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>
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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>
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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
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
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.\[u=-<\frac{ 1 }{ 3 },\frac{ 2 }{ 3 },\frac{ 2 }{ 3 }>\]