Problem 38
Question
A measurement error in \(x\) affects the accuracy of the value \(f(x) .\) In each case, determine an interval of the form $$[f(x)-\Delta f, f(x)+\Delta f]$$ that reflects the measurement error \(\Delta x .\) In each problem, the quantities given are \(f(x)\) and \(x=\) true value of \(x \pm|\Delta x|\). $$ f(x)=\sqrt{x}, x=10 \pm 0.5 $$
Step-by-Step Solution
Verified Answer
The interval is approximately [3.082, 3.240].
1Step 1: Understanding the Problem Context
We have a function \( f(x) = \sqrt{x} \), and a value \( x = 10 \pm 0.5 \), which means \( x \) can vary between 9.5 and 10.5. We need to find an interval for \( f(x) \), considering this variability in \( x \).
2Step 2: Calculate the Function at Bounds of Interval
To find the interval for \( f(x) \), start by calculating \( f(x) \) at the minimum and maximum possible values of \( x \). This involves computing \( \sqrt{9.5} \) and \( \sqrt{10.5} \).
3Step 3: Compute Values
Calculate \( \sqrt{9.5} \approx 3.082 \) and \( \sqrt{10.5} \approx 3.240 \). These are the function values at the lower and upper bounds of \( x \).
4Step 4: Establish the Interval
The interval for \( f(x) \) is determined as \([\sqrt{9.5}, \sqrt{10.5}] \), which translates to approximately \([3.082, 3.240]\).
5Step 5: Interpret the Results
The interval \([3.082, 3.240]\) represents the possible range for \( f(x) = \sqrt{x} \) given the error in \( x \). This reflects the uncertainty in \( f(x) \) due to the measurement error in \( x \).
Key Concepts
Measurement ErrorInterval EstimationSquare Root Function
Measurement Error
Measurement error refers to the inaccuracies or variations in the values obtained when measuring real-world quantities. In many practical scenarios, it is impossible to measure something with absolute precision, which is why small errors, known as measurement errors, are common. This error can affect subsequent calculations which depend on the measured values.
When working with mathematical functions, any error in measuring the input variable gets propagated through the function and affects the outcome. For instance, if we have a function like the square root function, even a tiny error in the input can lead to noticeable variations in the computed output. This is especially important in scientific and engineering contexts where precise calculations are crucial.
To combat this uncertainty, mathematicians use interval estimation, ensuring that any propagated error results in a range of possible outcomes rather than a single misleading number.
When working with mathematical functions, any error in measuring the input variable gets propagated through the function and affects the outcome. For instance, if we have a function like the square root function, even a tiny error in the input can lead to noticeable variations in the computed output. This is especially important in scientific and engineering contexts where precise calculations are crucial.
To combat this uncertainty, mathematicians use interval estimation, ensuring that any propagated error results in a range of possible outcomes rather than a single misleading number.
Interval Estimation
Interval estimation provides a range within which a true value is expected to lie, considering any possible measurement errors. Instead of giving a singular outcome, interval estimation offers lower and upper bounds that encompass all likely values.
This method is highly beneficial in error analysis since it allows us to understand the extent of variability due to measurement inaccuracies. In the context of the square root function, for instance, if the input value of \( x \) is given as \( 10 \pm 0.5 \), interval estimation helps us establish the corresponding interval for \( f(x) = \sqrt{x} \).
By computing the function at the lowest possible value (9.5) and the highest possible value (10.5) of \( x \), we achieve an interval for \( f(x) \). This approach provides robust information about the possible results of the function, and helps in making informed decisions based on likely outcomes, rather than precise but potentially inaccurate numbers.
This method is highly beneficial in error analysis since it allows us to understand the extent of variability due to measurement inaccuracies. In the context of the square root function, for instance, if the input value of \( x \) is given as \( 10 \pm 0.5 \), interval estimation helps us establish the corresponding interval for \( f(x) = \sqrt{x} \).
By computing the function at the lowest possible value (9.5) and the highest possible value (10.5) of \( x \), we achieve an interval for \( f(x) \). This approach provides robust information about the possible results of the function, and helps in making informed decisions based on likely outcomes, rather than precise but potentially inaccurate numbers.
Square Root Function
The square root function is a basic mathematical function defined as \( f(x) = \sqrt{x} \). It appears frequently in both basic and advanced mathematical calculations. This function pairs each non-negative input \( x \) with a number that, when multiplied by itself, produces \( x \).
For example, \( \sqrt{9} = 3 \) because \( 3^2 = 9 \). When working with errors in measurements, the square root function plays a crucial role because any errors in the input values are directly reflected in the output.
In our exercise with \( x = 10 \pm 0.5 \), the square root function helps gauge the influence of this small error on the result of \( \sqrt{x} \). By calculating the square roots of 9.5 and 10.5, we find that the outputs vary reasonably within the interval \([3.082, 3.240]\). This illustrates how sensitive functions can be to input changes and emphasizes the importance of accurate measurements in calculations that involve such functions. The square root function thus serves not only as a calculator of square roots but also as a tool for understanding the propagation of measurement errors.
For example, \( \sqrt{9} = 3 \) because \( 3^2 = 9 \). When working with errors in measurements, the square root function plays a crucial role because any errors in the input values are directly reflected in the output.
In our exercise with \( x = 10 \pm 0.5 \), the square root function helps gauge the influence of this small error on the result of \( \sqrt{x} \). By calculating the square roots of 9.5 and 10.5, we find that the outputs vary reasonably within the interval \([3.082, 3.240]\). This illustrates how sensitive functions can be to input changes and emphasizes the importance of accurate measurements in calculations that involve such functions. The square root function thus serves not only as a calculator of square roots but also as a tool for understanding the propagation of measurement errors.
Other exercises in this chapter
Problem 37
Differentiate the functions with respect to the independent variable. $$ f(x)=\ln \frac{x}{x+1} $$
View solution Problem 37
Differentiate $$ g(N)=N\left(1-\frac{N}{K}\right) $$
View solution Problem 38
The following limit represents the derivative of a function \(f\) at the point \((a, f(a))\) : $$ \lim _{h \rightarrow 0} \frac{\frac{1}{(2+h)^{2}+1}-\frac{1}{5
View solution Problem 38
Assume that \(f(x)\) and \(g(x)\) are differentiable. Find \(\frac{d}{d x} f\left[\frac{1}{g(x)}\right]\)
View solution