To solve the problem of approximating $\sqrt{16.2}$ using differentials, let's follow these steps:

Step 1: Choose a function and a point.

We know that the function for which we are trying to approximate is: $f(x) = \sqrt{x}$ We want to approximate $\sqrt{16.2}$. Notice that 16 is very close to 16.2, and we know the exact value of $\sqrt{16} = 4$. So we choose $a = 16$ as our base point, which is easy to work with.

Step 2: Calculate the differential.

The differential approximation formula is: $f(x) \approx f(a) + f'(a)(x - a)$ First, calculate the derivative of $f(x) = \sqrt{x}$. Using the power rule, we get: $f'(x) = \frac{1}{2\sqrt{x}}$ At $a = 16$, the derivative is: $f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{2 \times 4} = \frac{1}{8}$

Step 3: Use the approximation formula.

Now, apply the differential approximation: $\sqrt{16.2} \approx \sqrt{16} + f'(16) \cdot (16.2 - 16)$ Substitute the values: $\sqrt{16.2} \approx 4 + \frac{1}{8} \cdot (16.2 - 16) = 4 + \frac{1}{8} \cdot 0.2$ $\sqrt{16.2} \approx 4 + \frac{0.2}{8} = 4 + 0.025 = 4.025$

Final Answer:

Thus, the differential approximation of $\sqrt{16.2}$ is approximately 4.0250, rounded to four decimal places.

Would you like more details, or do you have any questions?

Here are 5 related questions you could explore:

1. How does the differential approximation method compare to using a calculator for square roots?
2. Can you use differentials to approximate cube roots as well?
3. How would the result change if we used a closer value to 16.2, like 16.25, for the approximation?
4. How do differentials apply to functions more complex than square roots, such as trigonometric functions?
5. What are the conditions for a good approximation when using differentials?

Tip: Always choose a base point that simplifies the calculation but is close to the actual value you're approximating.