To approximate the value of $\sqrt{2.012^2 + 5.972^2 + 8.982^2}$ using differentials, we'll follow these steps:

Step 1: Define the function

Let $f(x, y, z) = \sqrt{x^2 + y^2 + z^2}$.

We want to evaluate this function at $(x, y, z) = (2, 6, 9)$, since $(2.012, 5.972, 8.982)$ is close to these values.

Step 2: Find the partial derivatives

The differential approximation formula is: $df = \frac{\partial f}{\partial x} \, dx + \frac{\partial f}{\partial y} \, dy + \frac{\partial f}{\partial z} \, dz$

For $f(x, y, z) = \sqrt{x^2 + y^2 + z^2}$, the partial derivatives are: $\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \quad \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \quad \frac{\partial f}{\partial z} = \frac{z}{\sqrt{x^2 + y^2 + z^2}}$

Step 3: Evaluate the partial derivatives at $(2, 6, 9)$

First, calculate $f(2, 6, 9)$: $f(2, 6, 9) = \sqrt{2^2 + 6^2 + 9^2} = \sqrt{4 + 36 + 81} = \sqrt{121} = 11$

Now, the partial derivatives evaluated at $(2, 6, 9)$ are: $\frac{\partial f}{\partial x} \bigg|_{(2, 6, 9)} = \frac{2}{11}, \quad \frac{\partial f}{\partial y} \bigg|_{(2, 6, 9)} = \frac{6}{11}, \quad \frac{\partial f}{\partial z} \bigg|_{(2, 6, 9)} = \frac{9}{11}$

Step 4: Calculate the differentials $dx, dy, dz$

We have: $dx = 2.012 - 2 = 0.012, \quad dy = 5.972 - 6 = -0.028, \quad dz = 8.982 - 9 = -0.018$

Step 5: Calculate $df$

Substituting into $df$: $df = \frac{2}{11} \cdot 0.012 + \frac{6}{11} \cdot (-0.028) + \frac{9}{11} \cdot (-0.018)$ Let's calculate this accurately.

The value of $df$ is approximately $-0.02781818181818182$.

Step 6: Approximate $f(2.012, 5.972, 8.982)$

Using the differential approximation: $f(2.012, 5.972, 8.982) \approx f(2, 6, 9) + df = 11 + (-0.02781818181818182)$ Let's compute the final value.The approximated value of $\sqrt{2.012^2 + 5.972^2 + 8.982^2}$ using differentials is approximately $10.97218$ (rounded to five decimal places).

Do you want more details or have any questions?

Here are 5 related questions that might expand your understanding:

1. How would the differential approximation change if you chose a different point?
2. Why is it useful to approximate functions using differentials in calculus?
3. Can we use this method to approximate other functions beyond square roots?
4. How accurate is this differential method compared to using a calculator?
5. What are partial derivatives and how do they differ from regular derivatives?

Tip: The differential approximation works best when the values you're approximating are very close to the chosen point.