Directional Derivative

  • Thread starter Larrytsai
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Homework Statement


The temperature at a point (x,y,z) is given by ,T(x,y,z)=200e^((-x^2-y^2)/(4-z^2)/9) where 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 (1, -1, 1) in the direction toward the point (2, 4, -4).


The Attempt at a Solution



What I have done so far, is taken the partial derivative of each component. What I plan to do is, plug in the points 1,-1, 1 and dot it with the unit vector of (2,4,-4)
can anyone confirm if this is the right path?

So for some reason Im still not getting the right answer, I have

Tx = [200[-2x]e^((-x^2-y^2)/(4-z^2)/9)] / 9(4 - z^2)
Ty = 200(-2y)e^((-x^2-y^2)/(4-z^2)/9) / 9(4 - z^2)
Tz = -200[-x^2-y^2 ](-18z)[[36-9z^2]^-1]e^((-x^2-y^2)/(4-z^2)/9)

then i dot it with this unit vector
(2,4,-4)- (1,-1,1) = (1,5,-5)
|u| = sqrt(51)

(400e^(-2/27))/(sqrt(51))[-1-5]+[7200(-5)e^(-2/27)]/((27^2)sqrt(51))
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
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No, not "unit vector of (2, 4, -4)". That would be a vector from (0, 0, 0) to (2, 4, -4).
You want to use the unit vector in the same direction as a vector form (1, -1, 1) to (2, 4, -4).
 
  • #3
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ohhh yeah i see now, I assumed that (2,4,-4) was a vector not a point. Okay thanks =)!
 

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