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Thread: Directional Derivatives and gradient vector killing me

  1. #1
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    Directional Derivatives and gradient vector killing me

    The temperature at a point (x,y,z) is given by , where is measured in degrees Celsius and x,y, and z in meters. There are lots of places to make silly errors in this problem; just try to keep track of what needs to be a unit vector.
    1) Find the rate of change of the temperature at the point (1, -1, 2) in the direction toward the point (-1, -5, 5).


    2) In which direction (unit vector) does the temperature increase the fastest at (1, -1, 2)?

    3)What is the maximum rate of increase of at (1, -1, 2)?
    I got the second one right.
    $\displaystyle <(-400e^{-49/36})/ \sqrt(e^{-98/36}{(-400)^2+(100)^2+(-800/9)^2}), $$\displaystyle 100e^{-49/36}/ \sqrt(e^{-98/36}{(-400)^2+(100)^2+(-800/9)^2}), $$\displaystyle -(800/9)e^{-49/36}/ \sqrt(e^{-98/36}{(-400)^2+(100)^2+(-800/9)^2})>
    $
    but I do not get 1 and 3.

    I tried to do both and i have
    1) 0
    3) $\displaystyle \sqrt(e^{-98/36}{(-400)^2+(100)^2+(-800/9)^2})$

    But both are not right and I am totally out of idea how to solve it. Please someone help me
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  2. #2
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    Directional Derivatives

    To find the rate of change in f at point a in the direction v, we use the equation $\displaystyle D_vf(a)=\bigtriangledown f(a)\cdot v$

    Here, $\displaystyle a=(1, -1, 2), v=(-1, -5, 5)-(1, -1, 2)=(-2,-4,3)$, whose unit vector is $\displaystyle \hat{v}=\frac{1}{\sqrt{29}}(-2,-4,3)$

    Also, $\displaystyle \bigtriangledown T(a)=(T_x(a),T_y(a),T_z(a))$, so the first step in the problem is to find the three partial derivatives of T to give the gradient. Then, plug and chug: $\displaystyle D_vT(a)=\frac{1}{\sqrt{29}}(T_x(1, -1, 2)(-2)+T_y(1, -1, 2)(-4)+T_z(1, -1, 2)(3))$

    The maximum/minimum rates of change will occur when $\displaystyle \bigtriangledown^2 T(a)=(T_{xx}(a),T_{yy}(a),T_{zz}(a))=\vec{0}$

    See if this can get you started.
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  3. #3
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    Quote Originally Posted by Media_Man View Post
    The maximum/minimum rates of change will occur when $\displaystyle \bigtriangledown^2 T(a)=(T_{xx}(a),T_{yy}(a),T_{zz}(a))=\vec{0}$
    I thought the magnitude of the gradient vector gives the max rate of change, doesnt it? Why square of gradient vector gives the max/min rate of change?
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  4. #4
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    Good question

    "$\displaystyle \bigtriangledown^2$" actually means "the gradient of the gradient" not "the square of the gradient." Consider an equation modeling a car traveling down a road. The position is the function. The velocity is the first derivative of the function. The acceleration is the second derivative. Hence, when $\displaystyle f''(x)=0$, that is a max or min rate of change. Therefore, $\displaystyle \bigtriangledown^2 T(a)=\vec{0}$ gives the point(s) of maximum or minimum velocity.
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  5. #5
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    So how about taking magnitude of $\displaystyle
    \bigtriangledown T(a)$ ==> $\displaystyle
    ||\bigtriangledown T(a)||$ I thought solving for magnitude gives me the max rate of change.

    How does it different from $\displaystyle
    \bigtriangledown^{2} T(a)$?
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  6. #6
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    Ah, yes. Forgive me. I was doing it the hard way. There is indeed a theorem stating that the maximum rate of change occurs in the direction of the gradient vector. You are correct.
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