Question

The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 45/(6 + x2 + y2), where T is measured in °C and x, y in meters. Find the rate of change of temperature with respect to distance at the point (3, 1) in the x-direction and the y-direction. (a) the x-direction °C/m (b) the y-direction °C/m

Answer 1

Answer:

Change in x direction is $-1.054^{o}C$

Change in y direction is $-0.35^{o}C$

Explanation:

The temperature is given by we need to find gradient of temperature to obtain the rate of change thus

$T(x,y)=\frac{45}{6+x^{2}+y^{2}}\\\\\therefore \bigtriangledown T(x,y)=\frac{\partial T(x,y)}{\partial x} \widehat{i}+\frac{\partial T(x,y)}{\partial y} \widehat{j}\\\\\frac{\partial T(x,y)}{\partial x}=\frac{-45}{(6+x^{2}+y^{2})^{2}}\times 2x\widehat{i}\\\\\frac{\partial T(x,y)}{\partial y}=\frac{-45}{(6+x^{2}+y^{2})^{2}}\times 2y\widehat{j}$

Now the rate of change at (3,1) in

1) X direction is given by

$\frac{\partial T(3,1)}{\partial x}=\frac{-45}{(6+3^{2}+1^{2})^{2}}\times 2\times 3\widehat{i}=-1.054\widehat{i}$

2) Y direction is given by

$\frac{\partial T(3,1)}{\partial y}=\frac{-45}{(6+3^{2}+1^{2})^{2}}\times 2\times 1\widehat{j}=-0.35\widehat{j}$