Question

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>

Answer 1

You posted a good bit of information, but I feel like I'm missing something. Could you link to the pset please?

Answer 2

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

Answer 3

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

Answer 4

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