in \(f\) is the sum of the relative errors in \(x\) and \(y\), weighted by their partial derivatives. So with the measurement errors \(r\%\) and \(s\%\) in \(x\) and \(y\) respectively, the function's relative error essentially combines these individual uncertainties into a holistic estimate.

Understanding and calculating the relative error is critical in fields requiring precise measurements to evaluate how closely the results reflect true values, which is often vital in scientific, engineering, and statistical analyses.

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Most popular questions from this chapter

A point moves along the intersection of the elliptic paraboloid \(z=x^{2}+3 y^{2}\) and the plane \(x=2 .\) At what rate is \(z\) changing with respect to \(y\) when the point is at \((2,1,7) ?\)

The volume \(V\) of a right circular cone is given by $$V=\frac{\pi}{24} d^{2} \sqrt{4 s^{2}-d^{2}}$$ where \(s\) is the slant height and \(d\) is the diameter of the base. (a) Find a formula for the instantaneous rate of change of \(V\) with respect to \(s\) if \(d\) remains constant. (b) Find a formula for the instantaneous rate of change of \(V\) with respect to \(d\) if \(s\) remains constant. (c) Suppose that \(d\) has a constant value of \(16 \mathrm{cm},\) but \(s\) when \(s=10 \mathrm{cm}\). (d) Suppose that \(s\) has a constant value of \(10 \mathrm{cm},\) but \(d\) varies. Find the rate of change of \(V\) with respect to \(d\) when \(d=16 \mathrm{cm} .\)

The volume \(V\) of a right circular cylinder is given by the formula \(V=\pi r^{2} h\), where \(r\) is the radius and \(h\) is the height. (a) Find a formula for the instantaneous rate of change of \(V\) with respect to \(r\) if \(r\) changes and \(h\) remains constant. (b) Find a formula for the instantaneous rate of change of \(V\) with respect to \(h\) if \(h\) changes and \(r\) remains constant. (c) Suppose that \(h\) has a constant value of 4 in, but \(r\) varies. Find the rate of change of \(V\) with respect to \(r\) at the point where \(r=6 \mathrm{in}\). (d) Suppose that \(r\) has a constant value of \(8 \mathrm{in},\) but \(h\) varies. Find the instantaneous rate of change of \(V\) with respect to \(h\) at the point where \(h=10\) in.

Prove: If \(f, f_{x},\) and \(f_{y}\) are continuous on a circular region containing \(A\left(x_{0}, y_{0}\right)\) and \(B\left(x_{1}, y_{1}\right),\) then there is a point \(\left(x^{*}, y^{*}\right)\) on the line segment joining \(A\) and \(B\) such that \(f\left(x_{1}, y_{1}\right)-f\left(x_{0}, y_{0}\right)\) $$ =f_{x}\left(x^{*}, y^{*}\right)\left(x_{1}-x_{0}\right)+f_{y}\left(x^{*}, y^{*}\right)\left(y_{1}-y_{0}\right) $$ This result is the two-dimensional version of the Mean Value Theorem. [Hint: Express the line segment joining \(A\) and \(B\) in parametric form and use the Mean-Value Theorem for functions of one variable. \(]\)

Confirm that the mixed second-order partial derivatives of \(f\) are the same. $$ f(x, y)=\left(x^{2}-y^{2}\right) /\left(x^{2}+y^{2}\right) $$

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