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Math Help - Directional Derivative

  1. #1
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    Directional Derivative

    Hey guy I'm having a problem with directional derivative, can any one give me a clue about how to solve this

    f(x,y) = x^3*y^2 at (-1,2) in the direction parallel to a vector \frac{pi}{3} clockwise from the positive x-axis
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    Re: Directional Derivative

    Can you at least get a vector going in the right direction?
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    Re: Directional Derivative

    Hey I can get the derivative of y and x and put them to f (d/dx, d/dy) and sub (-1, 2) in. Im not sure what to do in regard to finding direction
    Last edited by junkwisch; October 24th 2013 at 03:36 AM.
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    Re: Directional Derivative

    Start by drawing a diagram. Draw a set of axes, and then draw a right angle triangle that makes an angle of 60 degrees to the positive x axis. Fill in the lengths. You should be able to get the direction vector from that...
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    Re: Directional Derivative

    The directional derivative of function f in direction \theta is the dot product of \nabla f with the unit vector in direction theta. In two dimensional Cartesian coordinates, \nabla f= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j} and the unit vector in the \theta direction is cos(\theta)\vec{i}+ sin(\theta)\vec{j} so that dot product is just
    cos(\theta)\frac{\partial f}{\partial x}+ sin(\theta)\frac{\partial f}{\partial y}.

    Surely you have seen that before?
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    Re: Directional Derivative

    Oh so that is what it mean, thank prove it.i have seen it before but as a vector instead of theta.
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    Re: Directional Derivative

    Quote Originally Posted by junkwisch View Post
    Oh so that is what it mean, thank prove it.i have seen it before but as a vector instead of theta.
    Well you still have to get a vector from that though...
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