tangibleLime
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Homework Statement
The temperature at a point (x, y, z) is given by the following equation where T is measured in °C and x, y, z in meters.
T(x, y, z) = 200e-x2-3y2-9z2
Find the rate of change of temperature at the point P(2, -1, 2) in the direction towards the point (3, -3, 3).
Homework Equations
The Attempt at a Solution
I calculated the partial derivatives of T(x,y,z) as...
T_{x} = -400xe^{-x^{2}-3y^{2}-9z^{2}}
T_{y} = -1200ye^{-x^{2}-3y^{2}-9z^{2}}
T_{z} = -3600ze^{-x^{2}-3y^{2}-9z^{2}}
Giving me the gradient...
\nabla T(x,y,z) = <-400xe^{-x^{2}-3y^{2}-9z^{2}}, -1200ye^{-x^{2}-3y^{2}-9z^{2}}, -3600ze^{-x^{2}-3y^{2}-9z^{2}}>
Plugging in the given values for the gradient, I get...
\nabla T(2,-1,2) = <\frac{-800}{e^{35}},\frac{1200}{e^{35}},\frac{-7200}{e^{35}}>
Meanwhile, I take the point (3,-3,3) and convert it into a direction unit vector (I think this is where I make a mistake).
v = <3,-3,3>
u = \frac{1}{\sqrt{27}} * <3,-3,3>
u = <\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}>
Combining all of this in DuT...
D_{u}T = \frac{-800}{\sqrt{3}*e^{35}} + \frac{-1200}{\sqrt{3}*e^{35}} + \frac{-7200}{\sqrt{3}*e^{35}}
This gives me a wrong answer of...
\frac{-9200}{\sqrt{3}*e^{35}}
Any help that isn't a purposefully cryptic message that leaves me no closer to solving my problem is appreciated.
Thanks.