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retrace
Group Title
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>
 one year ago
 one year ago
retrace Group Title
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>
 one year ago
 one year ago

This Question is Closed

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

retrace Group TitleBest ResponseYou've already chosen the best response.1
Yes, 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
 one year ago

retrace Group TitleBest ResponseYou've already chosen the best response.1
I 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
 one year ago

Waynex Group TitleBest ResponseYou've already chosen the best response.1
I 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.\[u=<\frac{ 1 }{ 3 },\frac{ 2 }{ 3 },\frac{ 2 }{ 3 }>\]
 one year ago
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