Answer
a) $(\dfrac{20}{3})^{o}C$
b) $( \dfrac{10}{3})^{o}C $
Work Step by Step
Given: $T(x,y)=60(1+x^{2}+y^{2})^{-1}$
a) Applying the chain rule, we have
$T_{x}(x,y)=-60(1+x^{2}+y^{2})^{-2}\cdot(2x)=\dfrac{-120x}{(1+x^{2}+y^{2})^{2}} \implies T_{x}(2, 1)= \dfrac{-(120)(2)}{(1+4+1)^{2}}=-\dfrac{20}{3}$
This implies that the temperature is decreasing at a rate of $(\dfrac{20}{3})^{o}C $ in the $x$-direction at the point $(2,1)$.
b) $T_{y}(x,y)=-60(1+x^{2}+y^{2})^{-2} \cdot(2y)=\dfrac{-120y}{(1+x^{2}+y^{2})^{2}}
\implies T_{y}(2, 1)= \dfrac{-120}{36}=-\dfrac{10}{3}$
This implies that the temperature is decreasing at a rate of $( \dfrac{10}{3})^{o}C $ in the $y$-direction at the point $(2,1)$.
