Use differentials to approximate the number sqrt(6.01^2 + 6.98^2 + 5.97^2)
(Round your answer to five decimal places.)


We are asked to use differentials to approximate the number \( \sqrt{6.01^2 + 6.98^2 + 5.97^2} \).

### Step 1: Define the function
We define the function:

\[
f(x, y, z) = \sqrt{x^2 + y^2 + z^2}
\]

We want to approximate \( f(6.01, 6.98, 5.97) \). Notice that this is close to \( f(6, 7, 6) \), so we will calculate the differential at \( (6, 7, 6) \).

### Step 2: Compute the partial derivatives
The partial derivatives of \( f(x, y, z) \) 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 at \( (x_0, y_0, z_0) = (6, 7, 6) \)
First, calculate \( f(6, 7, 6) \):

\[
f(6, 7, 6) = \sqrt{6^2 + 7^2 + 6^2} = \sqrt{36 + 49 + 36} = \sqrt{121} = 11
\]

Now, evaluate the partial derivatives at \( (6, 7, 6) \):

\[
\frac{\partial f}{\partial x}(6, 7, 6) = \frac{6}{11}, \quad \frac{\partial f}{\partial y}(6, 7, 6) = \frac{7}{11}, \quad \frac{\partial f}{\partial z}(6, 7, 6) = \frac{6}{11}
\]

### Step 4: Use differentials to approximate
Let \( \Delta x = 6.01 - 6 = 0.01 \), \( \Delta y = 6.98 - 7 = -0.02 \), and \( \Delta z = 5.97 - 6 = -0.03 \).

The differential approximation is given by:

\[
df \approx \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + \frac{\partial f}{\partial z} \Delta z
\]

Substitute the values:

\[
df \approx \frac{6}{11}(0.01) + \frac{7}{11}(-0.02) + \frac{6}{11}(-0.03)
\]

\[
df \approx \frac{6}{11}(0.01) - \frac{7}{11}(0.02) - \frac{6}{11}(0.03)
\]

\[
df \approx \frac{6}{110} - \frac{14}{110} - \frac{18}{110} = \frac{6 - 14 - 18}{110} = \frac{-26}{110} = -0.23636
\]

### Step 5: Final approximation
Now, approximate the value of \( f(6.01, 6.98, 5.97) \):

\[
f(6.01, 6.98, 5.97) \approx f(6, 7, 6) + df = 11 - 0.23636 = 10.76364
\]

Thus, the approximation of \( \sqrt{6.01^2 + 6.98^2 + 5.97^2} \) is approximately:

\[
\boxed{10.76364}
\]

### Do you have any questions or would you like more details on a specific part?

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Here are five related questions to deepen your understanding:
1. How can we use differentials to approximate functions of more than three variables?
2. How does the approximation change if one of the variables is much larger than the others?
3. How does the differential approximation relate to the tangent plane approximation?
4. Why is using the differential better than directly calculating small changes in some cases?
5. Can you explain the geometric interpretation of \( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} \)?

**Tip**: When using differentials, keep the changes (\( \Delta x \), \( \Delta y \), etc.) small for the best accuracy!

Use differentials to approximate the number sqrt(6.01^2 + 6.98^2 + 5.97^2) (Round your answer to five decimal places.)

Use differentials to approximate the value of sqrt(6.01^2 + 6.98^2 + 5.97^2). This solution involves multivariable calculus concepts such as partial derivatives and differential approximation. Ideal for college-level students or advanced high school learners.

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