Answer
$2^{\circ} C$ per second
Work Step by Step
Apply the chain rule, we have $\dfrac{dT(t)}{dt}=T_t[x(t) , y(t)]=T_x[x(t), y(t)] \cdot x'(t)+T_y[x(t), y(t)] \cdot y'(t)$
Re-write as:
$\dfrac{dT(t)}{dt}=T_x[x(t), y(t)] \cdot \dfrac{1}{2\sqrt {1+t}}+T_y[x(t), y(t)] \cdot (\dfrac{1}{3})$
We need to find the required rate of the temperature at $T=3$
$\dfrac{dT(3)}{dt}=T_x(2,3)(\dfrac{1}{4})+T_y(2,3) (\dfrac{1}{3})=(4)(\dfrac{1}{4})+(3) (\dfrac{1}{3})$
Hence, $\dfrac{dT(3)}{dt}=2^{\circ} C/s$
