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Chapter 4: Problem 28

Use linear approximations to estimate the following quantities. Choose a value of a to produce a small error. $$1 / \sqrt{119}$$

Short Answer

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Question: Use linear approximation to estimate the value of \(\frac{1}{\sqrt{119}}\). Answer: The linear approximation of \(\frac{1}{\sqrt{119}}\) is approximately \(\frac{243}{2662}\).

Step by step solution

01

Define the function and find its derivative

Our goal is to approximate the value of \(\frac{1}{\sqrt{119}}\). Let's define a function: $$f(x) = \frac{1}{\sqrt{x}}$$ Next, we should determine the derivative, \(f'(x)\): $$f'(x) = -\frac{1}{2\sqrt{x^3}}$$
02

Choose a suitable value for a to minimize the error

We know that the closer the value of a to the point we want to find, the smaller the error will be. Since we want to find an approximation for \(\frac{1}{\sqrt{119}}\), we can pick \(a=121\) because its square root is an integer value (11). Choosing a value with an integer square root will simplify our calculations.
03

Calculate the linear approximation

Now we can find the linear approximation using our chosen value for \(a\). Recall that the linear approximation formula is: $$L(x) =f(a) + f'(a)(x-a)$$ Substitute \(a=121\) and \(x=119\): $$L(119) = \frac{1}{\sqrt{121}} - \frac{1}{2\sqrt{121^3}} (119-121)$$
04

Simplify the expression

We can simplify the expression by calculating the values: $$L(119) = \frac{1}{11} + \frac{1}{(2 \times 11^3)} (2)$$ $$L(119) = \frac{1}{11} + \frac{1}{2 \times 11^2}$$ $$L(119) = \frac{1}{11} + \frac{1}{242}$$ $$L(119) = \frac{243}{2662}$$ Therefore, the linear approximation of \(\frac{1}{\sqrt{119}}\) is approximately \(\frac{243}{2662}\).

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