Use a linear approximation (or differentials) to estimate the given number. (Round your answer to five decimal places.)365

1. Break Down the Problem

We want to estimate the value of $365^{1/2}$ using linear approximation.

2. Relevant Concepts

Linear approximation uses the formula: $f(x) \approx f(a) + f'(a)(x - a)$ where:

• $f(x)$ is the function we want to estimate.
• $a$ is a point near $x$ where we know the function value.
• $f'(a)$ is the derivative of the function at $a$.

We'll let $f(x) = \sqrt{x}$ and choose $a = 361$ (since $361$ is a perfect square and close to $365$).

3. Analysis and Detail

1. Calculate $f(a)$ and $f'(a)$:

   • $f(a) = \sqrt{361} = 19$
   • The derivative $f'(x) = \frac{1}{2\sqrt{x}}$, so: $f'(361) = \frac{1}{2\sqrt{361}} = \frac{1}{2 \cdot 19} = \frac{1}{38}$

2. Calculate $x - a$:

   • Here, $x = 365$ and $a = 361$, so: $x - a = 365 - 361 = 4$

3. Plug the values into the approximation formula: $f(365) \approx f(361) + f'(361)(365 - 361)$ $f(365) \approx 19 + \left( \frac{1}{38} \right) \cdot 4$

4. Verify and Summarize

1. Perform the multiplication: $\frac{4}{38} = \frac{2}{19} \approx 0.10526315789$
2. Now combine with $f(361)$: $f(365) \approx 19 + 0.10526315789 \approx 19.10526315789$

Rounding to five decimal places: $f(365) \approx 19.10526$

Final Answer

The linear approximation of $\sqrt{365}$ is approximately $\boxed{19.10526}$.