Approximating Change using Differentials
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 \( -0.11 \).
