Question

The temperature at a point (x,y,z) is given by T(x,y,z)=200e^-x^2-3y^-9z^2, where T measured in degrees Celsius and x,y,z in meters. Find the rate of change of temperature at the point P(2, -1, 2) in the direction toward the point (3, -3, 3)

Answer 1

Answer:

Therefore the rate change of temperature at the point P(2,-1,2) in the direction toward the point (3,-3,3) is $-\frac{5200\sqrt6}{3}e^{-43}$  °C/m.

Step-by-step explanation:

Given that, the temperature at a point (x,y,z) is

$T(x,y,z)=200e^{-x^2-3y^2-9z^2}$.

Rate change of temperature at the point P(2,-1,2) in the direction toward the point Q (3,-3,3) is $D_uT(2,-1,2)$

$T_x= 200(-2x)e^{-x^2-3y^2-9z^2}$$=-400x200.2xe^{-x^2-3y^2-9z^2}$

$T_y = 200.(-6y)e^{-x^2-3y^2-9z^2}$$=-1200ye^{-x^2-3y^2-9z^2}$

$T_z=200(-18z)e^{-x^2-3y^2-9z^2}$$=-3600ze^{-x^2-3y^2-9z^2}$

The gradient of the temperature

$\bigtriangledown T(x,y,z)= (-400x200.2xe^{-x^2-3y^2-9z^2},-1200ye^{-x^2-3y^2-9z^2},-3600ze^{-x^2-3y^2-9z^2})$                   $=-400e^{-x^2-3y^2-9z^2}(x,3y,9z)$

$\bigtriangledown T(2,-1,2) = -400e^{-2^2-3(-1)^2-9.2^2}(2,3.(-1),9.2)$

                   $=-400 e^{-43}(2,-3,18)$

V=$\overrightarrow {PQ}= \vec{Q}-\vec{P}$=(3,-3,3)-(2,-1,2)=(1,-2,1)

The unit vector of V is $\frac{(1,-2,1)}{\sqrt{1^2+(-2)^2+1^2}}$

                                   $=\frac{1}{\sqrt6}(1,-2,1)$

Therefore,

$D_uT(2,-1,2) = \bigtriangledown T(2,-1,2).\frac{1}{\sqrt6}(1,-2,1)$

                       $=-400 e^{-43}(2,-3,18).\frac{1}{\sqrt6}(1,-2,1)$

                       $=-\frac{400}{\sqrt6}e^{-43}(2.1+(-3).(-2)+18.1)$

                       $=-\frac{10400}{\sqrt6}e^{-43}$  

                        $=-\frac{5200\sqrt6}{3}e^{-43}$ °C/m

Therefore the rate change of temperature at the point P(2,-1,2) in the direction toward the point (3,-3,3) is $-\frac{5200\sqrt6}{3}e^{-43}$  °C/m.