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Thread: Directional Derivatives

  1. #1
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    Directional Derivatives

    I need to find the directions in which the directional derivative $\displaystyle D_u f(x,y)$ of $\displaystyle f(x,y)=ye^{-xy}$ at the point $\displaystyle (0,2)$ has the value 1.

    I use the dot product formula for directional derivatives where $\displaystyle u$ is a unit vector; I set equal to one:

    $\displaystyle D_u f(x,y)=u\cdot\nabla f(x,y)=1$

    The components of the gradient vector:

    $\displaystyle \frac{\partial f}{\partial x}i=-y^2e^{-xy}i$

    $\displaystyle \frac{\partial f}{\partial y}j=-xye^{-xy}j$

    I plug in the coordinates:

    $\displaystyle \nabla f(0,2)=<-4,0>$

    $\displaystyle u\cdot<-4,0>$=1 , which is $\displaystyle u\cdot(-4i)=1$ This tells me that the $\displaystyle a4=1$ where $\displaystyle a$ is the component of the unit vector $\displaystyle u$ in $\displaystyle i$ direction. If I write the unit vector as $\displaystyle u=ai + bj$ I have $\displaystyle \mid u \mid= \sqrt{a^2+b^2}=1$.

    Now I have the elementary system of equations:

    $\displaystyle a4=1$

    $\displaystyle \sqrt{a^2+b^2}=1$

    I not sure if the solutions for a,b the correct components of the unit vector givng the direction required. It makes sense to me, but I don't have the answer in my book, so I can't check this.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Du can also be written |f||u|cos(t) = 1

    so 4cos(t) =1

    t = arccos(1/4)


    u = cos(t) i +sin(t) j
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