To find the linear approximation of the function $f(x, y, z) = x^2 + y^2 + z^2$ at the point $(4, 4, 7)$, we use the formula for the linear approximation (or the tangent plane approximation):

$f(x, y, z) \approx f(a, b, c) + f_x(a, b, c)(x - a) + f_y(a, b, c)(y - b) + f_z(a, b, c)(z - c)$

Where:

• $(a, b, c) = (4, 4, 7)$ is the point at which we are approximating.
• $f_x$, $f_y$, and $f_z$ are the partial derivatives of $f(x, y, z)$.

Step 1: Compute the function value at $(4, 4, 7)$

First, calculate $f(4, 4, 7)$:

$f(4, 4, 7) = 4^2 + 4^2 + 7^2 = 16 + 16 + 49 = 81$

Step 2: Find the partial derivatives

Next, we compute the partial derivatives of $f(x, y, z) = x^2 + y^2 + z^2$:

• $f_x = \frac{\partial}{\partial x}(x^2 + y^2 + z^2) = 2x$
• $f_y = \frac{\partial}{\partial y}(x^2 + y^2 + z^2) = 2y$
• $f_z = \frac{\partial}{\partial z}(x^2 + y^2 + z^2) = 2z$

Step 3: Evaluate the partial derivatives at $(4, 4, 7)$

Now, evaluate these partial derivatives at the point $(4, 4, 7)$:

• $f_x(4, 4, 7) = 2 \times 4 = 8$
• $f_y(4, 4, 7) = 2 \times 4 = 8$
• $f_z(4, 4, 7) = 2 \times 7 = 14$

Step 4: Write the linear approximation formula

Now, we can write the linear approximation formula: $f(x, y, z) \approx 81 + 8(x - 4) + 8(y - 4) + 14(z - 7)$

Step 5: Use the approximation to estimate $f(4.02, 3.99, 6.97)$

Substitute $x = 4.02$, $y = 3.99$, and $z = 6.97$ into the linear approximation formula: $f(4.02, 3.99, 6.97) \approx 81 + 8(4.02 - 4) + 8(3.99 - 4) + 14(6.97 - 7)$ Simplifying the differences: $f(4.02, 3.99, 6.97) \approx 81 + 8(0.02) + 8(-0.01) + 14(-0.03)$ $f(4.02, 3.99, 6.97) \approx 81 + 0.16 - 0.08 - 0.42$ $f(4.02, 3.99, 6.97) \approx 80.76$

Thus, the linear approximation gives: $f(4.02, 3.99, 6.97) \approx 80.76$

Final Answer:

$f(4.02, 3.99, 6.97) \approx 80.7600$

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Would you like further clarification or have any questions about the process?

Here are five related questions you might find interesting:

1. How can we use linear approximation in multi-variable optimization problems?
2. What is the significance of the tangent plane in linear approximations?
3. How does the accuracy of a linear approximation change as we move farther from the point of approximation?
4. How do higher-order approximations (like quadratic) improve accuracy?
5. What are the key differences between linear approximations and Taylor series expansions?

Tip: When approximating functions using linear approximation, the closer the point of interest is to the point of approximation, the more accurate the result will generally be!