Approximating Temperature Difference using Differentials
The problem asks us to use differentials to approximate the temperature difference between two points in an \( xyz \) coordinate system where the temperature \( T \) is given by the function \( T = xy + yz + zx \).
To approximate the change in \( T \) when moving from point \( (x_0, y_0, z_0) \) to \( (x_0 + \Delta x, y_0 + \Delta y, z_0 + \Delta z) \), we can use differentials:
\[ dT = \frac{\partial T}{\partial x}dx + \frac{\partial T}{\partial y}dy + \frac{\partial T}{\partial z}dz \]
First, we'll find the partial derivatives of \( T \) with respect to \( x \), \( y \), and \( z \).
\[ \frac{\partial T}{\partial x} = y + z \]
\[ \frac{\partial T}{\partial y} = x + z \]
\[ \frac{\partial T}{\partial z} = x + y \]
Now, we can substitute the coordinates of the first point (\( 2, -1, 3 \)) into the derivatives to get the rates of change at that point:
\[ \frac{\partial T}{\partial x} = -1 + 3 = 2 \]
\[ \frac{\partial T}{\partial y} = 2 + 3 = 5 \]
\[ \frac{\partial T}{\partial z} = 2 - 1 = 1 \]
Using the values of the partial derivatives at \( (2, -1, 3) \) and the differences in \( x \), \( y \), and \( z \) which are \( \Delta x = 1.98 - 2 = -0.02 \), \( \Delta y = -0.98 - (-1) = 0.02 \), and \( \Delta z = 3.02 - 3 = 0.02 \), the differential approximation of the change in \( T \) is:
\[ dT \approx (2)(-0.02) + (5)(0.02) + (1)(0.02) \]
\[ dT \approx -0.04 + 0.10 + 0.02 \]
\[ dT \approx 0.08 \]
So, according to the differential approximation, the temperature difference between the points \( (2, -1, 3) \) and \( (1.98, -0.98, 3.02) \) is approximately \( 0.08 \) units.
