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Temperature as a function of 3 variables...

  1. Apr 9, 2016 #1
    Mod note: moved from non-HW forum section
    Apparently doing math homework on Physics Forums helps me think more clearly. The d-part of this question gave me a headache, and I already wrote all of this LaTeX code on the homework side of this site, so I figured I'd post it even though I solved it.


    Parts a-c might have small mistakes as well, but duck it, I'm done with this question.

    1. The problem statement, all variables and given/known data

    Let [itex]T(x,y,z) = 10+xy+xz+yz[/itex]. A falcon is at the point [itex]P_0 = (x_0, y_0, z_0) = (3,2,1)[/itex].

    a) What is the temperature at the point [itex]P_0[/itex].
    b) What is the instantaneous rate of change of temperature experienced by the falcon as it flies towards the point [itex]P_1 = (2,4,4)[/itex]
    c) In what direction should the falcon fly in if it wanted to experience the largest (positive) rate of change in temperature?
    d) Give a direction in which the temperature does not change (there are infinitely many).

    2. Relevant equations

    The Gradient: [tex]\nabla T(P)= (\frac{\partial T}{\partial x},\frac{\partial T}{\partial y},\frac{\partial T}{\partial z})[/tex]

    Directional derivative: [tex]D_u T(P) = \nabla T(P) \cdot \hat{u}[/tex]

    3. The attempt at a solution

    To start with, [itex]\nabla T(P_0)= (\frac{\partial T}{\partial x_0},\frac{\partial T}{\partial y_0},\frac{\partial T}{\partial z_0}) = (y_0+z_0, x_0+z_0, x_0+y_0) = (3, 4, 5)[/itex]

    a) Simply plug in the values of x,y and z to get [itex]T(P_0) = 21[/itex]


    b) Here
    [tex]
    \vec{u} = P_1 - P_0 = (2,4,4) - (3,2,1) = (-1,2,3)
    [/tex]
    and
    [tex]
    \hat{u} = \frac{\vec{u}}{||\vec{u}||} = \frac{(-1, 2, 3)}{\sqrt{1^2 + 2^2 + 3^3}} = \frac{(-1, 2, 3)}{ \sqrt{14} }
    [/tex]

    Therefore the directional derivative:

    [tex]
    D_u T(P_0) = \nabla T(P_0) \cdot \hat{u} = (3, 4, 5) \cdot \frac{(-1, 2, 3) }{ \sqrt{14} } = \frac{20}{\sqrt{14}}
    [/tex]

    c) In the direction of the gradient.


    Here ##\hat{u} = \frac{\nabla T(P_0)}{||\nabla T(P_0)||} = \frac{1}{5 \sqrt{2}}(3,4,5)##.

    Therefore

    [tex]
    D_u T(P_0) = \nabla T(P_0) \cdot \hat{u} = (3,4,5) \cdot (\frac{3}{5 \sqrt{2}},\frac{4}{5 \sqrt{2}},\frac{5}{5 \sqrt{2}}) = \frac{9}{5 \sqrt{2}} + \frac{16}{5 \sqrt{2}} + \frac{25}{5 \sqrt{2}} =\frac{10}{\sqrt{2}}
    [/tex]

    d) This is where I ran into trouble. I knew that in the direction of the level curve (if such a thing exists in 4D), the directional derivative is equal to 0. Using this fact and the fact that ##||\hat{u}|| = 1##, I get two seemingly messy equations, that I was initially unable to solve:

    If we mark ##\hat{u} = (x,y,z)##
    [tex]
    D_u T(P_0) = \nabla T(P_0) \cdot \hat{u} = 3x+4y+5z = 0
    [/tex]
    and
    [tex]
    \sqrt{x^2+y^2+z^2} = 1 \iff x^2 + y^2 + z^2 = 1
    [/tex]
    More clearly:

    \begin{cases}
    3x+4y+5z = 0 \ || \ solve \ for \ z\\
    x^2 + y^2 + z^2 = 1
    \end{cases}

    \begin{cases}
    z = \frac{1}{5}(-3x - 4y) \\
    x^2 + y^2 + z^2 = 1 \ || \ Substitute \ z \ here
    \end{cases}

    \begin{cases}
    z = \frac{1}{5}(-3x - 4y) \\
    x^2 + y^2 + (\frac{1}{5}(-3x - 4y))^2 = 1 \ || \ Multiply\ (-3x - 4y))^2
    \end{cases}

    \begin{cases}
    z = \frac{1}{5}(-3x - 4y) \\
    x^2 + y^2 + \frac{1}{25}(9x^2 + 24xy + 16y^2) = 1 \ || \ \cdot 25
    \end{cases}

    \begin{cases}
    z = \frac{1}{5}(-3x - 4y) \\
    25x^2 + 25y^2 + (9x^2 + 24xy + 16y^2) = 25
    \end{cases}

    \begin{cases}
    z = \frac{1}{5}(-3x - 4y) \\
    34x^2 + 24xy + 41y^2 = 25 \iff 34x^2 + 24xy + 41y^2 -25 = 0 \ || \ Solve \ for \ x
    \end{cases}

    [tex]x = \frac {-24y \pm \sqrt{(24y)^2 - 4(34)(41y^2 - 25)}}{2(34)}[/tex]
    [tex]x = \frac {-24y \pm \sqrt{-200(25y^2-17)}}{68}[/tex]

    Now to make things simple, let the discriminant be equal to zero:
    [tex]
    -200(25y^2-17)=0 \iff 5000y^2 = 3400 \iff y= \pm \sqrt{\frac{3400}{5000}} = \pm \frac{\sqrt{17}}{5}
    [/tex]

    Again, to make things simple, plug positive y into x:
    \begin{array}{ll}
    x &= \frac {-24y \pm \sqrt{-200(25y^2-17)}}{68}\\
    x &= \frac {-24(\frac{\sqrt{17}}{5}) \pm \sqrt{0}}{68} \ || \ Remember \ the \ discriminant \ was \ zero \\
    x &= \frac {-24(\frac{\sqrt{17}}{5})}{68} = \frac{-6}{5 \sqrt{17}}
    \end{array}

    Then by pugging x and y into z:
    \begin{array}{ll}
    z &= \frac{1}{5}(-3x - 4y)\\
    &= \frac{1}{5}(-3({\frac{-6}{5 \sqrt{17}}) - 4(\frac{\sqrt{17}}{5}})\\
    &= \frac{-2}{\sqrt{17}}
    \end{array}

    Therefore
    \begin{cases}
    x = \frac{-6}{5 \sqrt{17}}\\
    y = \frac{\sqrt{17}}{5}\\
    z = \frac{-2}{\sqrt{17}}
    \end{cases}
    and the direction we want to head in is

    [tex]\hat{u} = (x,y,z) = (\frac{-6}{5 \sqrt{17}},\frac{\sqrt{17}}{5},\frac{-2}{\sqrt{17}})[/tex]
     
    Last edited by a moderator: Apr 9, 2016
  2. jcsd
  3. Apr 9, 2016 #2
    Looks good TheSodesa!
     
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