Question - Approximating Temperature Difference using Differentials
mathothertemperature difference approximationusing differentialsxyz coordinate systempartial derivatives approach
Solution:
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.
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