Problem 28 - Calculus Early Transcendentals

Question

Use linear approximations to estimate the following quantities. Choose a value of a that produces a small error. \(1 / \sqrt{119}\)

Step-by-Step Solution

Answer

The linear approximation used to estimate \(1 / \sqrt{119}\) is \(\frac{344}{2662}\).

Step 1: Choose a suitable base value a

Choose a base value \(a\) that is close to \(119\) and has an easy-to-calculate square root. The closest number to \(119\) with a known square root is \(121\), as \(\sqrt{121}=11\). Therefore, let \(a = 121\).

Step 2: Define the function f(x)

We want to estimate \(1 / \sqrt{119}\), so let's define the function \(f(x) = \frac{1}{\sqrt{x}}\).

Step 3: Calculate the derivative of f(x)

Using the product and chain rule, compute the derivative \(f'(x)\), where: \(f'(x) = -\frac{1}{2}x^{-\frac{3}{2}}\)

Step 4: Evaluate f(a) and f'(a)

Using the chosen base value \(a=121\) and the function\( f(x) =\frac{1}{\sqrt{x}}\), compute \(f(a)\) and \(f'(a)\): \(f(121) = \frac{1}{\sqrt{121}} = \frac{1}{11}\) \(f'(121) = -\frac{1}{2}121^{-\frac{3}{2}} = -\frac{1}{2} \left(\frac{1}{11}\right)^3 = -\frac{1}{2} \cdot\frac{1}{1331}\)

Step 5: Use the linear approximation formula

Apply the linear approximation formula to estimate \(\frac{1}{\sqrt{119}}\): \(f(x) \approx f(a) + f'(a) (x-a)\) Let \(x = 119\), thus: \(f(119) \approx f(121) + f'(121) (119-121)\) \(f(119) \approx \frac{1}{11} -\frac{1}{2} \cdot\frac{1}{1331} \cdot (-2)\) \(f(119) \approx \frac{1}{11} + \frac{1}{1331}\) Now, we can use this approximation for the desired estimate: \(\frac{1}{\sqrt{119}} \approx \frac{1}{11} + \frac{1}{1331} = \frac{344}{2662}\) Thus, the linear approximation to estimate \(1 / \sqrt{119}\) is \(\frac{344}{2662}\).

Key Concepts

Derivative Calculation

The process of finding the derivative is key in understanding the rate at which functions change. In this exercise, we have a function \( f(x) = \frac{1}{\sqrt{x}} \). To find its derivative, \( f'(x) \), we employ certain rules. First, recognize that \( \sqrt{x} = x^{1/2} \), which can be rewritten as a power.
To find the derivative, consider:

• The power rule: The derivative of \( x^n \) is \( nx^{n-1} \).
• The chain rule: This helps differentiate composite functions, expressed as \( f(g(x)) \), deriving as \( f'(g(x)) \times g'(x) \).

Applying these, since \( f(x) = x^{-1/2} \), we find:
- The outer function is \( u^{-1} \) and the inner function is \( u = x^{1/2} \).- The derivative of the outer function using power rule provides \( -1/2 x^{-3/2} \), simplified to \( -\frac{1}{2} x^{-3/2} \).Hence, the derivative of \( f(x) \) is \( f'(x) = -\frac{1}{2}x^{-\frac{3}{2}} \).
This derivative tells us how changes in \( x \) influence \( f(x) \).

Chain Rule

The Chain Rule is an essential tool in calculus that allows us to differentiate compositions of functions. It is particularly useful for expressions where one function is nested inside another. In our example, we have a function \( f(x) = \frac{1}{\sqrt{x}} \), where the square root can be considered a composition of functions.
Let's break it down:Suppose you have a function \( g(x) = \sqrt{x} \), rewritten as \( x^{1/2} \). If \( g(x) \) is embedded within another function, say \( h(u) = 1/u \), forming \( h(g(x)) \), you apply the chain rule as follows:

• First, determine the derivative of the outer function \( h(u) \), resulting in \( -u^{-2} \).
• Then, apply the derivative of the inner function \( g(x) \), yielding \( 1/2\times x^{-1/2} \).

Multiply these derivatives to obtain the composite's derivative:
Thus, \( h'(g(x)) \times g'(x) \) becomes \( -\frac{1}{2} x^{-3/2} \).
Using the chain rule simplifies the derivation of complex function combinations and is fundamental to calculus.

Function Estimation

Function estimation through linear approximation is a mathematical strategy for predicting values of a function near a specific point. This method is especially handy when direct calculation is difficult. In this problem, we're estimating \( \frac{1}{\sqrt{119}} \).
To achieve an accurate approximation, choose a center point \( a \) near the target value \( x \). For this scenario, \( a = 121 \) was selected, as it's easy to compute \( \sqrt{121} = 11 \).Linear approximation formula:
\[ f(x) \approx f(a) + f'(a)(x-a)\]Here's how it works:

• Calculate \( f(a) \) as \( \frac{1}{\sqrt{121}} = \frac{1}{11} \).
• Compute \( f'(a) \), which after evaluation becomes \( -\frac{1}{2} (\frac{1}{11})^3 \).
• Substitute these into the approximation formula where \( x = 119 \).

So the approximation for \( \frac{1}{\sqrt{119}} \) is achieved by:
\[f(119) \approx \frac{1}{11} + \frac{1}{1331}\]
This results in \( \frac{344}{2662} \), matching our earlier finding.
Ultimately, linear approximation helps us simplify and swiftly estimate values, leveraging local linearity of functions.