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anonymous
 3 years ago
Confused about a practice problem on the directional derivatives session.
The function T = x^2 + 2*y^2 + 2*z^2
gives the temperature at each point in space.
1. At the point P = (1,1,1), in which direction should you go to get the most rapid decrease in T? What is the directional derivative in this direction?
So I get ∇T = <2x,4y,4z> and ∇T = <2,4,4> at P=(1,1,1). So the direction opposite ∇T gives the greatest magnitude. Let u be a unit vector in this direction.
dT/dsu = ∇T*u*cosθ = 6*1*(1) = 6
But the solution says that ∇T=<1,1,2> at P=(1,1,1) and that dT/dsu = 3.
They also compute u=<1/3,2/3,2/3>
anonymous
 3 years ago
Confused about a practice problem on the directional derivatives session. The function T = x^2 + 2*y^2 + 2*z^2 gives the temperature at each point in space. 1. At the point P = (1,1,1), in which direction should you go to get the most rapid decrease in T? What is the directional derivative in this direction? So I get ∇T = <2x,4y,4z> and ∇T = <2,4,4> at P=(1,1,1). So the direction opposite ∇T gives the greatest magnitude. Let u be a unit vector in this direction. dT/dsu = ∇T*u*cosθ = 6*1*(1) = 6 But the solution says that ∇T=<1,1,2> at P=(1,1,1) and that dT/dsu = 3. They also compute u=<1/3,2/3,2/3>

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Waynex
 3 years ago
Best ResponseYou've already chosen the best response.1You posted a good bit of information, but I feel like I'm missing something. Could you link to the pset please?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, it's actually from the problems and solutions in the directional derivatives session: http://ocw.mit.edu/courses/mathematics/1802scmultivariablecalculusfall2010/2.partialderivatives/partbchainrulegradientanddirectionalderivatives/session38directionalderivatives/ Direct link to solutions PDF: http://ocw.mit.edu/courses/mathematics/1802scmultivariablecalculusfall2010/2.partialderivatives/partbchainrulegradientanddirectionalderivatives/session38directionalderivatives/MIT18_02SC_pb_45_comb.pdf

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think it may actually be an error. I just tried Wolfram Alpha and, assuming I phrased my query correctly, got my answer. http://www.wolframalpha.com/input/?i=derivative+of+x%5E2+%2B+2+y%5E2+%2B+2+z%5E2+in+the+direction+%281%2C2%2C2%29+evaluated+at+%281%2C1%2C1%29

Waynex
 3 years ago
Best ResponseYou've already chosen the best response.1I agree with you retrace. Your gradient at (1,1,1) is certainly correct, therefore, MIT should have a scalar multiplier in front of their answer, as such:\[2<1,2,2>\] The unit vector in the direction of the most rapid decrease is correct though.\[u=<\frac{ 1 }{ 3 },\frac{ 2 }{ 3 },\frac{ 2 }{ 3 }>\]
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