Answer:
[tex]D_uT(3,1)=-\frac{44}{9}*\frac{1}{\sqrt{2} } \approx-3.46[/tex]
Step-by-step explanation:
To find the rate of change of temperature with respect to distance at the point (3, 1) in the x-direction and the y-direction we need to find the Directional Derivative of T(x,y). The definition of the directional derivative is given by:
[tex]D_uT(x,y)=T_x(x,y)i+T_y(x,y)j[/tex]
Where i and j are the rectangular components of a unit vector. In this case, the problem don't give us additional information, so let's asume:
[tex]i=\frac{1}{\sqrt{2} }[/tex]
[tex]j=\frac{1}{\sqrt{2} }[/tex]
So, we need to find the partial derivative with respect to x and y:
In order to do the things easier let's make the next substitution:
[tex]u=2+x^2+y^2[/tex]
and express T(x,y) as:
[tex]T(x,y)=88*u^{-1}[/tex]
The partial derivative with respect to x is:
Using the chain rule:
[tex]\frac{\partial u}{\partial x}=2x[/tex]
Hence:
[tex]T_x(x,y)=88*(u^{-2})*\frac{\partial u}{\partial x}[/tex]
Symplying the expression and replacing the value of u:
[tex]T_x(x,y)=\frac{-176x}{(2+x^2+y^2)^2}[/tex]
The partial derivative with respect to y is:
Using the chain rule:
[tex]\frac{\partial u}{\partial y}=2y[/tex]
Hence:
[tex]T_y(x,y)=88*(u^{-2})*\frac{\partial u}{\partial y}[/tex]
Symplying the expression and replacing the value of u:
[tex]T_y(x,y)=\frac{-176y}{(2+x^2+y^2)^2}[/tex]
Therefore:
[tex]D_uT(x,y)=(\frac{1}{\sqrt{2} } )*(\frac{-176x}{(2+x^2+y^2)^2} -\frac{176y}{(2+x^2+y^2)^2})[/tex]
Evaluating the point (3,1)
[tex]D_uT(3,1)=(\frac{1}{\sqrt{2} } )*(\frac{-176(3)-176(1)}{(2+3^2+1^2)^2})=(\frac{1}{\sqrt{2} })* (-\frac{704}{144})=(\frac{1}{\sqrt{2} }) ( - \frac{44}{9})\approx -3.46[/tex]
