Find the linear approximation of the following function at (x, y, z) = (3, 2, 6). f(x, y) = √(x² + y² + z²)

Solution:

Given, f(x, y, z) = √(x² + y² + z²)

We have to find the linear approximation of the function at (3, 2, 6).

The function can also be written as

f(x) = (x² + y² + z²)1/2

f’(x) = \(\frac{1}{2}(x^{2}+y^{2}+z^{2})^{\frac{-1}{2}}(2x)\)

f’(x) \(=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}\)

f’(y) = \(\frac{1}{2}(x^{2}+y^{2}+z^{2})^{\frac{-1}{2}}(2y)\)

f’(y) \(=\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}}\)

f’(z) = \(\frac{1}{2}(x^{2}+y^{2}+z^{2})^{\frac{-1}{2}}(2z)\)

f’(z) \(=\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}\)

Now, f(3, 2, 6) = \(={\sqrt{(3)^{2}+(2)^{2}+(6)^{2}}}\\=\sqrt{9+4+36}\\=\sqrt{49}\\=7\)

f’(x)(3, 2, 6) \(=\frac{3}{\sqrt{(3)^{2}+(2)^{2}+(6)^{2}}}\\=\frac{3}{7}\)

f’(y)(3, 2, 6) \(=\frac{2}{\sqrt{(3)^{2}+(2)^{2}+(6)^{2}}}\\=\frac{2}{7}\)

f’(z)(3, 2, 6) \(=\frac{6}{\sqrt{(3)^{2}+(2)^{2}+(6)^{2}}}\\=\frac{6}{7}\)

Using the formula,

L(x) = f(a) + f’(a)(x - a)

So, L(x, y, z) = f(3, 2, 6) + f’(x)(3, 2, 6) + f’(y)(3, 2, 6) + f’(z)(3, 2, 6)

L(x, y, z) = 7 + \(\frac{3}{7}(x-3)+\frac{2}{7}(y-2)+\frac{6}{7}(z-6)\)

Therefore, L(x, y, z) = \(7+\frac{3}{7}(x-3)+\frac{2}{7}(y-2)+\frac{6}{7}(z-6)\).

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Find the linear approximation of the following function at (x, y, z) = (3, 2, 6). f(x, y) = √(x² + y² + z²)

Summary:

The linear approximation of the following function at (x, y, z) = (3, 2, 6). f(x, y) = √(x² + y² + z²) is L(x, y, z) = \(7+\frac{3}{7}(x-3)+\frac{2}{7}(y-2)+\frac{6}{7}(z-6)\).