Question

Use linear approximations to estimate the following quantities. Choose a value of a to produce a small error. $$\sqrt{5 / 29}$$

Short answer

Answer: The approximate value of \(\sqrt{5 / 29}\) using linear approximations is 0.414218.

Step-by-step solution

Choose a suitable value for a

For this problem, we want to find a suitable value of \(a\) to approximate \(\sqrt{5 / 29}\). Since \(5 / 29 \approx 0.1724\), we can choose \(a = \frac{1}{6}\), which is close to \(0.1724\) and easy to work with.

Find the function and its derivative

We need a function to approximate the given expression. Let's take the function \(f(x) = \sqrt{x}\), then \(f'(x) = \frac{1}{2 \sqrt{x}}\).

Use the linear approximation formula

The linear approximation formula is \(L(x) = f(a) + f'(a)(x-a)\). In our case, we have: - \(a = \frac{1}{6}\) - \(f(a) = \sqrt{\frac{1}{6}}\) - \(f'(a) = \frac{1}{2 \sqrt{\frac{1}{6}}}\) Now, we can plug these values into the linear approximation formula to estimate \(\sqrt{5 / 29}\): $$L(x) = \sqrt{\frac{1}{6}} + \frac{1}{2 \sqrt{\frac{1}{6}}} \left( \frac{5}{29} - \frac{1}{6} \right)$$

Calculate the approximation

We can simplify the expression for \(L(x)\): $$L\left(\frac{5}{29}\right) = \sqrt{\frac{1}{6}} + \frac{1}{2 \sqrt{\frac{1}{6}}} \left( \frac{5}{29} - \frac{1}{6} \right) \approx 0.414218$$ So, using linear approximations, we have estimated \(\sqrt{5 / 29} \approx 0.414218\).

Key concepts

Linear Approximation in Mathematics

Linear approximation is a technique used in calculus to estimate the value of a function near a given point. This method is particularly useful when dealing with complex functions that are difficult or time-consuming to calculate directly.
Linear approximation simplifies these problems using a straight line, called the tangent line, which closely represents the function's curve at a specific point. The formula for linear approximation is:

• \[ L(x) = f(a) + f'(a)(x-a) \]

where:

• \(L(x)\) is the linear approximation of the function,
• \(f(a)\) is the actual value of the function at point \(a\),
• \(f'(a)\) is the derivative of the function at point \(a\),
• \((x-a)\) is the distance from \(a\) to \(x\).

By using this formula, an approximation is made which is often close enough for practical purposes. Linear approximation is particularly useful in engineering and physics, where small changes need quick calculations.

Understanding Derivatives

Derivatives are a fundamental concept in calculus representing the rate at which a function is changing at any given point. They provide critical information about the behavior of a function.
When you compute the derivative of a function, you are finding a new function that gives the slope of the tangent line to the graph of the original function at any given point.
For a function \(f(x)\), its derivative \(f'(x)\) can be visually interpreted as:

• The steepness of the graph,
• Whether the function is increasing or decreasing,
• Behavior of the graph near specific points.

For example, in our linear approximation exercise, knowing the derivative \(f'(x) = \frac{1}{2 \sqrt{x}}\) allows us to determine how the function \(f(x) = \sqrt{x}\) is changing near the point of interest.

Estimating Functions with Approximations

Function estimation involves predicting the value of a function based on its behavior around a near point. This is particularly useful in real-world applications where exact values are not feasible due to time or complexity constraints.
Linear approximation serves as a handy tool by relying on tangent lines to estimate the precise function value near a point \(a\).
For instance, we approximate \(\sqrt{5/29}\) by choosing \(a = \frac{1}{6}\) as it is close to the function evaluation point and easy to manage.
Function estimation doesn't just simplify calculations; it also provides insights into the trends and patterns of such functions, which can be vital in scenarios requiring quick decisions.

The Role of Small Error Approximation

Small error approximation maximizes the accuracy of linear approximations by selecting a tangent point close enough to the value you want to estimate. The smaller the gap or error between the chosen point \(a\) and the actual value \(x\), the more accurate your approximation will be.
This concept is crucial when exact precision is needed, but calculation or resource limitations prevent detailed computation.
In the exercise, choosing \(a = \frac{1}{6}\) for the approximation was strategic because it minimizes the deviation from \(\frac{5}{29}\), thus leading to a more precise estimated value.
This technique ensures approximations are not only efficient but also reliable across various mathematical applications, making calculus highly applicable in real-world scenarios.