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