Question - Approximating Change using Differentials
Solution:
To approximate the change in $$ f(x, y, z) = 3x^2 - yz + 2 $$ when $$ (x, y, z) $$ changes from $$ (1, 2, 3) $$ to $$ (0.99, 2.01, 3.01) $$, we will use differentials. First, we need to compute the partial derivatives of $$ f $$ with respect to $$ x $$, $$ y $$, and $$ z $$:\[ \frac{\partial f}{\partial x} = 6x \]\[ \frac{\partial f}{\partial y} = -z \]\[ \frac{\partial f}{\partial z} = -y \]Now we evaluate these partial derivatives at the point $$ (1, 2, 3) $$:\[ \frac{\partial f}{\partial x}(1, 2, 3) = 6(1) = 6 \]\[ \frac{\partial f}{\partial y}(1, 2, 3) = -(3) = -3 \]\[ \frac{\partial f}{\partial z}(1, 2, 3) = -(2) = -2 \]The differential approximates the change in $$ f $$ as follows:\[ df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz \]Since $$ dx = 0.99 - 1 = -0.01 $$, $$ dy = 2.01 - 2 = 0.01 $$, and $$ dz = 3.01 - 3 = 0.01 $$, we can substitute these values into the equation to find $$ df $$:\[ df = (6)(-0.01) + (-3)(0.01) + (-2)(0.01) \]\[ df = -0.06 - 0.03 - 0.02 \]\[ df = -0.11 \]Thus, the differential $$ df $$ approximates the change in $$ f $$ when $$ (x, y, z) $$ changes from $$ (1, 2, 3) $$ to $$ (0.99, 2.01, 3.01) $$ to be approximately $$ -0.11 $$.
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