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Directional Derivative at an Angle from the Gradient

  1. Sep 25, 2016 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    (a) Find the directional derivative of z = x2y at (3,4) in the direction of 3π/4 with the x-axis. Give an exact answer.

    (b) Find the directional derivative of z = x2y at (3,4) in the direction that makes an angle of 3π/4 with the gradient vector at (3,4). Give an exact answer.

    (c) In which direction is the directional derivative the largest? Give your answer as a vector.

    2. Relevant equations
    ##F\vec u (a,b) = \nabla F(a,b) \cdot \vec u##

    Found the following two equations using the one above:
    ##\nabla F(3,4) = \langle 24,9 \rangle##
    ##\frac { \vec \nabla F(3,4)} {\left\|\vec \nabla F(3,4)\right\|} = \langle \frac {8}{\sqrt{73}},\frac {3} {\sqrt {73}} \rangle ##

    3. The attempt at a solution

    A. I found the directional derivative as ##\frac {-15}{\sqrt{2}}##.

    B. I'm stuck here. I can't seem to figure out how to get the unit vector I need to dot the gradient with.
    I've tried using the geometric and algebraic definitions of the dot product and the fact that##\sqrt{{u_1}^2 + {u_2}^2} = 1## to find each component of u, but I'm just getting garbage as an answer (1 = not 1 ?:)). Part of the problem is that I'm supposed to give an exact answer, which I haven't quite figured out how to do yet.

    I just tried to use some trig logic to get the angles I need, but it doesn't seem to be working either.
    Since ##θ = arccos (x)##, and ##θ = arcsin (y)##, I set my unit vector to ##\vec u = \langle cos(\frac {3π}{4} + arccos\frac {8}{\sqrt{73}}), sin(\frac {3π}{4} + arcsin\frac {3}{\sqrt{73}})\rangle## and dotted it with the gradient, but that's wrong as well.

    C. Haven't gotten to this part yet.

    The good news is that after 6 years on PF, I've finally learned the basics of LaTeX! :rolleyes:
     
    Last edited: Sep 25, 2016
  2. jcsd
  3. Sep 26, 2016 #2

    LCKurtz

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    Do you mean ##x^2y##?

    I will give you a hint on how to find that vector for (b). Let's start with ##\vec v = \langle 8,3\rangle## which is in the direction of your gradient. Two vectors perpendicular to ##\vec v## are ##\langle -3,8\rangle## and ##\langle 3,-8\rangle## (obvious from the dot product = 0). Draw these three vectors and also ##-\vec v## at the origin. Do you see in that picture two vectors you could add to give you a vector in the direction ##\frac{3\pi}{4}## from ##\vec v##? If so, make it a unit vector and proceed.
     
  4. Sep 26, 2016 #3

    Drakkith

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    Indeed.

    That worked perfectly.

    B. The answer is ##-\frac{657}{\sqrt{1314}}##

    C. The answer is ##\left\langle 24,9\right\rangle ##

    Thanks, Kurtz.
     
  5. Sep 26, 2016 #4

    Ray Vickson

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    I hope you realize that for any ##f(x_1, x_2, \ldots, x_n)## the directional derivative is largest in the direction ##\nabla f##, and the proof is extremely easy.
     
  6. Sep 26, 2016 #5

    Drakkith

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    Yep. There's several examples in my book that I've gone through.
     
  7. Sep 26, 2016 #6

    LCKurtz

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    Are you sure about that answer? What did you get for your unit vector in the new direction?
    [Edit, added]: That answer is correct but it can be reduced.

    Of course, the geometric trick I just showed you depended on the fact that the angle ##\frac{3\pi}{4}## is easy to construct geometrically. But what if you had wanted to rotate the vector clockwise by an arbitrary angle ##\theta##? If you have studied the complex plane, you know that if you represent a vector ##z=\langle a,b \rangle## as ##a + bi## and multiply it by the complex number ##e^{i\theta}##, the result rotates ##z## by the angle ##\theta##. In your problem you would multiply by ##e^{\frac {3\pi i} 4} = \cos(\frac {3\pi i} 4) + i \sin(\frac {3\pi i} 4) = -\frac 1 {\sqrt 2} + i\frac 1 {\sqrt 2}##. You could use this idea to check your work above.
     
    Last edited: Sep 26, 2016
  8. Sep 26, 2016 #7

    Drakkith

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    The online program accepted it as a correct answer.

    ##\langle \frac {-33}{\sqrt{1314}}, \frac{15}{\sqrt{1314}}\rangle##

    I have not studied the complex plane but I'll try to remember this. Thanks Kurtz.
     
  9. Sep 26, 2016 #8

    LCKurtz

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    I edited my answer about the same time that you replied noting that your answer is correct but that it can be reduced. You can factor a 3 out of both numerator and denominator. Ditto for your unit vector.
     
  10. Sep 26, 2016 #9

    Drakkith

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    Roger!
     
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