Problem 40 - Calculus for Biology and Medicine

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 \(x \pm|\Delta x| .\) \(f(x)=\sin x, x=-1 \pm 0.05\)

Step-by-Step Solution

Answer

The interval reflecting the measurement error is approximately \([-0.8696, -0.8134]\).

Step 1: Understanding the Problem

We need to find the interval \([f(x)-\Delta f, f(x)+\Delta f]\), where \(f(x) = \sin(x)\), given the error in \(x\) is \(-1 \pm 0.05\). This means \(x\) could be anywhere between \(-1.05\) and \(-0.95\).

Step 2: Calculate \(f(x)\) at the boundaries

Calculate \(f(x)\) at \(x = -1.05\) and \(x = -0.95\):\[\sin(-1.05) \approx -0.8674\]\[\sin(-0.95) \approx -0.8134\]

Step 3: Determine \(f(x)\) at the nominal value

Find \(f(x)\) at the nominal \(x = -1\):\[\sin(-1) \approx -0.8415\]

Step 4: Calculate Measurement Error in \(f(x)\)

Determine the maximum deviation \(\Delta f\) from \(f(x)\) at the nominal value:\[\Delta f = \max(|\sin(-1.05) + 0.8415|, |\sin(-0.95) + 0.8415|)\]\[\Delta f = \max(|-0.8674 + 0.8415|, |-0.8134 + 0.8415|)\]\[\Delta f = \max(0.0259, 0.0281)\]\[\Delta f \approx 0.0281\]

Step 5: Formulate the Interval

Using \(\Delta f \approx 0.0281\), find the interval:\[[f(x) - \Delta f, f(x) + \Delta f] = [-0.8415 - 0.0281, -0.8415 + 0.0281]\]\[[-0.8696, -0.8134]\]This is the interval for \(f(x)\) reflecting the measurement error.

Key Concepts

Measurement Error

Measurement error is a crucial concept in calculus error analysis. It involves the uncertainty or deviation when measuring a quantity. In the exercise, the true value of \( x \) is given as

• \(-1 \pm 0.05\)

This means the measured value can vary between \(-1.05\) and \(-0.95\). The goal is to understand how this variability in \( x \) affects the function \(f(x) = \sin(x)\).

The measurement error in \( x \) translates into an error in \( f(x) \) as sin is sensitive to changes in its input. Even small changes around \( x \) lead to measurable shifts in \( \sin(x) \). This requires us to calculate the output interval of the function that reflects the measurement uncertainty.

Function Approximation

Function approximation involves evaluating how well a function approximates values over certain intervals. In this exercise, we approximate \(f(x) = \sin(x)\) near the point \(x = -1 \).

To approximate \( f(x) \), we calculate its values at the extremes of the interval influenced by the measurement error. This involves non-exact calculations because

• \(\sin(-1.05)\) and \(\sin(-0.95)\)

are computed using a calculator or numerical methods, not algebraic expressions.

This stretch in \( x \)'s value is necessary to capture how functions behave over small changes, reflecting approximation's role in determining realistic outcomes where some error is present.

Interval Error Estimation

Interval error estimation helps define the range in which the actual function value lies, considering the input errors. After calculating \( \sin(-1.05) \) and \( \sin(-0.95) \), we can see both variations deviate from \( \sin(-1) \).

The error interval

• \([-0.8696, -0.8134]\)

provides the range where \( f(x) \) might fall due to the measurement error. We calculate this by determining the maximum deviations from the nominal value \( \sin(-1) \).

With \( \Delta f = 0.0281 \), the interval is formed to encompass all possible outcomes, ensuring our results reflect the variability introduced by the initial measurement error.