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larisa86 [58]
3 years ago
10

The last option is D: 13 Please help im stuck

Mathematics
1 answer:
Eva8 [605]3 years ago
8 0
I believe that the answer to that problem is d
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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 deg
yulyashka [42]
First find the gradient by using partial derivatives.

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)}

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

Next, find unit vector from directional vector (0,-1,2) --> (-3,4,-3)
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}})

Finally, use dot product on gradient and unit vector to get directional rate of change.
Answer is approx 61.398 
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Step-by-step explanation:

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Thats the answer
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