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Find the linear approximation of the function sqrt(x² y² z²)

User Rymo
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Final answer:

The linear approximation of the function sqrt(x² y² z²) can be found using the concept of the tangent plane.

Step-by-step explanation:

The linear approximation of the function sqrt(x² y² z²) can be found using the concept of the tangent plane. The linear approximation at a point (a, b, c) is given by f(x, y, z) ≈ f(a, b, c) + (x-a) * fx(a, b, c) + (y-b) * fy(a, b, c) + (z-c) * fz(a, b, c), where fx, fy, and fz are the partial derivatives of f(x, y, z) with respect to x, y, and z respectively. In this case, we have f(x, y, z) = sqrt(x² y² z²), so the linear approximation becomes L(x, y, z) ≈ sqrt(a² b² c²) + (x-a) * (2a b² c²)/sqrt(a² b² c²) + (y-b) * (2a² b c²)/sqrt(a² b² c²) + (z-c) * (2a² b² c)/sqrt(a² b² c²).

User Angad Arora
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