Problem 42 - Calculus for Biology and Medicine

Question

Assume that the measurement of \(x\) is \(a c-\) curate within \(2 \% .\) In each case, determine the error \(\Delta f\) in the calculation of \(f\) and find the percentage error \(100 \frac{\Delta f}{f} .\) The quantities \(f(x)\) and the true value of \(x\) are given. \(f(x)=x^{1 / 4}, x=10\)

Step-by-Step Solution

Answer

The error \( \Delta f \) is approximately 0.0089, and the percentage error is approximately 0.5%.

Step 1: Determine the Error in x

Given that the measurement of \( x \) is 2% accurate, calculate the absolute error \( \Delta x \) by finding 2% of \( x = 10 \). Use the formula \( \Delta x = \frac{2}{100} \times x = 0.2 \).

Step 2: Calculate f(x)

Compute \( f(x) = x^{1/4} \) for the given \( x = 10 \). Use the expression \( f(10) = 10^{1/4} \). By evaluating, we find \( f(10) \approx 1.778 \).

Step 3: Derive the Derivative of f(x)

Determine the derivative \( f'(x) \) needed for the error analysis. For \( f(x) = x^{1/4} \), use the power rule to find \( f'(x) = \frac{1}{4} x^{-3/4} \).

Step 4: Evaluate f'(x) at x = 10

Substitute \( x = 10 \) into \( f'(x) = \frac{1}{4} x^{-3/4} \) to find \( f'(10) = \frac{1}{4} \times 10^{-3/4} \). Simplify to get \( f'(10) \approx 0.0445 \).

Step 5: Calculate the Change in f(x)

Use the formula \( \Delta f = f'(10) \times \Delta x \) to find the change in \( f(x) \). Substitute \( f'(10) \approx 0.0445 \) and \( \Delta x = 0.2 \) to get \( \Delta f \approx 0.0445 \times 0.2 = 0.0089 \).

Step 6: Calculate the Percentage Error

Finally, calculate the percentage error using \( 100 \times \frac{\Delta f}{f(10)} \). Use the values \( \Delta f \approx 0.0089 \) and \( f(10) \approx 1.778 \). Therefore, the percentage error is \( 100 \times \frac{0.0089}{1.778} \approx 0.5\% \).

Key Concepts

Percentage Error

Percentage error helps us understand how much our calculated value deviates from the actual or true value, expressed as a percentage. It's a useful way to compare the precision of different measurements or calculations.
In our exercise, the true value of the measurement was evaluated, and we wanted to find out how accurately it was determined. Here is how this can be done:

• We started by estimating the error in our function value, denoted as \( \Delta f \).
• The percentage error formula is given by \( 100 \times \frac{\Delta f}{f(x)} \).
• In our example, \( \Delta f \) was approximately \( 0.0089 \), and \( f(10) \) was approximately \( 1.778 \).
• Therefore, the percentage error was calculated as \( 100 \times \frac{0.0089}{1.778} \) which is approximately \( 0.5\% \).

This tells us that our calculation is very accurate, having a small deviation from the true value.

Derivative Calculation

Derivatives are essential tools in calculus for analyzing changes in functions. They indicate how a function behaves as its input changes, providing a rate of change or slope at any point.
For our exercise:

• The given function was \( f(x) = x^{1/4} \), where the derivative describes how our function value changes as \( x \) changes.
• We applied the power rule for differentiation, resulting in the derivative \( f'(x) = \frac{1}{4} x^{-3/4} \).
• Next, we evaluated this derivative at the specified point, \( x = 10 \), yielding \( f'(10) \approx 0.0445 \).

This derivative value reflects the rate at which \( f(x) \) changes near \( x=10 \), and it is crucial for subsequent error analysis.

Function Approximation

Function approximation is a primary use of derivatives, allowing us to estimate the function's behavior around a specific point. It is vital in situations where it's difficult or impossible to directly measure changes.
In the given problem:

• We approximated the error in the function \( f(x) = x^{1/4} \) using the derivative \( f'(x) \).
• The approximation used was \( \Delta f = f'(10) \times \Delta x \), meaning the change in \( f \), \( \Delta f \), is approximately the outcome of multiplying the derivative at the point with the change in \( x \), \( \Delta x = 0.2 \).
• Computing this gave us \( \Delta f \approx 0.0089 \), which is the estimated error in \( f(x) \) for a small change in \( x \).

Function approximation illustrates how derivatives provide insights into the responsive nature of functions, especially close to specific points, making them indispensable in practical applications.