Answer
$\dfrac{dC}{dt} \approx -0.33$ (m/s) per minute
Work Step by Step
We need to use chain rule as follows: $\dfrac{dC}{dt}=(\dfrac{\partial C}{\partial T}) \ (\dfrac{dT}{ dt})+(\dfrac{\partial C}{\partial D}) \
(\dfrac{dD}{dt})$
Now, $\dfrac{dT}{dt} \approx \dfrac{14.00-12.00}{0.00-25.00} \approx -\dfrac{2}{25}^{\circ}C/min$
and
$\dfrac{dD}{dt} \approx \dfrac{15.00-0.0}{35.0-7.5}=\dfrac{6}{11}m/min$
When $T\approx 12. 7^{\circ}C$, then we have:
$\dfrac{dC}{dt} \approx -0.33$ (m/s) per minute
