Question

In Exercises \(11-14,\) approximate the root by using a linearization centered at an appropriate nearby number. \(\sqrt{101}\)

Short answer

The linear approximation of \(\sqrt{101}\) is \(10.05\).

Step-by-step solution

Select Appropriate Nearby Number

Here, 100 is a perfect square and it is near by 101. So it should be the center for the linearization.

Apply the formula of linear approximation

The formula of linear approximation or linearization is \(f(a) + f'(a)(x-a)\). The function is \(f(x)=\sqrt{x}\) here. So the derivative \(f'(x)= \frac{1}{2\sqrt{x}}\). Now to find the approximation for \(\sqrt{101}\), replace \(a\) with \(100\) (the nearby number) and \(x\) with \(101\) in the linearization equation.

Compute the Linear Approximation

The equation then becomes: \(\sqrt{100} + \frac{1}{2\sqrt{100}}*(101-100)\), which simplifies to: \(10 + \frac{1}{2*10}\) = \(10+\frac{1}{20}\), which simplifies to: \(10+0.05\).

Get the Final Answer

Finally, add the numbers obtained in previous step to get the root of 101 which results in \(10.05\).

Key concepts

Linearization of Functions

Linearization is a method used to approximate the value of a function near a given point, which is particularly useful for complex equations that are challenging to solve directly. Think of it like using a tiny straight-edge ruler to measure a curve; you're simplifying the curve into a straight line over a very short distance. Mathematically, the process creates a linear function, or a tangent line, that closely follows the curve of the original function at a specific point—this is the 'center' of the linearization.

To carry out linearization, we first need a function that is differentiable at the center point. Differentiable means that the function has a derivative at that point. We then use the function's value and its derivative at the center point to construct the linear approximation. This method is powerful because derivatives reflect how the function changes, which helps us to create the 'best-fit' straight line near the center.

The general formula for linear approximation is: \(f(a) + f'(a)(x-a)\), where \(f(x)\) is the original function, \(a\) is the center point, and \(x\) is the point we're approximating. By applying this technique, we can often obtain a very close approximation to the function's true value at certain points with minimal computation.

Derivative Application

The derivative of a function at a point is the heart of linear approximation. It gives us the slope of the tangent line at that point. In simpler terms, the derivative tells us how steep the 'hill' or 'valley' of our function is at the point of interest, which in turn dictates the angle of our approximating line. This rate of change is critical since it dictates the direction and steepness of our linear function.

When we apply the derivative to our approximation formula, we're effectively capturing the function’s immediate rate of change. For example, the derivative of \(f(x)=\sqrt{x}\), is \(f'(x)=\frac{1}{2\sqrt{x}}\). This reveals how rapidly the square root function is changing at any given point. For \(x=100\), the square root function changes at a slope of \(\frac{1}{2*10}\), or \(\frac{1}{20}\). Inserting this derivative into our linear approximation formula helps us project the value of the function at a nearby point, such as \(\sqrt{101}\), which is the basis of our approximation problem.

Square Root Approximation

Approximating the square root of a number that is not a perfect square can prove to be a tricky task without a calculator. However, by using linear approximation, we're equipped to estimate it with remarkable accuracy. The square root approximation is a specific application of linearization where we work with the function \(f(x)=\sqrt{x}\).

Since perfect squares are both familiar and easy to calculate, they make ideal center points for linear approximation. In the provided exercise, \(100\) was a natural choice for approximating \(\sqrt{101}\) since it's a perfect square close to \(101\). Using linearization, we simplified our task to a basic arithmetic operation: \(10 + \frac{1}{2*10} * 1\), resulting in \(10.05\) as the approximate value for \(\sqrt{101}\).

This approach is exceptionally useful in situations where precision is less critical or when resources are limited. The approximation provides an easily computed, yet sufficiently accurate, estimate of the square root for the given number.