anonymous
  • anonymous
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/ds|u = |∇T|*|u|*cosθ = 6*1*(-1) = -6 But the solution says that -∇T=-<1,1,2> at P=(1,1,1) and that dT/ds|u = -3. They also compute u=-<1/3,2/3,2/3>
OCW Scholar - Multivariable Calculus
jamiebookeater
  • jamiebookeater
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Waynex
  • Waynex
You posted a good bit of information, but I feel like I'm missing something. Could you link to the pset please?
anonymous
  • anonymous
Yes, it's actually from the problems and solutions in the directional derivatives session: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-b-chain-rule-gradient-and-directional-derivatives/session-38-directional-derivatives/ Direct link to solutions PDF: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-b-chain-rule-gradient-and-directional-derivatives/session-38-directional-derivatives/MIT18_02SC_pb_45_comb.pdf
anonymous
  • anonymous
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+%28-1%2C-2%2C-2%29+evaluated+at+%281%2C1%2C1%29

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Waynex
  • Waynex
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.\[u=-<\frac{ 1 }{ 3 },\frac{ 2 }{ 3 },\frac{ 2 }{ 3 }>\]

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