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Thread: Directional derivatives

  1. #1
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    Directional derivatives

    Find the directional derivative of $\displaystyle f(x,y) = \sqrt{xy}$ at $\displaystyle P(2,8)$ in the direction of $\displaystyle Q(5,4)$?

    I'm not sure exactly what to do.

    I have differentiated the equation, which gives me

    $\displaystyle f_x(x,y) = \frac{y}{2 \sqrt{xy}}, \ \ \mbox{and} \ \ f_y(x,y) = \frac{x}{2 \sqrt{xy}}$ with $\displaystyle \Delta x = 3, \Delta y = -4$ and from this point I have no clue on what to do.
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  2. #2
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    Quote Originally Posted by lllll View Post
    Find the directional derivative of $\displaystyle f(x,y) = \sqrt{xy}$ at $\displaystyle P(2,8)$ in the direction of $\displaystyle Q(5,4)$?

    I'm not sure exactly what to do.

    I have differentiated the equation, which gives me

    $\displaystyle f_x(x,y) = \frac{y}{2 \sqrt{xy}}, \ \ \mbox{and} \ \ f_y(x,y) = \frac{x}{2 \sqrt{xy}}$ with $\displaystyle \Delta x = 3, \Delta y = -4$ and from this point I have no clue on what to do.
    The directional derivative is given by the formula:

    $\displaystyle \frac{df}{d\vec{l}} = \nabla f \cdot \vec{\hat{l}}$

    where $\displaystyle \vec{\hat{l}}$ is a unit vector in the given direction.

    In your case $\displaystyle \vec{\hat{l}} = \frac{5 i + 4 j}{\sqrt{41}}$ and $\displaystyle \nabla f = \frac{y}{2 \sqrt{xy}} i + \frac{x}{2 \sqrt{xy}} j$.

    Now substitute the given point.
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