With respect tho the $Oxyz$ axes, the temperature of a media is given by: $T=T_0 (1+ax+by)e^{cz}$

Question

With respect tho the $Oxyz$ axes, the temperature of a media is given by: $$T=T_0 (1+ax+by)e^{cz}$$ Where $a, b, c$ and $T (>0)$ are constants. At the origin O, find the direction in which the temperature changes more quickly.

Answer 1

This is not an intricate problem of strategy, rather it is just the straightforward application of a single concept. The concept, and it is an important one, is that the gradient $\nabla T = \left({\partial T\over\partial x},{\partial T\over\partial y},{\partial T\over\partial z}\right)$ points in the direction of fastest increase in $T$.

So just compute the gradient, evaluate it at the origin, and also perhaps normalize the vector to a unit vector.

Derivative calculations to feed into the above: For ${\partial T\over \partial x}$, regard $T_0$ and $e^{cz}$ as constants, pull them to the front, then $${\partial T\over \partial x} = {\partial \over \partial x}\left[(T_0e^{cz})(1+ax+by)\right] = (T_0e^{cz}){\partial\over \partial x}(1+ax+by) = (T_0e^{cz})(a) = aT_0e^{cz}\,.$$

The calculation of ${\partial T\over\partial y}$ is similar. For ${\partial T\over\partial z}$, $${\partial \over \partial z}\left[T_0(1+ax+by)e^{cz}\right] = T_0(1+ax+by){\partial\over \partial z}(e^{cz}) = T_0(1+ax+by)(ce^{cz}) = cT_0(1+ax+by)e^{cz}\,.$$