## Calculus 8th Edition

a) $(\dfrac{20}{3})^{o}C$ b) $( \dfrac{10}{3})^{o}C$
(a) We will use chain rule. $T_{x}(x,y)=-60(1+x^{2}+y^{2})^{-2}\cdot(2x)=\dfrac{-120x}{(1+x^{2}+y^{2})^{2}}$ This gives: $T_{x}(2, 1)= \dfrac{-(120) \times (2)}{(1+4+1)^{2}}=-\dfrac{240}{36}=-\dfrac{20}{3}$ We can see that the temperature will be decreasing at a rate of $(\dfrac{20}{3})^{o}C$in the $x$-direction at point $(2,1)$. (b) Here, w ehave $T_{y}(x,y)=-\dfrac{60}{(1+x^{2}+y^{2})^{-2} }\times (2y)=\dfrac{-120y}{(1+x^{2}+y^{2})^{2}}$ and $T_{y}(2, 1)= \dfrac{-120}{36}=-\dfrac{10}{3}$ We can see that the temperature will be decreasing at a rate of $( \dfrac{10}{3})^{o}C$in the $y$-direction at point $(2,1)$.