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Chapter 1: Problem 1

Use linear approximation to approximate \(\sqrt{25.3}\) as follows. Let \(f(x)=\sqrt{x}\). The equation of the tangent line to \(f(x)\) at \(x=25\) can be written in the form \(y=m x+b\). Compute \(m\) and \(b\). \(m=\) \(\square\) \(b=\) \(\square\) Using this find the approximation for \(\sqrt{25.3}\). Answer: \(\square\)

Short Answer

Expert verified
Approximation of \(\sqrt{25.3}\) is 5.03, with \(m = \frac{1}{10}\) and \(b = 2.5\).

Step by step solution

01

Define the Function

Define the function as given: \(f(x) = \sqrt{x}\).
02

Find the Derivative of the Function

Compute the derivative of \(f(x)\) using the power rule. Since \(f(x) = x^{1/2}\), the derivative \( f'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2 \sqrt{x}} \).
03

Evaluate the Derivative at \(x = 25\)

Substitute \(x = 25\) into the derivative to find the slope of the tangent line: \(m = f'(25) = \frac{1}{2 \sqrt{25}} = \frac{1}{2 \cdot 5} = \frac{1}{10}\).
04

Find the y-Coordinate of the Tangent Point

Evaluate the original function at \(x = 25\) to find the y-coordinate of the point of tangency: \(f(25) = \sqrt{25} = 5\).
05

Write the Equation of the Tangent Line

Using the slope \(m = \frac{1}{10}\) and the point of tangency \((25, 5)\), write the equation of the tangent line in point-slope form: \(y - 5 = \frac{1}{10}(x - 25)\).
06

Convert the Equation to Slope-Intercept Form

Rearrange the equation to get slope-intercept form \(y = mx + b\): \(y - 5 = \frac{1}{10}x - 2.5\), therefore, \(y = \frac{1}{10}x + 2.5\). So, \(m = \frac{1}{10}\) and \(b = 2.5\).
07

Approximate \(\sqrt{25.3}\)

Use the tangent line equation to approximate \(\sqrt{25.3}\): \(y \approx \frac{1}{10} \cdot 25.3 + 2.5\). Simplifying this gives us \(y \approx 2.53 + 2.5 = 5.03\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent Line
The tangent line to a curve at a particular point gives us a linear approximation of the function near that point. This line just 'touches' the curve at one specific point without crossing it. For the function given in the exercise, we need to find the tangent line at a particular point by using its derivative. The derivative represents the slope of this tangent line. By knowing the slope and a point on the curve, you can construct the equation of the tangent line.
Derivative
A derivative helps us understand how a function changes as its input changes. It gives us the rate of change, or the slope, at any given point. In this exercise, the function is \( f(x) = \sqrt{x} \). By applying the power rule, we get \( f'(x) = \frac{1}{2 \sqrt{x}} \). Finding the derivative allows us to determine the slope of the tangent line at any given point. This slope is crucial for constructing the linear approximation.
Slope-Intercept Form
The slope-intercept form of a line is one of the most straightforward ways to represent a linear equation. It is represented as \( y = mx + b \), where \( m \) is the slope of the line and \( b \) is the y-intercept. In this exercise, once we found the slope (derivative) \( m = \frac{1}{10} \) and a point on the tangent line \( (25, 5) \), we could write the equation in slope-intercept form. This equation helps us approximate \sqrt{25.3}\.
Approximation
Approximations are valuable because they allow us to estimate values that might be difficult to compute exactly. By using the tangent line we constructed, we can approximate the value of \(鈭�25.3 \). The equation of the tangent line \( y = \frac{1}{10} x + 2.5 \) was used, substituting \( x = 25.3 \). This gave us the approximate value of \( y \), which is \( 鈭�25.3 鈮� 5.03 \). This close estimate demonstrates how linear approximations can simplify complex calculations.

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Most popular questions from this chapter

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