Problem 41 - 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. \(f(x)=4 x^{3}, x=1.5\)

Step-by-Step Solution

Answer

The error \( \Delta f = 0.81 \) and the percentage error is approximately 6\%.

Step 1: Calculate Change in x

Given the measurement of \(x\) is accurate within 2%, we first determine \( \Delta x \), which is \( 2\% \) of the true value of \( x \). Here, \( x = 1.5 \), so \( \Delta x = 0.02 \times 1.5 = 0.03 \).

Step 2: Determine the Formula for \(f(x)\) and Derivative

The given function is \( f(x) = 4x^3 \). We need to find the derivative of \( f(x) \) with respect to \( x \). Thus, \( f'(x) = \frac{d}{dx}(4x^3) = 12x^2 \).

Step 3: Compute Change in f (\(\Delta f\))

Using the derivative, the error in \( f \), \( \Delta f \), can be approximated by \( \Delta f = f'(x) \cdot \Delta x \). Substitute \( x = 1.5 \), \( f'(x) = 12(1.5)^2 = 27 \), and \( \Delta x = 0.03 \): \( \Delta f = 27 \cdot 0.03 = 0.81 \).

Step 4: Calculate the Percentage Error

The percentage error is given by \( 100 \frac{\Delta f}{f} \). First, compute \( f(x) = 4(1.5)^3 = 13.5 \). Substitute \( \Delta f = 0.81 \) and \( f(x) = 13.5 \) into the formula: \( 100 \cdot \frac{0.81}{13.5} \approx 6\% \).

Key Concepts

Percentage Error

In measurement and calculations, understanding the percentage error is crucial to estimate how much the error affects the output. Percentage error shows the extent of error in relation to the actual value.
It provides a sense of perspective on how significant the error is.

To calculate the percentage error, we use the formula:

• Firstly, calculate the absolute error in your result, which is the difference between the observed and true values.
• Then, divide this error by the true value.
• Finally, multiply the result by 100 to convert it into a percentage.

In the exercise above, the percentage error is calculated for the function \( f(x) = 4x^3 \) when \( x = 1.5 \), leading us to a percentage error of about 6%.
This means the error contributes to 6% of the total calculated value.

Derivatives in Calculus

Derivatives play a key role in calculus, representing the rate of change of a function with respect to a variable, often denoted as \( \frac{dy}{dx} \) when considering a function \( y = f(x) \).
Simply put, derivatives help us understand how one quantity changes as another quantity is varied.

This has numerous practical applications, including physics for velocity and acceleration, economics for cost functions, and engineering.

• For a power function like \( f(x) = 4x^3 \), the derivative is found by multiplying the exponent by the coefficient and decreasing the exponent by one.
• This results in \( f'(x) = 12x^2 \), a simplified representation showing how \( f(x) \) changes as \( x \) changes.

Understanding derivatives is essential because it allows us to predict how small changes in one variable affect another, which is crucial for error analysis and optimization.

Applications of Calculus

The applications of calculus extend to many real-world problems and contexts. Calculus helps us comprehend dynamics in systems where change is a persistent factor.
It's used to optimize processes in physics, engineering, and even in computer graphics to create smooth motion.

In error analysis like in this scenario, we used calculus primarily to approximate the error in the calculation.

• By applying derivatives, we estimated how small changes (like a measurement error) in an input impact the output significantly.
• Such calculations help assess the relative impact of inaccuracies, providing a critical perspective for improving precision in measurements or predictions.

Engaging with these concepts offers broader insights into processes, enhancing problem-solving and critical thinking skills.