To evaluate the change in temperature,
, experienced by the bee with respect to time at time 2, we need to evaluate the full derivative of
with respect to
and substitute in the value
. Writing the expression for the full time derivative gives:
![{\displaystyle {\begin{aligned}{\frac {dT}{dt}}&={\frac {\partial T}{\partial t}}+{\frac {\partial T}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial T}{\partial y}}{\frac {dy}{dt}}+{\frac {\partial T}{\partial z}}{\frac {dz}{dt}}\\&={\frac {\partial T}{\partial t}}+\nabla T\cdot \mathbf {v} \end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/ef44bca81df5cb7e462403c81717adf0794759ff)
where
is the velocity vector of the bee. Evaluating the terms in the above equation we find:
![{\displaystyle {\frac {\partial T}{\partial t}}=2y,\quad \nabla T=\left[y-3,x+2t,1\right]^{T}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/233e2f3c53bee28f51716ad797793e01b297cd7d)
Note that we don't need to re-evaluate
since we can just use our results from part (a) here. At time
, the bee is at
, so the above terms satisfy,
![{\displaystyle {\frac {\partial T}{\partial t}}=2,\quad \nabla T=\left[-2,5,1\right]^{T}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/a32877c3471e91e9d1d64af4706d2e3496201d50)
at
. Using this and the fact that
, we evaluate the change in temperature:
![{\displaystyle {\begin{aligned}{\frac {dT}{dt}}&=2+[-2,5,1]^{T}\cdot [-2,-4,4]^{T}=2+4-20+4=-10.\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/079df172b90ff19d922bc7f78620776c7ccf6bac)
Therefore, the bee is experiencing a change in temperature of -10 temperature units per second at t = 2.