The temperature at a point (x,y,z) is given by t(x,y,z)=200e−x2−y2/4−z2/9t(x,y,z)=200e−x2−y2/4−z2/9, where tt is measured in degrees celsius and x,y, and z in meters. there are lots of places to make silly errors in this problem; just try to keep track of what needs to be a unit vector. find the rate of change of the temperature at the point (0, -1, 2) in the direction toward the point (-3, 4, -3).

First find the gradient by using partial derivatives.

[tex]T_x = (-2x)200 e^{(-x^2 -y^2/4 -z^2/9)} \\ \\ T_y = (-\frac{1}{2}y)200 e^{(-x^2 -y^2/4 -z^2/9)} \\ \\ T_z = (-\frac{2}{9}z)200 e^{(-x^2 -y^2/4 -z^2/9)}[/tex]

Sub in given point (0,-1,2) for x,y,z
∨[tex]T = (T_x, T_y,T_z) = (0,100e^{-25/36, \frac{-800}{9} e^{-25/36})[/tex]

Next, find unit vector from directional vector (0,-1,2) --> (-3,4,-3)
[tex]v = (-3-0, 4-(-1), -3-2) = (-3,5,-5) \\ \\ ||v|| = \sqrt{(-3)^2 + 5^2 + (-5)^2} = \sqrt{59} \\ \\ u = (\frac{-3}{\sqrt{59}}, \frac{5}{\sqrt{59}},\frac{-5}{\sqrt{59}})[/tex]

Finally, use dot product on gradient and unit vector to get directional rate of change.
Answer is approx 61.398