Science:Math Exam Resources/Courses/MATH200/December 2011/Question 03 (b)/Solution 1

To evaluate the change in temperature, $T$, experienced by the bee with respect to time at time 2, we need to evaluate the full derivative of $\displaystyle T=T(x(t),y(t),z(t),t)$ with respect to $t$ and substitute in the value $t=2$. Writing the expression for the full time derivative gives:

${\begin{aligned}{\frac {dT}{dt}}&={\frac {\partial T}{\partial t}}+{\frac {\partial T}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial T}{\partial y}}{\frac {dy}{dt}}+{\frac {\partial T}{\partial z}}{\frac {dz}{dt}}\\&={\frac {\partial T}{\partial t}}+\nabla T\cdot \mathbf {v} \end{aligned}}$

where $\mathbf {v}$ is the velocity vector of the bee. Evaluating the terms in the above equation we find:

${\frac {\partial T}{\partial t}}=2y,\quad \nabla T=\left[y-3,x+2t,1\right]^{T}.$

Note that we don't need to re-evaluate $\mathbf {v}$ since we can just use our results from part (a) here. At time $t=2$, the bee is at $[1,1,0]$, so the above terms satisfy,

${\frac {\partial T}{\partial t}}=2,\quad \nabla T=\left[-2,5,1\right]^{T}$

at $t=2$. Using this and the fact that $\mathbf {v}$ $=[-2,-4,4]$, we evaluate the change in temperature:

${\begin{aligned}{\frac {dT}{dt}}&=2+[-2,5,1]^{T}\cdot [-2,-4,4]^{T}=2+4-20+4=-10.\end{aligned}}$

Therefore, the bee is experiencing a change in temperature of -10 temperature units per second at t = 2.