Answer:
Change in x direction is [tex]-1.054^{o}C[/tex]
Change in y direction is [tex]-0.35^{o}C[/tex]
Explanation:
The temperature is given by we need to find gradient of temperature to obtain the rate of change thus
[tex]T(x,y)=\frac{45}{6+x^{2}+y^{2}}\\\\\therefore \bigtriangledown T(x,y)=\frac{\partial T(x,y)}{\partial x} \widehat{i}+\frac{\partial T(x,y)}{\partial y} \widehat{j}\\\\\frac{\partial T(x,y)}{\partial x}=\frac{-45}{(6+x^{2}+y^{2})^{2}}\times 2x\widehat{i}\\\\\frac{\partial T(x,y)}{\partial y}=\frac{-45}{(6+x^{2}+y^{2})^{2}}\times 2y\widehat{j}\\\\[/tex]
Now the rate of change at (3,1) in
1) X direction is given by
[tex]\frac{\partial T(3,1)}{\partial x}=\frac{-45}{(6+3^{2}+1^{2})^{2}}\times 2\times 3\widehat{i}=-1.054\widehat{i}[/tex]
2) Y direction is given by
[tex]\frac{\partial T(3,1)}{\partial y}=\frac{-45}{(6+3^{2}+1^{2})^{2}}\times 2\times 1\widehat{j}=-0.35\widehat{j}\\\\[/tex]
