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Math Help - Gradient question

  1. #1
    Junior Member
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    Gradient question

    Assume you are located at a point $\vec{a}=(a,b)$ on a function $f(x,y)$. If you want to travel in a direction $\vec{u}=(u_1,u_2)$ corresponding to an angle $\alpha$ measured from the positive x-axis, show that
    $D_{\vec{u}}f(\vec{a})=f_x(a,b)cos\alpha + f_y(a,b)sin\alpha$

    My attempt at the solution.
    $D_{\vec{u}}f(\vec{a}) = \nabla f(\vec{a})*\vec{u} = |\nabla f(\vec{a})||\vec{u}|cos\alpha$
    I'm not sure where to go from here, or if I'm even using the correct angle, I think the angle for dot product should be the one between the two vectors and $\alpha$ is from the x axis.
    Some help would be appreciated.

    Thanks
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  2. #2
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    Dallas, Texas
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    Re: Gradient question

    Quote Originally Posted by nubshat View Post
    Assume you are located at a point $\vec{a}=(a,b)$ on a function $f(x,y)$. If you want to travel in a direction $\vec{u}=(u_1,u_2)$ corresponding to an angle $\alpha$ measured from the positive x-axis, show that
    $D_{\vec{u}}f(\vec{a})=f_x(a,b)cos\alpha + f_y(a,b)sin\alpha$

    My attempt at the solution.
    $D_{\vec{u}}f(\vec{a}) = \nabla f(\vec{a})*\vec{u} = |\nabla f(\vec{a})||\vec{u}|cos\alpha$
    I'm not sure where to go from here, or if I'm even using the correct angle, I think the angle for dot product should be the one between the two vectors and $\alpha$ is from the x axis.
    Some help would be appreciated.

    Thanks
    Think about decomposing $\vec{u}=(u_1,u_2)$ into the Cartesian basis vectors. Hint: how else could you write $\nabla f(\vec{a})\cdot\vec{u}$
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